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Add and Subtract Three Integers | KS3 Maths Resources
What you need to know
Things to remember:
• Addition and subtraction can be done in any order (this is called the commutative law)
• When you subtract a negative, it is the same as adding the number part.
• Because the order of additions doesn’t matter, we can just add and subtract going from left to right.
Consider the following:
$$5+2=7$$
$$2+5 = 7$$
$$12-4=8$$
$$-4+12=8$$
Looking at these, we can see that the order we add and subtract doesn’t matter. This is called the “commutative law”. This is true for any number of additions and subtractions.
$$2-4+5=3$$
$$5-4+2=3$$
$$-4+2+5=3$$
$$5+2-4=3$$
We need to remember two key points:
• Adding a negative number is the same as subtracting.
• Subtracting a negative number is the same as adding.
$$5+-2=5-3=3$$
$$5 - -2= 5+2 = 7$$
These points work for any number of additions and subtractions, but we will focus on when we have three. We can do these in two steps.
Find the value of $12- - 13-6$
Step 1: Change the double negatives into additions.
$$12 - - 13 -6 = 12 + 13 -6$$
Step 2: Do the additions and subtractions starting on the left.
$$12 + 13 -6=25-6=19$$
Find the value of $-4 -9 --17$
Step 1: Change the double negatives into additions.
$$-4 -9 --25 = -4-9+17$$
Step 2: Do the additions and subtractions starting on the left.
$$-4-9+17=-13+17=4$$
KS3 Maths Revision Cards
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Example Questions
Step 1: Change the double negatives into additions.
$$- -7 +3-2=+7+3-2=7+3-2$$
Step 2: Do the additions and subtractions starting on the left.
$$7+3-2=10-2=8$$
Step 1: Change the double negatives into additions.
$$- -11- -15 - -9=+11+15+9=11+15+9$$
Step 2: Do the additions and subtractions starting on the left.
$$11+15+9=26+9=35$$
KS3 Maths Revision Cards
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• All of the major KS2 Maths SATs topics covered
• Practice questions and answers on every topic
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# Chapter 1 - Section 1.1 - Propositional Logic - Exercises - Page 15: 27
a) Converse: I will ski tomorrow only if it snows today. Contrapositive: If I do not ski tomorrow, then it will not have snowed today. Inverse: If it does not snow today, then I will not ski tomorrow. ________________________________________________ b) Converse: If I come to class, then there will be a quiz. Contrapositive: If I do not come to class, then there will not be a quiz. Inverse: If there is not going to be a quiz, then I don’t come to class. ______________________________________________________ c) Converse: A positive integer is a prime if it has no divisors other than 1 and itself. Contrapositive: If a positive integer has a divisor other than 1 and itself, then it is not prime. Inverse: If a positive integer is not prime, then it has a divisor other than 1 and itself.
#### Work Step by Step
For each step, we first come up with the contrapositive, ¬q → ¬p, of a conditional statement p → q that has the same truth value as p → q. Knowing that the contrapositive is false when ¬p is false, and ¬q is true, that is, only when p is true, and q is false. We are currently sure that neither the converse, q → p, nor the inverse, ¬p → ¬q, has the same truth value as p → q for overall possible truth values of p and q. Understanding that when p is true, and q is false, the original conditional statement is false, but the converse and the inverse are both true. ################################# For (a) : Because “q whenever p” can be considered either way to show the conditional statement, we consider that the conditional statement, p → q, where p is “it snows today” and for q is “I will ski tomorrow”: The original statement can be rewritten as “If it snows today, I will ski tomorrow.” The contrapositive of this conditional statement is “If I do not ski tomorrow, then it will not have snowed today. The converse is “I will ski tomorrow only if it snows today.” The inverse is “If it does not snow today, then I will not ski tomorrow.” ################################################# For (b): Same as (a), Let p to be “There is going to be a quiz” and q to be “I come to class” The original statement can be rewritten either as “If there is going to be a quiz, then I come to class.” Or “I come to class whenever there is going to be a quiz.” For Converse, “If I come to class, then there will be a quiz.” For Contrapositive, “If I do not come to class, then there will not be a quiz.” For Inverse, “If there is not going to be a quiz, then I don’t come to class.” ############################ For (c): Same as (a) and (b), p as “A positive integer is a prime” and q as “Has no divisor other than 1 and itself.” The original statement can be rewritten either as “A positive integer is a prime only if it has no divisors other than 1 and itself.” For Converse, “A positive integer is a prime if it has no divisors other than 1 and itself.” For Contrapositive, “If a positive integer has a divisor other than 1 and itself, then it is not prime.” For Inverse, “If a positive integer is not prime, then it has a divisor other than 1 and itself.”
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Question Video: Solving Simultaneous Equations Where One Is an Ellipse and the Other Is Linear | Nagwa Question Video: Solving Simultaneous Equations Where One Is an Ellipse and the Other Is Linear | Nagwa
# Question Video: Solving Simultaneous Equations Where One Is an Ellipse and the Other Is Linear Mathematics • Third Year of Preparatory School
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Solve the simultaneous equations π¦ = π₯ β 4 and (π₯Β²/5) + (π¦Β²/3) = 4, giving your answers to two decimal places.
05:58
### Video Transcript
Solve the simultaneous equations π¦ equals π₯ minus four and π₯ squared over five plus π¦ squared over three equals four, giving your answers to two decimal places.
There are several techniques that we can use to solve systems of linear equations. When the equations are of a different order β in this case, we have a linear and a quadratic β we generally tend to use the method of substitution. In this case, weβre going to substitute the equation π¦ equals π₯ minus four into the second equation. Now we might also notice that weβre told to give our answers correct to two decimal places. When weβre dealing with a quadratic equation, we tend to try and solve by factoring to find exact answers. However, since weβre told to give our answers to two decimal places, thatβs a good indication that weβre going to need to use the quadratic formula to solve any results in quadratics.
So, what does it look like when we substitute π¦ equals π₯ minus four into our second equation? Weβre going to replace π¦ squared with π₯ minus four squared. So, our second equation becomes π₯ squared over five plus π₯ minus four squared over three equals four. Now, because these fractions are making everything a little bit nasty, letβs multiply by the lowest common multiple of five and three. That will have the effect of getting rid of any fractions in this problem. π₯ squared over five times 15 is three π₯ squared, and π₯ minus four squared over three times 15 is five times π₯ minus four squared. Then, four times 15 is equal to 60.
Next, letβs distribute the parentheses π₯ minus four squared. Remember, thatβs not as simple as simply squaring the π₯ and squaring the negative four. We, in fact, need to multiply π₯ minus four by π₯ minus four. And to do so, we multiply each term in the first expression by each term in the second. π₯ times π₯ is π₯ squared; then π₯ times negative four and negative four times π₯ gives us negative four π₯ minus four π₯. Finally, negative four times negative four is 16. So, distributing the parentheses and we have π₯ squared minus four π₯ minus four π₯ plus 16, which simplifies to π₯ squared minus eight π₯ plus 16. We then replace π₯ minus four squared in our earlier equation with this new expression. And we have three π₯ squared plus five times π₯ squared minus eight π₯ plus 16 equals 60.
Letβs distribute the parentheses one more time. When we do, the left-hand side becomes three π₯ squared plus five π₯ squared minus 40π₯ plus 80. Simplifying further and our equation is eight π₯ squared minus 40π₯ plus 80 equals 60. If we then subtract 60 from both sides, we get a quadratic equation of the form we need to use the quadratic formula. Itβs eight π₯ squared minus 40π₯ plus 20 equals zero. And in fact, we could then divide everything through by four to get two π₯ squared minus 10π₯ plus five equals zero.
Letβs clear some space and use the quadratic formula to solve this quadratic equation. The quadratic formula says that the solutions to the equation ππ₯ squared plus ππ₯ plus π equals zero, if they exist, are given by π₯ equals negative π plus or minus the square root of π squared minus four ππ over two π. π is the coefficient of π₯ squared, whilst π is the coefficient of π₯, and π is the constant. So, in our quadratic equation, π is two, π is negative 10, and π is five. Substituting these values into the quadratic formula and we get π₯ equals negative negative 10 plus or minus the square root of negative 10 squared minus four times two times five over two times two. That simplifies to 10 plus or minus the square root of 60 all over four.
And so, there are two solutions to our quadratic equation, one where weβre adding 10 and the square root of 60 and one where weβre subtracting. Letβs calculate both of these values using our calculator. Correct to two decimal places, 10 plus the square root of 60 over four is 4.44 and 10 minus the square root of 60 over four is 0.56. Now it might be tempting to think weβre finished, but weβre not quite. We need to work out the corresponding values of π¦ for these two π₯-values, so we substitute each of them into the equation π¦ equals π₯ minus four. When π₯ equals 4.44 then, π¦ is 4.44 minus four, which is 0.44. Similarly, when π₯ is 0.56, π¦ is 0.56 minus four, which is negative 3.44. So, the solutions to our simultaneous equations are π₯ equals 4.44 and π¦ equals 0.44 and π₯ equals 0.56 when π¦ equals negative 3.44.
And itβs worth noting at this stage that these are pairs of solutions. The π₯-values will only work with their corresponding π¦-values and vice versa. And at this stage itβs worth noting that we can check our answer. To check, weβll substitute each pair of values into the second equation π₯ squared over five plus π¦ squared over three equals four. Weβll just demonstrate this using the first pair. When π₯ equals 4.44 and π¦ equals 0.44, the left-hand side of our equation is 4.44 squared over five plus 0.44 squared over three. And remember, if these values satisfy our equations, weβd be expecting theyβre equal to four. But we know weβve rounded our answers, so we might get a value just a little bit away from this. In fact, we get 4.007 and so on. So, we can assume that our answers are likely to be correct and weβre done.
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## College Algebra 7th Edition
convergent; $S_{\infty}=2$
RECALL: (1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula: $S_{\infty}=\dfrac{a}{1-r}$ (2) An infinite geometric series is divergent if $|r|\ge1$. (3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it: $r = \dfrac{a_n}{a_{n-1}}$ Solve for $r$ to obtain: $\require{cancel} r = \dfrac{-\frac{3}{2}}{3} \\r=-\dfrac{3}{2} \cdot \dfrac{1}{3} \\r=-\dfrac{\cancel{3}}{2} \cdot \dfrac{1}{\cancel{3}} \\r=-\dfrac{1}{2}$ Since $|-\frac{1}{2}|<1$, then the series is convergent. Solve for the sum using the formula above, with $a=3$ and $r=-\frac{1}{2}$, to obtain: $\require{cancel} S_{\infty}=\dfrac{3}{1-(-\frac{1}{2})} \\S_{\infty}=\dfrac{3}{\frac{2}{2}+\frac{1}{2}} \\S_{\infty}=\dfrac{3}{\frac{3}{2}} \\S_{\infty} = 3 \cdot \frac{2}{3} \\S_{\infty} = \cancel{3} \cdot \dfrac{2}{\cancel{3}} \\S_{\infty}=2$
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# How do you differentiate y=tanx/(2x^3)?
Dec 12, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} {\sec}^{2} x - 3 \tan x}{2 {x}^{4}}$
#### Explanation:
If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$
I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.
So with $y = \frac{\tan x}{2 {x}^{3}}$ Then
$\left\{\begin{matrix}\text{Let "u=tanx & => & (du)/dx=sec^2x \\ "And } v = 2 {x}^{3} & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 6 {x}^{2}\end{matrix}\right.$
$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 {x}^{3}\right) \left({\sec}^{2} x\right) - \left(\tan x\right) \left(6 {x}^{2}\right)}{2 {x}^{3}} ^ 2$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{3} {\sec}^{2} x - 6 {x}^{2} \tan x}{4 {x}^{6}}$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} {\sec}^{2} x - 3 \tan x}{2 {x}^{4}}$
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# How to Calculate 1/1 Divided by 22/1
Are you looking to work out and calculate how to divide 1/1 by 22/1? In this really simple guide, we'll teach you exactly what 1/1 ÷ 22/1 is and walk you through the step-by-process of how to divide fractions.
Before we dive into the calculation, let's recap on some fraction basics. The number above the dividing line is called the numerator, while the number above the dividing line is called the denominator.
For dividing fractions it's also useful to know that the first fraction (1/1) is called the dividend and the second fraction (22/1) is called the divisor.
Let's set up 1/1 and 22/1 side by side so they are easier to see:
1 / 1 / 22 / 1
Here is a really quick way to divide fractions. In the divisor (the second fraction) we flip the numerator and the denominator. This is known as the reciprocal and basically it means the reverse of the fraction. When we find the reciprocal, we also have to change the division sign to a multiplication sign:
1 / 1 x 1 / 22
Once you've flipped the second fraction and changed the symbol from divide to multiply, we can multiply the numerators together and the denominators together and we have our solution:
1 x 1 / 1 x 22 = 1 / 22
You're done! You now know exactly how to calculate 1/1 - 22/1. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
1/22
## Convert 1/1 times 22/1 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
1 / 22 = 0.0455
### Cite, Link, or Reference This Page
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# We are learning to: - Enhance Mathematical basic skills knowledge. (Which PLT skills?) -Accurately draw probability trees for dependent events and work.
## Presentation on theme: "We are learning to: - Enhance Mathematical basic skills knowledge. (Which PLT skills?) -Accurately draw probability trees for dependent events and work."— Presentation transcript:
We are learning to: - Enhance Mathematical basic skills knowledge. (Which PLT skills?) -Accurately draw probability trees for dependent events and work out probabilities from this. (Grade A*) Always aim high! LESSON OBJECTIVES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner PLT Skills Which ones are you using? AUTHOR Where are we in our journey? Real life cross/curricular links? www.mistrymaths.co.uk
LEARNING JOURNEY Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills
STARTER Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills EXTENSION TASK 1) 2) 3) 4) x - 7 When x = -6 Find the median of: 2, 4, 7, 11 Expand 5(-3x + 2) Write the value 5 cubed. 5) Calculate 6) Find the HCF of 18 and 30. 7) 25% of 36 miles 8) Simplify: 5e + 8y – 9e – 7y 9) Solve: 8(x + 4) = 56 2 11m 7m 4m Calculate the area of the trapezium = 5.5 (-6) - 7 2 = 29 = = -15x + 10 =125 =7 and -7 =6 = = 9 = -4e + y 8x + 32 = 56 8x = 24 x = 3 Area = (a + b)h Area = (7 + 11)4 Area = 36m 2
Hubert has 10 coloured counters in a bag. Three of the counters are blue and seven of the counters are green. He removes a counter at random from the bag and notes the colour and does not replace it. He then chooses a second counter. Record the information on a tree diagram and work out the probabilities of the different outcomes. PROBABILITY TREES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills EXAMPLE Blue Green 3 10 FIRST PICK 7 10 SECOND PICK Blue Green Blue Green COUNTERS HAVE NOT BEEN REPLACED SO THE PROBABILITY DOES CHANGE ON THE SECOND PICK 2 9 7 9 3 9 6 9 P(B,B) = ALWAYS MULTIPLY THE BRANCHES 10 3 X 9 2 = 90 6 P(B,G) = 10 3 X 9 7 = 90 21 P(G,B) = 10 7 X 9 3 = 90 21 P(G,G) = 10 7 X 9 6 = 90 42 P(The same colour) = P(B,B) + P(G,G) = 90 6 + 42 = 90 48 = 15 8 P(BG in any order) = P(B,G) + P(G,B) = 90 21 + 90 21 = 90 42 = 15 7
PROBABILITY TREES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills TASK (GRADE A*) 1) Q1 beads Lucy has a box of 30 chocolates. 18 are milk chocolate and the rest are dark chocolate. She takes a chocolate at random from the box and eats it. She then chooses a second. Draw a tree diagram to show all the possible outcomes. Milk First Pick Second Pick Dark P(Milk and Milk) = P(Milk and Dark) = P(Dark and Milk) = P(Dark and Dark) = Dark Milk Dark Milk
PROBABILITY TREES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills TASK (GRADE A*) 2) Q1 beads Jimmy has five coloured counters in a bag. Two of the counters are blue and the rest are red. He removes a counter at random from the bag and notes the colour but does not replace it. He then chooses a second counter. Draw a tree diagram showing all possible outcomes. (a) P(Blue,Blue) = (b) P(Red,Red) = (c) P(One blue and one red in any order) = Blue Red 2 5 FIRST PICK 3 5 SECOND PICK Blue Red Blue Red COUNTERS HAVE NOT BEEN REPLACED SO THE PROBABILITY DOES CHANGE ON THE SECOND PICK 1 4 3 4 2 4 2 4 P(B,B) = ALWAYS MULTIPLY THE BRANCHES 5 2 X 4 1 = 20 2 P(B,R) = 5 2 X 4 3 = 20 6 P(R,B) = 5 3 X 4 2 = 20 6 P(R,R) = 5 3 X 4 2 = 20 6 2 = 10 1 20 6 = 10 3 20 6 + 6 = 12 = 5 3
PROBABILITY TREES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills TASK (GRADE A*) 3) Q1 beads The probability Manchester City will win the Premiership is 0.7. If they do, the probability they will go on to win the F.A. Cup is 0.6. But if they dont, its only 0.5. Draw a tree diagram to represent this information. Use a tree diagram to find the probability that they win (a) the double (b) only one of them Win Lose 0.7 PREMIERSHIP FA CUP Win Lose Win Lose P(W,W) = ALWAYS MULTIPLY THE BRANCHES P(W,L) = P(L,W) = P(L,L) = 0.3 0.6 0.4 0.5 0.7 x0.6= 0.42 0.7 x0.4= 0.28 0.3 x0.5= 0.15 0.3 x0.5= 0.15 P(Double) = 0.42 P(Only one W) = 0.28 + 0.15 = 0.43
PROBABILITY TREES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills EXTENSION (GRADE A*) 1) Q1 beads A shop offers a package where customers can choose either a video player or a DVD player or a Hi-fi system and either a flat screen or a normal screen television. Over a period of time, 35% of the customers choose a video player and 60% choose a DVD player. The rest choose a Hi-fi system. Independently 75% of the customers choose flat screen televisions. (a) Complete the tree diagram below: (b) All customers were entered into a prize draw. What is the probability that the winner bought a Hi-Fi system with a normal screen television? 0.05 Probability adds up to one whole 0.25 P(H,N) = 0.05 x 0.25 = 0.0125
PROBABILITY TREES Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills EXTENSION (GRADE A*) 2) Q1 beads Sports activities are held after school. Zoe attends these classes on Monday and Wednesday. She can choose one of badminton, dance or netball on each day. The probability she chooses badminton is 0.5. The probability she chooses dance is 0.4. Assume Zoes choices are independent. (a) Complete the tree diagram below: (b) What is the probability Zoe will choose the same sports activity on both days? 0.1 Probability adds up to one whole 0.1 P(B,B) = 0.5 x 0.5 = 0.25 P(D,D) = 0.4 x 0.4 = 0.16 P(N,N) = 0.1 x 0.1 = 0.01 P(Same sport) = 0.25+ 0.16 + 0.01 = 0.42
DISCOVERY Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner Which ones are you using?PLT Skills LINK BACK TO OBJECTIVES - Accurately draw probability trees for dependent events and work out probabilities from this. What grade are we working at?
Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner PLT Skills Which ones are you using? PLENARY ACTIVITY – ROCK, PAPER AND SCISSORS PROBABILITY TREES Complete a fully labelled tree diagram to shows the outcomes of a rock, paper and scissors game between two players. TASK
P(Draw) = Effective Participator Self Manager Independent Enquirer Creative Thinker Team Worker Reflective Learner PLT Skills Which ones are you using? PLENARY ACTIVITY PROBABILITY TREES Player A Scissors Paper Stone Scissors Paper Stone Scissors Paper Stone Scissors Paper Stone Player B Draw A Wins B Wins Draw A Wins B Wins Draw 1/9 1/3 x 1/3 = 1/9 P(A Wins) = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 P(B Wins) = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 1/3
Draw your brain What have you learnt? In your brain, write or draw everything you can remember about drawing probability trees for dependent events. It can be a skill or a reflection, or something else that might be prominent in your brain. What grade are we working at? Where are we in our journey?
. How well do you understand the task? I dont understand I nearly understand I fully understand Creative Entrepreneur Responsible Citizen Independent Learner Positive Thinker Team Worker Reflective Learner Enterprise Skills Which ones are you using? SELF ASSESSMENT Plenary Activity www.mistrymaths.co.uk
SELF ASSESSMENT Plenary Activity On your post it notes… Think about how you can improve your work. On your post it notes… Think about how you can improve your work. WWW (What Went Well) EBI (Even Better If) Creative Entrepreneur Responsible Citizen Independent Learner Positive Thinker Team Worker Reflective Learner Enterprise Skills Which ones are you using? www.mistrymaths.co.uk
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## Properties of Circle
Introduction
A circle is a simple, beautiful and symmetrical shape.
When a circle is rotated through any angle about its centre, its orientation remains the same.
When any straight line is drawn through its centre, it divides the circle into two identical semicircles.
The line is known as the diameter.
The common distance of the points of a circle from its centre is called radius.
The perimeter or length of the circle is also known as the circumference.
Diameter
The diameter of a circle is the length of a line segment whose endpoints lie on the circle and which passes through the centre of the circle.
This is the largest distance between any two points on the circle.
The diameter of a circle is twice its radius.
In other words, two radius make one diameter.
Chords
A chord of a circle is a line segment whose two endpoints lie on the circle.
The diameter, passing through the circle’s centre, is the largest chord in a circle.
Tangent
A tangent to a circle is a straight line that touches the circle at a single point.
Secant
A secant is an extended chord: a straight line cutting the circle at two points.
Arc
An arc of a circle is any connected part of the circle’s circumference.
Sector
A sector is a region bounded by two radius and an arc lying between the radius.
Segment
A segment is a region bounded by a chord and an arc lying between the chord’s endpoints.
__________________________________________________________________________________________________________________________
Angle Properties in a circle
Angles in the same segment are equal.
∠A and ∠B are equal.
Proof:
AôB = 2×1 = 2×2 (Ð at centre= 2Ðat ⊙ce)
x1 = x2
ÐAPB = ÐAQB
Property can be abbreviated as Ðs in the same segment.
Angles in a semicircle = 90°.
∠B = 90°
Proof:
AôB = 2AĈB ( at centre = 2 at )
But AôB = 180°
AĈB = 90°
Angle at the centre = twice the angle subtended by the same arc at the circumference.
∠O = 2∠A
Proof:
In the figure below, the angles are subtended by the minor arc AB. Since OA = OD (radii of circle), a = b (base angles of isos. triangle)
But angle AôE if the exterior angle of triangle AOD
AôE = 2a
Similarly, c = d (base angles of isos. triangle)
BôE = 2c
Hence, AôB = 2a + 2c = 2(a + c) = 2 angle ADB
__________________________________________________________________________________________________________________
Angle Properties of a Cyclic Quadrilateral
Angles in opposite segments are supplementary.
∠A + ∠C = 180° ; ∠B + ∠D = 180°
∠A + ∠C = 180° ; ∠B + ∠D = 180°
Proof:
Let b = 50
2d = 360 – 100 = 260
d = 260/2 = 30
b + d = 30 + 50 = 180.
Property can be abbreviated as opp. Ðs of cyclic quad.
Exterior angle of a cyclic quadrilateral = the interior opposite angle.
a° = b°
Proof:
b + d = 180°(opp. angles of cyclic quad.)
x + d = 180°(adj. angles on a str. l)
b + d = x + d
b = x
angle ABC = angle CDE
Property can be abbreviated as ext.Ð of a cyclic quad.
__________________________________________________________________________________________________________________________
Symmetrical Properties of Circle
Property 1:
A circle is symmetrical about every diameter. Hence any chord AB perpendicular to a diameter is bisected by the diameter.
Also, any chord bisected by a diameter is perpendicular to the diameter.
Proof:
Given a circle, centre O and a chord, AB, with a mid-point D, we are required to show that OĈB = 90°.
Join OA and OB. In triangle OAC and OBC,
OA = OB (radii of circle)
AC = BC (given)
OC is common.
Triangle OCD is congruent to triangle OBC (SSS property)
OĈA = OĈB.
Since these are adjacent angles on a straight line, OĈA = OĈB = 90°
Property 2:
In equal circles or in the same circle, equal chords are equidistant from the centre. Chords which are equidistant from the centre are equal.
Proof:
In the figure, triangle OAB is rotated through an angle AOA’ to triangle OA’B’ about O.
Since rotation preserves shape and size, AB = A’B’ and OG = OH.
__________________________________________________________________________________________________________________________
Alternate Segment Theorem
The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ÐPTB = ÐPQT.
Proof:
ÐSPT = 90° (rt. Ð in a semicircle)
Ðx + Ðy = 90° ( Ð sum of ∆)
Ðy + Ðz = 90° ( tan rad.)
Ðx = Ðz
ÐPTB = ÐPST
ÐPST = ÐPQT ( Ðs in same segment)
ÐPTB = ÐPQT
____________________________________________________________________________________________
Tags
## One comment
• Also, a circle is a simple closed shape.
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# Unit J.1-J.2 Trigonometric Ratios
## Presentation on theme: "Unit J.1-J.2 Trigonometric Ratios"— Presentation transcript:
Unit J.1-J.2 Trigonometric Ratios
Tuesday, Feb 28 Trigonometry Ratios Find the tangent, sine and cosine ratios of acute angles. Use tangents, sines and cosines to find side lengths in right triangles and to solve real-world problems. Geometry Unit J: Trigonometry
Similar Right Triangles
Every right triangle with a given acute angle must be similar. How do we know this is true? The AA Similarity Postulate. Each of these three triangles have two congruent angles which makes them similar. 32° 32° 32° Unit J
Trigonometric Ratios When two right triangles are similar, then the ratios of their sides are the same. In a right triangle, these ratios are called trigonometric ratios. Let’s see a demonstration of these ratios. They are the same for each acute angle. Similar Right Triangles Unit J
The Trigonometric Ratios
Trigonometric ratios are ratios of the sides concerning one acute angle in a right triangle. In this triangle, let’s talk about the ratios for angle A. The side BC is the opposite side (the side directly across from angle A). B AC is the adjacent side AB is the hypotenuse A C Unit J
The Sine Ratio The sine of an angle is the ratio of the length of the leg opposite the angle to the length of the hypotenuse. B sin A = c a b A C Unit J
The Cosine Ratio The cosine of an angle is the ratio of the length of the leg adjacent the angle to the length of the hypotenuse. B cos A = c a b A C Unit J
Unit J.1-J.2 Trigonometric Ratios
Tuesday, Feb 28 The Tangent Ratio The tangent of an angle is the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle. B tan A = c Insert examples 1 and 2 here. a b A C Unit J Geometry Unit J: Trigonometry
Get Organized: Memory Device
Memory Device for Trigonometric Functions sin = S = cos = C = tan = T = S O H C A H T O A Unit J
Unit J.1-J.2 Trigonometric Ratios Tuesday, Feb 28 Using your graphing calculator You will be using your calculator to find the sine, cosine, or tangent functions. Before you start, you need to make sure that your calculator is set on degrees and not radians. All angles we will be using here are to be in degrees. First push and then make sure that the “degree” is highlighted. To check your calculator, clear screen, push and then “30)”. You should get 0.5 MODE Insert example 3 here SIN Unit J Geometry Unit J: Trigonometry
How do we use this information?
We can use these ratios to find missing measures in a right triangle. B x 40° A C 10 Unit J
Examples of Using Trigonometric Ratios
Unit J.1-J.2 Trigonometric Ratios Tuesday, Feb 28 Examples of Using Trigonometric Ratios To find x, use the sine ratio. To find y, use the cosine ratio. A x 25 Insert examples 4 and 5 here. 63° C y B Unit J Geometry Unit J: Trigonometry
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# How do you rewrite 2 \times 37 as the sum of two products?
Jul 6, 2018
Set 37 as 100-63 giving: $2 \times 37 \to 2 \times \left(100 - 63\right)$
$\left[200\right] + \left[- 126\right]$
#### Explanation:
$2 \times \left(37\right) = 74$
'Split' the 37 (partition it) into the sum of any two numbers.
Just to be different I select such that one of them is negative.
Set $37 = \textcolor{red}{100 - 63}$ so we have:
$\textcolor{g r e e n}{2 \times \left(\textcolor{red}{37}\right) \textcolor{w h i t e}{\text{d")-> color(white)("d}} 2 \times \left(\textcolor{red}{100 - 63}\right)}$
Multiply everything inside the brackets by the 2 that is outside.
$\textcolor{g r e e n}{\left[2 \times \textcolor{red}{100}\right] + \left[2 \times \left(\textcolor{red}{- 63}\right)\right]}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$
$\left[200\right] - \left[126\right] = 74 \leftarrow \text{ As required}$
Jul 6, 2018
color(brown)(=> 2 * 30 + 2 * 7
#### Explanation:
$2 \cdot 37 = \left(2 \cdot \left(30 + 7\right)\right)$ Distributive Law
$\implies 2 \cdot 30 + 2 \cdot 7$
We can write many such sum of two products.
$\implies 60 + 14 = 74$
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# How do you simplify the rational expression: (x+2)/(4x-8)(3x-9)/(x+4)(2x-21)/(x^2-x-6)?
Oct 14, 2015
$\frac{6 x - 63}{4 {x}^{2} + 8 x - 32}$
#### Explanation:
The idea is to factor out numbers is the first-degree polynomial (if possible), and to factor the highest-degree, finding their roots (if possible). So, let's work separately on the three pieces:
• First fraction:
Numerator: $x + 2 \to$ nothing to do;
**Denominator: $4 x - 8 \to$ can factor a $4$, obtaining $4 \left(x - 2\right)$.
• Second fraction:
Numerator: $3 x - 9 \to$ can factor a $3$, obtaining $3 \left(x - 3\right)$;
**Denominator: $x + 4 \to$ nothing to do.
• First fraction:
Numerator: $2 x - 21 \to$ nothing to do;
**Denominator: ${x}^{2} - x - 6 \to$ its roots are $3$ and $- 2$, so we can write it as $\left(x - 3\right) \left(x + 2\right)$.
Writing back the whole expression with this changes gives
$\frac{\textcolor{red}{\cancel{x + 2}}}{4 \left(x - 2\right)} \cdot \frac{3 \textcolor{b l u e}{\cancel{\left(x - 3\right)}}}{x + 4} \cdot \frac{2 x - 21}{\textcolor{b l u e}{\cancel{\left(x - 3\right)}} \textcolor{red}{\cancel{\left(x + 2\right)}}}$
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# KS2 Maths Quiz - Number Sequences (Year 5) (Questions)
In KS2 Maths as children advance to Year Five their knowledge of number sequences will increase. By now they should be able to recognise patterns that are increasing or decreasing by whatever amount in each step. They will also have a firm understanding of the properties of odd and even numbers and they will now be exposed to number sequences involving decimal numbers and fractions.
Number sequences involve finding a pattern in order to ascertain what the next number will be. Some patterns are straightforward, such as add 10, but others, like ones involving decimal numbers or fractions, can seem quite difficult to find. But once you know how, sequences with decimal numbers and fractions are just as easy to spot patterns in as ones made of whole numbers. It just takes a bit of practise.
Can you find the rules of these number sequences in this quiz for 9 -10 year olds?
1. Can you find the rule of this sequence? 6, 12, 18, 24, 30
[ ] The numbers increase in 6s [ ] The numbers increase in 3s [ ] The numbers increase in 5s [ ] The numbers increase in 4s
2. Can you find the rule of this sequence? 0, 0.1, 0.2, 0.3, 0.4
[ ] The numbers increase in 0.1s [ ] The numbers increase in 0.5s [ ] The numbers increase in 1s [ ] The numbers increase in 0.2s
3. Can you find the rule of this sequence? 14, 21, 28, 35
[ ] The numbers increase in 6s [ ] The numbers increase in 7s [ ] The numbers increase in 5s [ ] The numbers increase in 9s
4. What is the next number of the following sequence? 18, 27, 36, 45
[ ] 50 [ ] 52 [ ] 54 [ ] 56
5. What is the next number of the following sequence? 64, 56, 48, 40, 32
[ ] 35 [ ] 28 [ ] 26 [ ] 24
6. What will be the tenth number of this sequence? 22, 33, 44, 55, 66
[ ] 108 [ ] 110 [ ] 115 [ ] 121
7. The sum of three even numbers will be what?
[ ] Even [ ] Odd [ ] It will end in 0 [ ] It will end in 5
8. The sum of three odd numbers will be what?
[ ] Even [ ] Odd [ ] It will end in 0 [ ] It will end in 5
9. The difference between an odd number and an even number is what?
[ ] Sometimes odd and sometimes even [ ] Always odd [ ] Always even [ ] It always ends in 0
10. The difference between two odd numbers is what?
[ ] Sometimes odd and sometimes even [ ] Always odd [ ] Always even [ ] It always ends in 0
KS2 Maths Quiz - Number Sequences (Year 5) (Answers)
1. Can you find the rule of this sequence? 6, 12, 18, 24, 30
[x] The numbers increase in 6s [ ] The numbers increase in 3s [ ] The numbers increase in 5s [ ] The numbers increase in 4s
This sequence is the 6 times table
2. Can you find the rule of this sequence? 0, 0.1, 0.2, 0.3, 0.4
[x] The numbers increase in 0.1s [ ] The numbers increase in 0.5s [ ] The numbers increase in 1s [ ] The numbers increase in 0.2s
The sequence would continue: 0.5, 0.6, 0.7, 0.8... etc
3. Can you find the rule of this sequence? 14, 21, 28, 35
[ ] The numbers increase in 6s [x] The numbers increase in 7s [ ] The numbers increase in 5s [ ] The numbers increase in 9s
All of the numbers in this sequence are found in the 7 times table
4. What is the next number of the following sequence? 18, 27, 36, 45
[ ] 50 [ ] 52 [x] 54 [ ] 56
The numbers are increasing by 9
5. What is the next number of the following sequence? 64, 56, 48, 40, 32
[ ] 35 [ ] 28 [ ] 26 [x] 24
The numbers are decreasing by 8
6. What will be the tenth number of this sequence? 22, 33, 44, 55, 66
[ ] 108 [ ] 110 [ ] 115 [x] 121
The numbers are increasing by 11
7. The sum of three even numbers will be what?
[x] Even [ ] Odd [ ] It will end in 0 [ ] It will end in 5
The sum of 2 even numbers will be even. If you add 2 of the 3 numbers and then add another even it is the same as adding 2 even numbers
8. The sum of three odd numbers will be what?
[ ] Even [x] Odd [ ] It will end in 0 [ ] It will end in 5
The sum of two odd numbers is even but add another odd number and the total will be odd
9. The difference between an odd number and an even number is what?
[ ] Sometimes odd and sometimes even [x] Always odd [ ] Always even [ ] It always ends in 0
An odd + an odd = an even so an even - an odd = an odd
10. The difference between two odd numbers is what?
[ ] Sometimes odd and sometimes even [ ] Always odd [x] Always even [ ] It always ends in 0
An odd + an even = an odd so an odd - an odd = an even
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# What does vertices of a polygon mean?
Contents
## What does vertices of a polygon mean?
A point at which two polygon edges of a polygon meet.
## How do you find the vertices of a polygon?
Use this equation to find the vertices from the number of faces and edges as follows: Add 2 to the number of edges and subtract the number of faces. For example, a cube has 12 edges. Add 2 to get 14, minus the number of faces, 6, to get 8, which is the number of vertices.
How many vertices are in a polygon?
In any polygon a pair of sides that are inclined to each other meet at a point. This point is called a vertex of the polygon. In the polygon there are as many vertices as there are sides. The triangle has 3 sides and 3 vertices.
What are vertices and sides of a polygon?
The segments of a polygonal circuit are called its edges or sides. The points where two edges meet are the polygon’s vertices (singular: vertex) or corners. The interior of a solid polygon is sometimes called its body.
### What is the difference between vertex and vertices?
vertex – a single point defined in space. vertices – the plural of vertex. Note: 1 point is a vertex, 2 or more points are vertices.
### What do you mean by vertices?
In geometry, a vertex (in plural form: vertices or vertexes), often denoted by letters such as , , , , is a point where two or more curves, lines, or edges meet. As a consequence of this definition, the point where two lines meet to form an angle and the corners of polygons and polyhedra are vertices.
How many vertices does a Heptagon have?
7
Heptagon/Number of vertices
What is a polygon with 3 vertices?
POLYGONS—TRIANGLES A triangle has three sides and three vertices.
## Can a polygon have only 2 vertices?
In geometry, a digon is a polygon with two sides (edges) and two vertices.
## What is the difference between sides and vertices?
The two parts of a flat shape are its sides and its vertices. Sides are lines. Vertices are the points where the sides meet. You’re in good shape for this lesson!
Which is considered a four sided polygon?
Definition: A quadrilateral is a polygon with 4 sides. Definition: A parallelogram is a quadrilateral where both pairs of opposite sides are parallel.
Do polygon have more vertices than sides?
Every polygon has an equal number of sides and vertices . Thus, every polygon has an equal number of vertices and angle. For any simple polygon, the sum of its interior angles can be determined by the formula:
### How many faces edges vertices does a polygon have?
For a polygon an edge is a line segment on the boundary joining one vertex (corner point) to another. This Tetrahedron. Has 6 Edges. For a polyhedron an edge is a line segment where two faces meet.
### What are the segments of a polygon called?
The line segments of a polygon are called sides or edges. The point where two line segments meet is called vertex or corners, henceforth an angle is formed. An example of a polygon is a triangle with three sides.
What are the corners of a polygon?
A polygon is a closed geometric figure whose sides are simple line segments. Each corner of a polygon where two sides intersect is called a vertex of the polygon. For example, a triangle is a polygon with 3 sides. There are also three vertices, one at each point.
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# The Easiest Way to Round to the Nearest Hundred
Listen to the “talking digit” to quickly round numbers
Did you ever see numbers like ${\displaystyle 473}$ and ${\displaystyle 231}$ and wish you could instantly know about how much they added up to? Rounding numbers gives you a way to do that. Your number won't be exact, but it'll be close! This skill really comes in handy if you're trying to figure out how far away a place is or how much money you need to buy something you want. Read on to learn how to round to the nearest hundred, then see how you do on some practice problems.
## Things You Should Know
• Round down if the tens digit is ${\displaystyle 0,1,2,3,}$ or ${\displaystyle 4}$.
• Round up if the tens digit is ${\displaystyle 5,6,7,8,}$ or ${\displaystyle 9}$.
• Add ${\displaystyle 1}$ to the hundreds digit to round up. Leave the hundreds digit the same when rounding down.
• Change all the digits to the right of the hundreds place to ${\displaystyle 0}$ after rounding.
Method 1
Method 1 of 3:
### Finding the Nearest Hundred
1. 1
Look at the digit in the tens place. As you probably guessed by "nearest" hundred, rounding is all about distance. The tens digit, the one just to the right of the hundreds digit, tells you which hundreds digit your number is closest to—that's why it's sometimes called the "talking digit."[1]
• For example, in ${\displaystyle 478}$, the "talking digit" in the tens place is a ${\displaystyle 7}$.
• Using a number line can help you easily visualize which hundred your number is closest to.
2. 2
Round down if the tens digit is or ${\displaystyle 4}$. When the tens digit is ${\displaystyle 0,1,2,3,}$ or ${\displaystyle 4}$, the number is closer to the lower hundred than it is to the higher hundred. To round these numbers, just drop the extra digits and stay with the original hundreds digit.[2]
• For example, in ${\displaystyle 423}$, the tens digit is the talking digit, and it's a ${\displaystyle 2}$. So you'd round down to ${\displaystyle 400}$.
3. 3
Round up if the tens digit is or . The ${\displaystyle 5}$ is a little weird—technically it's equally close to both hundreds! But it's rounded up just to keep the rule simple.[3]
• For example, in ${\displaystyle 478}$, the tens digit is a ${\displaystyle 7}$, so you'd round up to ${\displaystyle 500}$.
4. 4
Add to the hundreds digit if you rounded up. When you round up, you basically give the number credit for another full hundred, even though it's not quite there. It's close enough! So you just add another ${\displaystyle 1}$.[4]
• For example, as you saw with ${\displaystyle 478}$, the ${\displaystyle 7}$ in the tens place means you need to round up, so you'd add ${\displaystyle 1}$ to ${\displaystyle 400}$ to get ${\displaystyle 500}$.
• If you round down, the hundreds digit stays the same. For example, ${\displaystyle 423}$ gets rounded down to ${\displaystyle 400}$ because the digit in the tens place is ${\displaystyle 2}$.
5. 5
Change all the digits to the right of the hundreds place to . The rounding takes care of all the numbers to the right—that's what makes it easier to do math with rounded numbers. Any numbers to the left of the hundreds place, on the other hand, are going to stay the same.[5]
• For example, if you have ${\displaystyle 478}$ and you're rounding to the nearest hundred, you get ${\displaystyle 500}$.
• But what if you have ${\displaystyle 2,478}$? The numbers to the left stay the same, so if you're rounding it to the nearest hundred, you get ${\displaystyle 2,500}$.
Method 2
Method 2 of 3:
### Practice Problems
1. 1
Round , , and to the nearest hundred. Remember to look at the number in the tens place to decide whether to round up or round down. You've got this![6]
2. 2
Round and to the nearest hundred. Decide whether to round up or down based on the number in the tens place. Remember to change the numbers to the right of the hundreds digit to ${\displaystyle 0}$.[7]
3. 3
Car shopping word problem. Your mom is shopping for a new car and has a budget of \$20,000. The car she's looking at has a starting retail price of \$18,000. By rounding to the nearest hundred, determine if she can afford to add the following options:[8]
• Power sunroof: \$725
• Rear spoiler: \$250
Method 3
Method 3 of 3:
### Practice Problem Solutions
1. 1
Your rounded numbers are , , and . Did the last one throw you? The number in the tens place is ${\displaystyle 8}$, which means you need to round up. Just add ${\displaystyle 1}$ to the digit in the hundreds place, which is ${\displaystyle 9}$, and you get ${\displaystyle 10}$, so your answer is ${\displaystyle 1,000}$.[9]
• Remember that you're rounding to the nearest hundred. With ${\displaystyle 147}$, it would be incorrect to first round the ${\displaystyle 4}$ up to ${\displaystyle 5}$ because of the ${\displaystyle 7}$ in the ones place, and then round up to ${\displaystyle 200}$.[10]
2. 2
Your rounded numbers are and . There's another one that carries over to the thousands! But here, you already have a number in the thousands place, ${\displaystyle 8}$. So you'd need to add a ${\displaystyle 1}$ to it to get ${\displaystyle 99,000}$.[11]
3. 3
Car shopping word problem. Yes, she can afford the power sunroof and the rear spoiler. Your mom's budget of \$20,000 gives her an extra \$2,000 over the base price of the car, which is \$18,000. You'd find this by subtracting the base price from your mom's budget.[12]
• Round the cost of each of the two options. The power sunroof is \$725, which rounds to \$700. The rear spoiler is \$250, which rounds to \$300.
• Add your rounded numbers together and you get \$1,000. Since you've already established that your mom has an extra \$2,000, and \$1,000 is less, you can conclude that she can afford the two options she wants.
## Expert Q&A
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Co-authored by:
Doctor of Law, Indiana University
This article was co-authored by wikiHow staff writer, Jennifer Mueller, JD. Jennifer Mueller is a wikiHow Content Creator. She specializes in reviewing, fact-checking, and evaluating wikiHow's content to ensure thoroughness and accuracy. Jennifer holds a JD from Indiana University Maurer School of Law in 2006. This article has been viewed 2,042 times.
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# ZingPath: Volume of Prisms and Pyramids
Searching for
## Volume of Prisms and Pyramids
Learn in a way your textbook can't show you.
Explore the full path to learning Volume of Prisms and Pyramids
### Lesson Focus
#### Problem Solving Involving Volumes of Prisms
Math Foundations
Students will apply the mathematical problem solving process to solve a real life problem about volumes of rectangular prisms.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Solve problems involving the volume of rectangular prisms.
• Calculate the volume of rectangular prisms.
### Everything You'll Have Covered
The volume of a three-dimensional figure is a measure of the space it occupies. Volume is measured in cubic units. A rectangular prism is a three-dimensional figure that has rectangular bases.
Volume of a prism V is the product of the area of its base B (l x w) and the height. The formula for the volume of a rectangular prism is
V = B x h or V = l x w x h
In this Activity Object, students will use V = l x w x h. Provide students with a several example problems using whole numbers less than ten for each of the dimensions to ensure that students can solve volume of rectangular prism problems.
A cube is used in this Activity Object. A cube is a rectangular prism in the same way that a square is a rectangle. Be sure students realize that a cube is also a rectangular prism whose dimensions are the same length. Therefore, its volume is found by multiplying the length of any dimension by itself twice.
• base (b) - any side or flat surface of the object
• cube - a rectangular prism whose length, width, and height are the same number of units
• dimension - a measure of length, width or height
• height (h) - the perpendicular distance to the base
• length (l)- the distance along a line or shape from one point to another
• prism - a figure made of two parallel faces that are polygons of the same shape and sides that are parallelograms
• rectangular prism - a prism whose base is a rectangle
• volume - the measure of the amount of space inside of a solid figure
• width (w)- the horizontal measurement to the length
### Tutorial Details
Approximate Time 30 Minutes Pre-requisite Concepts attributes of rectangular prisms and cubes, volume of rectangular prisms and cubes Course Math Foundations Type of Tutorial Problem Solving & Reasoning Key Vocabulary problem solving involving volumes of prisms, volume of a cube, volume of a rectangular prism
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FREE 8th Grade STAAR Math Practice Test
Welcome to our FREE 8th Grade STAAR Math practice test, with answer key and answer explanations. This practice test’s realistic format and high-quality practice questions can help your student succeed on the STAAR Math test. Not only does the test closely match what students will see on the real STAAR, but it also comes with detailed answer explanations.
For this practice test, we’ve selected 20 real questions from past exams for your student’s STAAR Practice test. Your student will have the chance to try out the most common STAAR Math questions. For every question, there is an in-depth explanation of how to solve the question and how to avoid mistakes next time.
Use our free STAAR Math practice tests and study resources (updated for 2022) to help your students ace the STAAR Math test! Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
$39.99 Satisfied 95 Students 10 Sample 8th Grade STAAR Math Practice Questions 1- What is the slope of a line that is perpendicular to the line $$4x-2y=12$$ ? A. 2 B. 1 C. $$-2$$ D. $$-\frac{1}{2}$$ 2- The diagonal of a rectangle is 10 inches long and the height of the rectangle is 8 inches. What is the perimeter of the rectangle in inches? 3- You can buy 5 cans of green beans at a supermarket for$3.40. How much does it cost to buy 35 cans of green beans?
A. $17 B.$23.80
C. $34.00 D.$119
4- Which of the following is the solution to the following inequality?
$$2x+4>11x-12.5-3.5x$$
A. $$x<3$$
B. $$x>3$$
C. $$x≤4$$
D. $$x≥4$$
5- What is the perimeter of a square that has an area of 595.36 feet?
6- A tree 32 feet tall casts a shadow 12 feet long. Jack is 6 feet tall. How long is Jack’s shadow?
A. 2.25 ft
B. 4 ft
C. 4.25 ft
D. 8 ft
7- The perimeter of the trapezoid below is 54 cm. What is its area?
8- Which graph does not represent $$y$$ as a function of $$x$$?
A.
B.
C.
D.
9- Which of the following is equivalent to $$13<-3x-2<22$$?
A. $$-8 < x < -5$$
B. $$5 < x < 8$$
C. $$\frac{11}{3} < x < \frac{20}{3}$$
D. $$\frac{-20}{3} < x < \frac{-11}{3}$$
10- In a certain bookshelf of a library, there are 35 biology books, 95 history books, and 80 language books. What is the ratio of the number of biology books to the total number of books on this bookshelf?
A. $$\frac{1}{4}$$
B. $$\frac{1}{6}$$
C. $$\frac{2}{7}$$
D. $$\frac{3}{8}$$
11- A bank is offering $$3.5\%$$ simple interest on a savings account. If you deposit $12,000, how much interest will you earn in two years? A.$420
B. $840 C.$4200
D. $8400 12- The area of a circle is $$64 π$$. What is the circumference of the circle? A. $$8 π$$ B. $$16 π$$ C. $$32 π$$ D. $$64 π$$ 13- A shirt costing$200 is discounted $$15\%$$. After a month, the shirt is discounted another $$15\%$$. Which of the following expressions can be used to find the selling price of the shirt?
A. $$(200) (0.70)$$
B. $$(200) – 200 (0.30)$$
C. $$(200) (0.15) – (200) (0.15)$$
D. $$(200) (0.85) (0.85)$$
14- Joe scored 20 out of 25 marks in Algebra, 30 out of 40 marks in science, and 68 out of 80 marks in mathematics. In which subject his percentage of marks is best?
A. Algebra
B. Science
C. Mathematics
D. Algebra and Science
15- What is the volume of the following triangular prism?
16- The marked price of a computer is D dollar. Its price decreased by $$20\%$$ in January and later increased by $$10\%$$ in February. What is the final price of the computer in D dollars?
A. 0.80 D
B. 0.88 D
C. 0.90 D
D. 1.20 D
17- Triangle ABC is graphed on a coordinate grid with vertices at A $$(–3, –2)$$, B $$(–1, 4)$$ and C $$(7, 9)$$. Triangle ABC is reflected over$$x$$ axes to create triangles A’ B’ C’.
Which order pair represents the coordinate of C’?
A. $$(7, 9)$$
B. $$(–7, –9)$$
C. $$(–7, 9)$$
D. $$(7, –9)$$
18-
What’s the maximum ratio of women to men in the four cities?
A. 0.98
B. 0.97
C. 0.96
D. 0.95
19-
What’s the ratio of the percentage of men in city A to the percentage of women in city C?
A. 0.9
B. 0.95
C. 1
D. 1.05
20-
A container holds 3.5 gallons of water when it is $$\frac{7}{24}$$ full. How many gallons of water does the container hold when it’s full?
A. 8
B. 12
C. 16
D. 20
Best 8th Grade STAAR Math Prep Resource for 2022
1- D
The equation of a line in slope intercept form is:$$y=mx+b$$
Solve for $$y$$.
$$4x-2y=12 {\Rightarrow} -2y=12-4x {\Rightarrow} y=(12-4x){\div}(-2) {\Rightarrow} y=2x-6$$
The slope of this line is 2.
The product of the slopes of two perpendicular lines is$$-1$$.
Therefore, the slope of a line that is perpendicular to this line is:
$$m_{1} {\times} m_{2} = -1 {\Rightarrow} 2 {\times} m_{2} = -1 {\Rightarrow} m_{2} = \frac{-1}{2}$$
2- 28
Let$$x$$ be the width of the rectangle. Use Pythagorean Theorem:
$$a^2 + b^2 = c^2$$
$$x^2 + 8^2 = 10^2 {\Rightarrow} x^2 + 64 = 100 {\Rightarrow} x^2 = 100 – 64 = 36 ⇒ x = 6$$
Perimeter of the rectangle =$$2 (length + width) = 2 (8 + 6) = 2 (14) = 28$$
3- B
Let $$x$$ be the number of cans. Write the proportion and solve for $$x$$.
$$\frac{5 \space cans}{ 3.40}=\frac{35 \space cans}{x}$$
$$x =\frac{3.40×35}{5}⇒x=23.8$$
4- A
$$2x+4>11x-12.5-3.5x$$→ Combine like terms:
$$2x+4>7.5x-12.5→$$ Subtract $$2x$$ from both sides: $$4>5.5x-12.5$$
Add 12.5 both sides of the inequality.
$$16.5>5.5x,$$
Divide both sides by 5.5.
$$\frac{16.5}{5}>x→x<3$$
5- 97.6
Area of a square: $$S = a^2 ⇒ 595.36 = a^2 ⇒ a = 24.4$$
Perimeter of a square: $$P = 4a ⇒ P = 4 × 24.4 ⇒ P = 97.6$$
6- A
Write the proportion and solve for the missing number.
$$\frac{32}{12}=\frac{6}{x}→ 32x=6×12=72$$
$$32x=72→x=\frac{72}{32}=2.25$$
7- 130
The perimeter of the trapezoid is 54 cm.
Therefore, the missing side (high) is $$54 – 18 – 12 – 14 = 10$$
Area of a trapezoid: $$A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2} (10) (12 + 14) = 130$$
8- C
A graph represents $$y$$ as a function of $$x$$ if
$$x_1=x_2→y_1=y_2$$
In choice C, for each $$x$$, we have two different values for $$y$$.
9- A
$$13<-3x-2<22$$→ Add 2 to all sides. $$13+2<-3x-2+2<22+2$$
$$→15<-3x<24$$→ Divide all sides by $$- 3$$. (Remember that when you divide all sides of an inequality by a negative number, the inequality sing will be swapped. < becomes >)
$$\frac{15}{-3} > \frac{-3x}{3} >\frac{24}{-3}$$
$$-8 < x < -5$$
10- B
Number of biology book: 35
Total number of books; $$35+95+80=210$$
The ratio of the number of biology books to the total number of books is: $$\frac{35}{210}=\frac{1}{6}$$
11- B
Use simple interest formula:
I=prt
(I = interest, p = principal, r = rate, t = time)
$$I=(12000)(0.035)(2)=840$$
12- B
Use the formula for area of circles.
Area $$= πr^2 ⇒ 64 π = πr^2 ⇒ 64 = r^2 ⇒ r = 8$$
Radius of the circle is 8. Now, use the circumference formula:
Circumference$$= 2πr = 2π (8) = 16 π$$
13- D
To find the discount, multiply the number by ($$100\% –$$ rate of discount).
Therefore, for the first discount we get: $$(200) (100\% – 15\%) = (200) (0.85)$$
For the next $$15\%$$ discount: $$(200) (0.85) (0.85)$$
14- C
Compare each mark:
In Algebra Joe scored 20 out of 25 Algebra. It means Joe scored $$80\%$$ of the total mark.
$$\frac{20}{25}=\frac{x}{100}⇒x= 80%$$
Joe scored 30 out of 40 in science. It means Joe scored $$75\%$$ of the total mark.
$$\frac{30}{40}=\frac{x}{100}⇒x= 75%$$
Joe scored 68 out of 80 in mathematics that which means $$85\%$$ of the total mark.
$$\frac{68}{80}=\frac{x}{100}⇒x= 85%$$
Therefore, his score in mathematic is higher than his other scores.
15- 12
Use the volume of the triangular prism formula.
$$V =\frac{1}{2} (length) (base) (high)$$
$$V = \frac{1}{2} × 4 × 3 × 2 ⇒ V = 12 \space m^3$$
16- B
To find the discount, multiply the price by ($$100\% –$$ rate of discount).
Therefore, for the first discount we get: $$(D) (100\% – 20\%) = (D) (0.80) = 0.80 D$$
To increase the $$10 \%: (0.80 D) (100\% + 10\%) = (0.85 D) (1.10) = 0.88 D = 88\%$$ of $$D$$
17- D
When a point is reflected over $$x$$ axes, the $$(y)$$ coordinate of that point changes to $$(-y)$$ while its $$x$$ coordinate remains the same.
$$C (7, 9) → C’ (7, -9)$$
18- B
Ratio of women to men in city A: $$\frac{570}{600}=0.95$$
Ratio of women to men in city B: $$\frac{291}{300}=0.97$$
Ratio of women to men in city C: $$\frac{665}{700}=0.95$$
Ratio of women to men in city D: $$\frac{528}{550}=0.96$$
19- D
Percentage of men in city $$A = \frac{600}{1170}×100=51.28%$$
Percentage of women in city $$C = \frac{665}{1365}×100=48.72%$$
Percentage of men in city $$A$$ to percentage of women in city $$C =\frac{51.28}{48.72}=1.05$$
20- B
let $$x$$ be the number of gallons of water the container holds when it is full.
Then;$$\frac{7}{24}x=3.5→x=\frac{24×3.5}{7}=12$$
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# Appendix E Complex Numbers E1 E Complex Numbers - Middlebury College
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9781285057095 AppE.qxp 2/18/13 8:25 AM Page E1 Appendix E E Complex Numbers E1 Complex Numbers Use the imaginary unit i to write complex numbers, and add, subtract, and multiply complex numbers. Find complex solutions of quadratic equations. Write the trigonometric forms of complex numbers. Find powers and nth roots of complex numbers. Operations with Complex Numbers Some equations have no real solutions. For instance, the quadratic equation x2 1 0 Equation with no real solution has no real solution because there is no real number x that can be squared to produce 1. To overcome this deficiency, mathematicians created an expanded system of numbers using the imaginary unit i, defined as i 冪 1 Imaginary unit where i 1. By adding real numbers to real multiples of this imaginary unit, you obtain the set of complex numbers. Each complex number can be written in the standard form a bi. The real number a is called the real part of the complex number a bi, and the number bi (where b is a real number) is called the imaginary part of the complex number. 2 Definition of a Complex Number For real numbers a and b, the number a bi is a complex number. If b 0, then a bi is called an imaginary number. A number of the form bi, where b 0, is called a pure imaginary number. To add (or subtract) two complex numbers, you add (or subtract) the real and imaginary parts of the numbers separately. Addition and Subtraction of Complex Numbers If a bi and c di are two complex numbers written in standard form, then their sum and difference are defined as follows. Sum: 共a bi兲 共c di兲 共a c兲 共b d兲i Difference: 共a bi兲 共c di兲 共a c兲 共b d兲i
9781285057095 AppE.qxp 2/18/13 8:25 AM Page E3 Appendix E Complex Numbers E3 Multiplying Complex Numbers a. 共3 2i兲共3 2i兲 3共3 2i兲 2i共3 2i兲 9 6i 6i 4i 2 9 6i 6i 4共 1兲 9 4 13 b. 共3 2i兲2 共3 2i兲共3 2i兲 3共3 2i兲 2i共3 2i兲 9 6i 6i 4i 2 9 6i 6i 4共 1兲 9 12i 4 5 12i Distributive Property Distributive Property i 2 1 Simplify. Write in standard form. Square of a binomial Distributive Property Distributive Property i 2 1 Simplify. Write in standard form. In Example 2(a), notice that the product of two complex numbers can be a real number. This occurs with pairs of complex numbers of the form a bi and a bi, called complex conjugates. 共a bi兲共a bi兲 a2 abi abi b2i 2 a2 b2共 1兲 a2 b2 To write the quotient of a bi and c di in standard form, where c and d are not both zero, multiply the numerator and denominator by the complex conjugate of the denominator to obtain a bi a bi c di c di c di c di 共ac bd 兲 共bc ad 兲i . c2 d 2 冢 冣 Multiply numerator and denominator by complex conjugate of denominator. Write in standard form. Writing Complex Numbers in Standard Form 2 3i 2 3i 4 2i 4 2i 4 2i 4 2i 冢 冣 8 4i 12i 6i 2 16 4i 2 8 6 16i 16 4 2 16i 20 4 1 i 10 5 Multiply numerator and denominator by complex conjugate of denominator. Expand. i 2 1 Simplify. Write in standard form.
9781285057095 AppE.qxp E4 2/18/13 Appendix E 8:25 AM Page E4 Complex Numbers Complex Solutions of Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, you often obtain a result such as 冪 3, which you know is not a real number. By factoring out i 冪 1, you can write this number in standard form. 冪 3 冪3共 1兲 冪3冪 1 冪3i The number 冪3i is called the principal square root of 3. REMARK The definition of principal square root uses the rule 冪ab 冪a冪b for a 0 and b 0. This rule is not valid when both a and b are negative. For example, 冪 5冪 5 冪5共 1兲冪5共 1兲 冪5i冪5i 冪25i 2 5i 2 5 whereas 冪共 5兲共 5兲 冪25 5. To avoid problems with multiplying square roots of negative numbers, be sure to convert to standard form before multiplying. Principal Square Root of a Negative Number If a is a positive number, then the principal square root of the negative number a is defined as 冪 a 冪ai. Writing Complex Numbers in Standard Form a. 冪 3冪 12 冪3i冪12i 冪36i 2 6共 1兲 6 b. 冪 48 冪 27 冪48i 冪27i 4冪3i 3冪3i 冪3i 2 2 c. 共 1 冪 3 兲 共 1 冪3i 兲 2 共 1兲2 2冪3i 共冪3 兲 共i 2兲 1 2冪3i 3共 1兲 2 2冪3i Complex Solutions of a Quadratic Equation Solve 3x2 2x 5 0. Solution 共 2兲 冪共 2兲2 4共3兲共5兲 2共3兲 2 冪 56 6 2 2冪14i 6 1 冪14 i 3 3 x Quadratic Formula Simplify. Write 冪 56 in standard form. Write in standard form.
9781285057095 AppE.qxp 2/18/13 8:25 AM Page E5 Appendix E Just as real numbers can be represented by points on the real number line, you can represent a complex number (3, 2) or 3 2i 3 ( 1, 3) or 2 1 3i z a bi 1 1 E5 Trigonometric Form of a Complex Number Imaginary axis 2 Complex Numbers 1 2 3 Real axis ( 2, 1) or 2 i as the point 共a, b兲 in a coordinate plane (the complex plane). The horizontal axis is called the real axis and the vertical axis is called the imaginary axis, as shown in Figure E.1. The absolute value of a complex number a bi is defined as the distance between the origin 共0, 0兲 and the point 共a, b兲. The Absolute Value of a Complex Number The absolute value of the complex number z a bi is given by Figure E.1 ⱍa biⱍ 冪a2 b2. When the complex number a bi is a real number 共that is, b 0兲, this definition agrees with that given for the absolute value of a real number. ⱍa 0iⱍ 冪a2 02 ⱍaⱍ To work effectively with powers and roots of complex numbers, it is helpful to write complex numbers in trigonometric form. In Figure E.2, consider the nonzero complex number a bi. By letting be the angle from the positive real axis (measured counterclockwise) to the line segment connecting the origin and the point 共a, b兲, you can write Imaginary axis (a, b) a r cos r b θ a and b r sin where r 冪a2 b2. Consequently, you have Real axis a bi 共r cos 兲 共r sin 兲i from which you can obtain the trigonometric form of a complex number. Figure E.2 Trigonometric Form of a Complex Number The trigonometric form of the complex number z a bi is given by z r共cos i sin 兲 where a r cos , b r sin , r 冪a2 b2, and tan b兾a. The number r is the modulus of z, and is called an argument of z. The trigonometric form of a complex number is also called the polar form. Because there are infinitely many choices for , the trigonometric form of a complex number is not unique. Normally, is restricted to the interval 0 2 , although on occasion it is convenient to use 0.
9781285057095 AppE.qxp E6 2/18/13 Appendix E 8:25 AM Page E6 Complex Numbers Trigonometric Form of a Complex Number Write the complex number z 2 2冪3i in trigonometric form. Solution The absolute value of z is r 2 2冪3i 冪共 2兲2 共 2冪3 兲2 冪16 4 ⱍ ⱍ and the angle is given by Imaginary axis 3 2 4π 3 1 1 2 z 4 3 z 2 2 3i Figure E.3 4 b a 2冪3 2 冪 3. tan Real axis Because tan共 兾3兲 冪3 and because z 2 2冪3i lies in Quadrant III, choose to be 兾3 4 兾3. So, the trigonometric form is z r 共cos i sin 兲 4 4 4 cos i sin . 3 3 冢 冣 See Figure E.3. The trigonometric form adapts nicely to multiplication and division of complex numbers. Consider the two complex numbers z1 r1共cos 1 i sin 1 兲 and z2 r2共cos 2 i sin 2 兲. The product of z1 and z2 is z1z2 r1r2共cos 1 i sin 1 兲共cos 2 i sin 2 兲 r1r2关共cos 1 cos 2 sin 1 sin 2 兲 i 共sin 1 cos 2 cos 1 sin 2 兲兴. Using the sum and difference formulas for cosine and sine, you can rewrite this equation as z1z2 r1r2关cos共 1 2 兲 i sin共 1 2 兲兴. This establishes the first part of the rule shown below. The second part is left for you to verify (see Exercise 109). Product and Quotient of Two Complex Numbers Let z1 r1共cos 1 i sin 1 兲 and z2 r2共cos 2 i sin 2 兲 be complex numbers. z1z2 r1r2关cos共 1 2 兲 i sin共 1 2 兲兴 z1 r1 关cos共 1 2 兲 i sin共 1 2 兲兴, z2 0 z2 r2 Product Quotient
9781285057095 AppE.qxp 2/18/13 8:25 AM Page E7 Appendix E Complex Numbers E7 Note that this rule says that to multiply two complex numbers you multiply moduli and add arguments, whereas to divide two complex numbers you divide moduli and subtract arguments. Multiplying Complex Numbers Find the product z1z2 of the complex numbers. 冢 z1 2 cos 2 2 i sin , 3 3 冣 冢 z2 8 cos 11 11 i sin 6 6 冣 Solution 2 2 11 11 i sin 8 cos i sin 3 3 6 6 2 11 2 11 16 cos i sin 3 6 3 6 5 5 16 cos i sin 2 2 16 cos i sin 2 2 16关0 i共1兲兴 16i 冢 冣 冢 z1z2 2 cos 冤 冢 冤 冤 冣 冢 冣冥 冣 Multiply moduli and add arguments. 冥 冥 5 and are coterminal. 2 2 Check this result by first converting to the standard forms z1 1 冪3i and z2 4冪3 4i and then multiplying algebraically. Dividing Complex Numbers Find the quotient z1兾z2 of the complex numbers. z1 24共cos 300 i sin 300 兲, z2 8共cos 75 i sin 75 兲 Solution z1 24共cos 300 i sin 300 兲 z2 8共cos 75 i sin 75 兲 24 关cos共300 75 兲 i sin共300 75 兲兴 8 3关cos 225 i sin 225 兴 冤冢 冣 冢 冣冥 3 冪2 2 i 3冪2 3冪2 i 2 2 冪2 2 Divide moduli and subtract arguments.
9781285057095 AppE.qxp E8 2/18/13 Appendix E 8:25 AM Page E8 Complex Numbers Powers and Roots of Complex Numbers To raise a complex number to a power, consider repeated use of the multiplication rule. z r 共cos i sin 兲 z2 r 2 共cos 2 i sin 2 兲 z3 r 3共cos 3 i sin 3 兲 This pattern leads to the next theorem, which is named after the French mathematician Abraham DeMoivre (1667–1754). THEOREM E.1 DeMoivre’s Theorem If z r共cos i sin 兲 is a complex number and n is a positive integer, then zn 关r 共cos i sin 兲兴 n r n共cos n i sin n 兲. Finding Powers of a Complex Number Use DeMoivre’s Theorem to find 共 1 冪3i 兲 . 12 Solution First convert the complex number to trigonometric form. 冢 1 冪3i 2 cos 2 2 i sin 3 3 冣 Then, by DeMoivre’s Theorem, you have 共 1 冪3i兲12 冤 2冢cos 23 i sin 23 冣冥 12 2 i sin 12 3 4096 共cos 8 i sin 8 兲 4096. 冤 冢 212 cos 12 冣 冢 2 3 冣冥 Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial equation of degree n has n solutions in the complex number system. Each solution is an nth root of the equation. The nth root of a complex number is defined below. Definition of nth Root of a Complex Number The complex number u a bi is an nth root of the complex number z when z un 共a bi兲n.
9781285057095 AppE.qxp 2/18/13 8:25 AM Page E9 Appendix E Complex Numbers E9 To find a formula for an n th root of a complex number, let u be an n th root of z, where u s共cos i sin 兲 and z r共cos i sin 兲. By DeMoivre’s Theorem and the fact that un z, you have sn共cos n i sin n 兲 r共cos i sin 兲. Taking the absolute value of each side of this equation, it follows that s n r. Substituting r for s n in the previous equation and dividing by r, you get cos n i sin n cos i sin . So, it follows that cos n cos and sin n sin . Because both sine and cosine have a period of 2 , these last two equations have solutions if and only if the angles differ by a multiple of 2 . Consequently, there must exist an integer k such that n 2 k 2 k . n By substituting this value for in the next theorem. into the trigonometric form of u, you get the result stated THEOREM E.2 n th Roots of a Complex Number For a positive integer n, the complex number z r 共cos i sin 兲 has exactly n distinct n th roots given by 冢 n r cos 冪 2 k 2 k i sin n n 冣 where k 0, 1, 2, . . . , n 1. For k n 1, the roots begin to repeat. For instance, when k n, the angle 2 n 2 n n is coterminal with 兾n, which is also obtained when k 0. The formula for the n th roots of a complex number z has a nice geometric interpretation, as shown in Figure E.4. Note that because the n nth roots of z all have the same magnitude冪 r, n they all lie on a circle of radius冪r with center at the origin. Furthermore, because successive n r nth roots have arguments that differ by 2 兾n, the n roots are equally spaced along the circle. Figure E.4 Imaginary axis 2π n 2π n Real axis
9781285057095 AppE.qxp E10 2/18/13 Appendix E 8:25 AM Page E10 Complex Numbers Finding the n th Roots of a Complex Number Find the three cube roots of z 2 2i. Solution Because z lies in Quadrant II, the trigonometric form for z is z 2 2i 冪8 共cos 135 i sin 135 兲. By the formula for n th roots, the cube roots have the form 冢 6 8 cos 冪 135 360 k 135 360 k i sin . 3 3 冣 Finally, for k 0, 1, and 2, you obtain the roots 冪2 共cos 45 i sin 45 兲 1 i 兲 1.3660 0.3660i 冪2 共cos 285 i sin 285 兲 0.3660 1.3660i. 冪2 共cos 165 i sin 165 E Exercises Performing Operations In Exercises 1–24, perform the operation and write the result in standard form. 1. 共5 i兲 共6 2i兲 Writing a Complex Conjugate In Exercises 25–32, write the complex conjugate of the complex number. Then multiply the number by its complex conjugate. 25. 5 3i 26. 9 12i 3. 共8 i兲 共4 i兲 27. 2 冪5i 28. 4 冪2i 4. 共3 2i兲 共6 13i兲 29. 20i 30. 冪 15 31. 冪8 32. 1 冪8 6. 共8 冪 18 兲 共4 3冪2i兲 Writing in Standard Form In Exercises 33–42, write the 7. 13i 共14 7i兲 quotient in standard form. 2. 共13 2i兲 共 5 6i兲 5. 共 2 冪 8 兲 共5 冪 50 兲 8. 22 共 5 8i兲 10i 3 5 5 11 9. 共2 2i兲 共3 3 i兲 10. 共1.6 3.2i兲 共 5.8 4.3i兲 11. 冪 6 冪 2 12. 冪 5 冪 10 13. 共冪 10 兲2 14. 共冪 75 兲2 15. 共1 i兲共3 2i兲 16. 共6 2i兲共2 3i兲 10 2i 33. 6 i 34. 35. 4 4 5i 36. 3 1 i 37. 2 i 2 i 38. 8 7i 1 2i 39. 6 7i i 40. 8 20i 2i 41. 1 共4 5i兲2 42. 共2 3i兲共5i兲 2 3i 17. 6i共5 2i兲 Performing Operations In Exercises 43–46, perform the 18. 8i共9 4i兲 operation and write the result in standard form. 19. 共冪14 冪10i兲共冪14 冪10i兲 20. 共3 冪 5 兲共7 冪 10 兲 43. 2 3 1 i 1 i 44. 2i 5 2 i 2 i 21. 共4 5i兲2 45. i 2i 3 2i 3 8i 46. 3 1 i i 4 i 22. 共2 3i兲 2 23. 共2 3i兲2 共2 3i兲2 24. 共1 2i兲2 共1 2i兲2
9781285057095 AppE.qxp 2/18/13 8:25 AM Page E11 Appendix E Complex Numbers E11 Using the Quadratic Formula In Exercises 47–54, use the Performing Operations In Exercises 83–86, perform the Quadratic Formula to solve the quadratic equation. operation and leave the result in trigonometric form. 47. x2 2x 2 0 48. x2 6x 10 0 49. 4x2 16x 17 0 50. 9x2 6x 37 0 51. 4x2 16x 15 0 52. 9x2 6x 35 0 53. 16t2 4t 3 0 83. 冤3冢cos 3 i sin 3 冣冥冤4冢cos 6 i sin 6 冣冥 84. 冤 2 冢cos 2 i sin 2 冣冥冤6冢cos 4 i sin 4 冣冥 3 85. 关53共cos 140 86. cos共5 兾3兲 i sin共5 兾3兲 cos i sin Writing in Standard Form In Exercises 55–62, simplify i sin 140 兲兴关23共cos 60 i sin 60 兲兴 DeMoivre’s Theorem to find the indicated power of the complex number. Write the result in standard form. the complex number and write it in standard form. 87. 共1 i兲5 55. 6i3 i 2 88. 共2 2i兲6 56. 4i 2 2i 3 89. 共 1 i兲10 57. 5i5 90. 共1 i兲12 58. 共 i兲3 91. 2共冪3 i兲7 60. 共冪 2 兲6 93. 92. 4共1 冪3i兲3 59. 共冪 75 兲3 1 i3 62. 1 共2i兲3 Using DeMoivre’s Theorem In Exercises 87–94, use 54. 5s2 6s 3 0 61. 冢cos 54 i sin 54 冣 94. 冤 2冢cos i sin 冣冥 2 2 10 8 Absolute Value of a Complex Number In Exercises 63–68, plot the complex number and find its absolute value. 63. 5i 64. 5 65. 4 4i 66. 5 12i 67. 6 7i 68. 8 3i Finding n th Roots In Exercises 95–100, (a) use Theorem E.2 to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write each of the roots in standard form. 95. Square roots of 5共cos 120 i sin 120 兲 96. Square roots of 16共cos 60 i sin 60 兲 冢 97. Fourth roots of 16 cos Writing in Trigonometric Form In Exercises 69–76, 冢 represent the complex number graphically, and find the trigonometric form of the number. 98. Fifth roots of 32 cos 69. 3 3i 70. 2 2i 99. Cube roots of 71. 冪3 i 72. 1 冪3i 73. 2共1 冪3i兲 74. 75. 6i 76. 4 5 2 共冪3 i兲 4 4 i sin 3 3 5 5 i sin 6 6 冣 冣 125 共1 冪3i兲 2 100. Cube roots of 4冪2共1 i兲 Writing in Standard Form In Exercises 77–82, represent Solving an Equation In Exercises 101–108, use Theorem E.2 to find all the solutions of the equation and represent the solutions graphically. the complex number graphically, and find the standard form of the number. 101. x 4 i 0 102. x3 1 0 103. x5 243 0 104. x 4 81 0 105. x3 64i 0 106. x6 64i 0 77. 2共cos 150 i sin 150 兲 78. 5共cos 135 i sin 135 兲 3 79. 2共cos 300 i sin 300 兲 3 80. 4共cos 315 i sin 315 兲 冢 81. 3.75 cos 冢 82. 8 cos 3 3 i sin 4 4 i sin 12 12 冣 107. x3 共1 i兲 0 109. Proof 冣 108. x 4 共1 i兲 0 Given two complex numbers z1 r1共cos 1 i sin 1兲 and z2 r2共cos 2 i sin 2兲 show that z1 r1 关cos共 1 2兲 i sin共 1 2兲兴, z2 r2 z2 0.
To add (or subtract) two complex numbers, you add (or subtract) the real and imaginary parts of the numbers separately. a bi, bi b a bi. a i2 1. i 1 1. x x2 1 0 Appendix E Complex Numbers E1 E Complex Numbers Definition of a Complex Number For real numbers and the number is a complex number.If then is called an imaginary number.
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Definition:
A quadratic equation in which the term containing x without an exponent is not present is called a pure quadratic equation. In other words, a quadratic equation in which the term containing x raised to the power of 1 is not present is called a pure quadratic equation.
A pure quadratic equation can also be described as a quadratic equation in which there are only two terms, one is the term containing x-squared, and one is the constant number.
Sometimes, as you would notice in the below examples, pure quadratic equations behave like difference of two squares, which can be very useful to solve them quickly.
Examples:
• x2 = 25
• x2 = 36
• x2 = -81
• x2 - 121 = 0
• x2 - 25 = 0
• x2 + 4 = 0
The above are six examples of pure quadratic equations. In the above examples, the fourth and fifth examples are pure quadratic equations that are in the form of difference of two squares.
Solving:
Pure quadratic equations are easy to solve. You don't need to factor a pure quadratic equation to solve it, although you can. There are two methods to solve pure quadratic equations:
1. Taking square roots on both sides
2. Difference of two squares method
The first method, that of taking square roots, will help you solve all types of pure quadratic equations. The second method will allow you to solve only those pure quadratic equations that are in the form of difference of two squares. Let us learn how to do each method:
Taking square roots on both sides:
Let us learn this method by solving the pure quadratic equation x2 - 50 = 0
Step 1: Move the number/constant term to the right hand side of the equation, if not already.
x2 = 50
Step 2: Take square roots on both sides
x^2 = √50
x = + √50 or x = - √50
If required, you can simplify the above radicals to get x = 5√2, x = -5√2
Difference of two squares method:
This method can only be applied when the pure quadratic equation to be solved is in the form of difference of two squares. For example, the pure quadratic x2 - 25 = 0 can also be written as x2 - 52 = 0, because 5 squared equals 25.
Applying the factorization formula for difference of two squares, we obtain:
(x + 5) (x - 5) = 0
By applying the zero product rule to obtain the roots, we get:
Either x = -5 or x = 5.
You can learn more about the difference of two squares method here, and see more solved examples here.
1. Correct! Both of the examples (along with other ones in the post) are of pure quadratic expressions. The post explains what are pure quadratic equations. Although the equation x^2 = -81 does not have any real solutions, its still a pure quadratic equation.
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# What is the difference between ratio and proportions?
## What is the difference between ratio and proportions?
A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0. A proportion is an equation in which two ratios are set equal to each other. For example, if there is 1 boy and 3 girls you could write the ratio as: 1 : 3 (for every one boy there are 3 girls)
## What is the difference between ratio and proportion ks2?
Ratio tells us how much of one thing there is in relation to another thing. For example, ‘For every 2 apples we have 3 bananas’. Proportion tells us about how much of one thing there is in relation to the whole amount of something.
What is a proportion problem?
Proportion problems are word problems where the items in the question are proportional to each other. In these lessons, we will learn the two main types of proportional problems: Directly Proportional Problems and Inversely Proportional Problems.
### What is difference between ratio and proportion for Class 6?
Ratio is defined as the comparison of sizes of two quantities of the same unit. Proportion, on the other hand, refers to the equality of two ratios.
### How do you teach ratio and proportion?
To introduce proportions to students, give them tables of equivalent rates to fill in, such as the one below. This will help them learn proportional reasoning. Work with these tables (first using easy numbers) until the students get used to them. You can tie in some of them with real-life situations.
What is ratio proportion method?
Ratio-Proportion Method allows us the ability to compare numbers, units of measurement, or values. Clinicians must define a ratio and proportion. Ratios, often expressed in fraction format, are mathematical works of art designed in relationship patterns that explore comparisons between units, words, numbers.
#### How do you know if a problem is proportional?
Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.
#### What is ratio and proportion in maths for Class 6?
If two ratios are equal we say that they are in proportion and use the symbol to equate the two ratios. and say that 2, 4, 60 and 120 are in proportion. We say that 2, 5, 60 and 15 are not in proportion. So, if two ratios are not equal, then we say that they are not in proportion.
Why are ratios and proportions important?
Ratios and proportions are foundational to student understanding across multiple topics in mathematics and science. In mathematics, they are central to developing concepts and skills related to slope, constant rate of change, and similar figures, which are all fundamental to algebraic concepts and skills.
## How do you explain proportions?
A proportion is a comparison of two numbers that each represent the parts of a whole. Essentially, a proportion says that two fractions are the same, even if the amount is different. For example, 1/2 of 10 marbles is the same proportion as 1/2 of 50 marbles.
## How do you calculate ratio and proportion?
We are sharing\$48 between three people in the ratio 3:1:2.
• The total number of parts in the ratio is 3+1+2 = 6 parts.
• To find the value of each part in the ratio we divide\$48 by the total number of parts.
• \$48 รท 6 =\$8.
• To find how much each person receives we multiply the ratio by the value of each part.
• 13:1:2 multiplied by\$8 is\$24:\$8:\$16.
• How do you turn ratios into proportions?
The sum of the parts in the ratio is the denominator The number on the bottom of a fraction,below the line.
• 4+2 = 6,so the denominators of both fractions will be 6.
• The separate parts in the ratio make the numerators The number on the top of a fraction,above the line.
• ### Is a ratio the same thing as a proportion?
The main difference between ratio and proportion is that ratio defined as the comparison of sizes of two quantities of the same unit and proportion refers to the equality of two ratios.
### What are some of the problems with ratio analysis?
Investors. The ratio analysis is useful to the investors as it will help them in evaluating the financial condition and status of the company and use profitability ratio analysis to
• Creditors.
• Employees.
• Management.
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# Volume of a Sphere
## Step 6: Applications of the Volume formula.
We're out there in the real world now. With a handy little toolbox and some useful tips, lets try and tackle some problems...
PROBLEM 1)
Imagine you are in the 16th century. Italy. Sitting at the desk of Galileo Galilei. Galileo has given you the task of finding out the volumes of certain planets. He has provided you with the radii of those planets. They are as follows:
Mercury-----2,440,000 metres.
Venus------ 6,051,000 metres.
Mars ----- 11145013.12 feet.
Jupiter----- 78184601.92 yards.
Galileo wants all answers in terms of cubed metres . What answers will you give Galileo?
SPOILER: Watch out for Mars and Jupiter!
ANSWER:
Well, since we know the radii ...the problem becomes very simple!
The Answers are:
Mercury---------6.085 multiplied by 10 raised to the power 19.
Venus----------9.28 multiplied by 10 raised to the power 20.
Mars------------1.642 multiplied by 10 raised to the power 20.
Jupiter--------- 1.5306 multiplied by 10 raised to the power 24.
(All values are in cubed metres.)
PROBLEM 2)
I have to build a semi- spherical planetarium inside a huge room of volume 63,000 cu.metres. The diameter of the planetarium should be 50 metres. Will my planetarium fit inside the room?
ANSWER: Remember that the planetarium is SEMI-spherical. So it's volume is half the volume that we calculate for a sphere. Half the sphere will quite easily fit inside the room.
Volume of the planetarium=32724.92 cu.metres.
PROBLEM 3) Imagine that one day Mr. Gates, the millionaire goes mad and decides to fund a project in which the moon is to be opened up and completely filled with marbles having circumference 2metres. Now the radius of the moon is 1,738,000 metres. Approximate how many marbles will fit into the moon till the moon fills up completely!
ANSWER:
The no. of marbles =(volume of the moon) divided by (volume of each marble).
And radius = (circumference)/ 2*PI
Volume of each marble= 0.1351 cu.metres
Volume of the moon =2.199 multiplied by 10 raised to the power 19.
The no. of marbles =1.6277 multiplied by 10 raised to the power 20!!!
NOTE: Given below, are the tables which summarize all that we have already examined in the steps prior to this one. They may prove to be useful in solving the above problems.
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mattduvall5 years ago
for measuring the volume of things like golfballs or a small sphere its easier to use water displacement, measure how much water u have in a measuring cup, put in the sphere, measure how much you have after. the difference = the volume, more accurate for things like golfballs which are not perfect spheres
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# How do you find the centroid in integration?
## How do you find the centroid in integration?
We divide the complex shape into rectangles and find x (the x-coordinate of the centroid) and y (the y-coordinate of the centroid) by taking moments about the y- and x-coordinates respectively. Because they are thin plates with a uniform density, we can just calculate moments using the area.
## What is centroid of integration?
The centroid of an area can be thought of as the geometric center of that area. The location of the centroid is often denoted with a ‘C’ with the coordinates being x̄ and ȳ, denoting that they are the average x and y coordinate for the area.
How do you find the centroid of a curve?
Centroid of a Curve
1. Find the length of the curve: L = \int\, dL , where dL is the arclength parameter, dL=\sqrt {\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt .
2. Find the x-coordinate of the centroid: \bar x= \displaystyle \frac 1 L \int_0^1 x \sqrt { 1 + 9x^4} \, dx .
### What is the formula for a centroid?
Then, we can calculate the centroid of the triangle by taking the average of the x coordinates and the y coordinates of all the three vertices. So, the centroid formula can be mathematically expressed as G(x, y) = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3).
### What is the difference between Centre and centroid?
The center of gravity of any object is termed to the point where gravity acts on the body. Where on the other hand, the centroid is referred to as the geometrical center of a uniform density object….
Difference Between Center of Gravity and Centroid
Center of Gravity Centroid
What is a centroid statics?
A centroid is a weighted average like the center of gravity, but weighted with a geometric property like area or volume, and not a physical property like weight or mass. This means that centroids are properties of pure shapes, not physical objects. They represent the coordinates of the “middle” of the shape.
## What is the centroid of a right triangle?
The centroid of a right angle triangle is the point of intersection of three medians, drawn from the vertices of the triangle to the midpoint of the opposite sides.
## What is a centroid curve?
The center of mass or centroid of a region is the point in which the region will be perfectly balanced horizontally if suspended from that point. So, let’s suppose that the plate is the region bounded by the two curves f(x) and g(x) on the interval [a,b] .
How to find the centroid of an area using integration?
Determining the centroid of a area using integration involves finding weighted average values ˉx and ˉy, by evaluating these three integrals, A = ∫dA, Qx = ∫ˉyel dA Qy = ∫ˉxel dA,
### How are center of gravity and centroids related?
For center of gravity, the weighting factor is the weight, for center of mass, it is the mass, for 3d Centroids it is the volume, and for 2d Centroids it is area. To understand how these equations relate to one another consider a plate with a cross-sectional area , A, divided into n pieces with volume .
### How are centroids measured from the indicated origin?
In this table, all centroids are measured from the indicated origin. You must make the appropriate adjustments when the origin of your coordinate system is located elsewhere. The equations we have been discussing (7.2.2), (7.3.1), (7.4.1) and (7.4.2) are all variations on the general weighted average formula (7.1.2).
Which is the centroid along the coordinate axis?
Centroids By common practice, we refer to the centroidal axis as the centroid but to keep the confusion down we will often speak of a x-centroid or a y-centroid referring to the coordinate along that axis where the centroidal axis intersects the coordinate axis. 5 Centroids by Integration
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# The Sides of Triangle is Given Below. Determine It is Right Triangle Or Not. a = 8 Cm, B = 10 Cm and C = 6 Cm - Mathematics
Course
#### Question
The sides of triangle is given below. Determine it is right triangle or not.
a = 8 cm, b = 10 cm and c = 6 cm
#### Solution
In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides.
Here, the larger side is b = 10 cm.
Hence, we have to prove that a2 + c2 = b2.
Let solve the left hand side of the above equation.
a2 + c2 = 82 + 62
= 64 + 36
= 100
Now we will solve the right hand side of the equation,
b2 = 102 = 100
Here we can observe that left hand side is equal to the right hand side that is a2 + c2 = b2 .
Therefore, the given sides of a certain triangle form a right angled triangle.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 7: Triangles
Ex. 7.7 | Q: 2.4 | Page no. 119
RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 7: Triangles
Ex. 7.7 | Q: 2.4 | Page no. 119
#### Video TutorialsVIEW ALL [2]
Solution The Sides of Triangle is Given Below. Determine It is Right Triangle Or Not. a = 8 Cm, B = 10 Cm and C = 6 Cm Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.
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# Find if a Number is Divisible by 22
Given here is the tutorial explaining the shortcut method to determine if the given number is divisible by 22. Using divisibility rule for 22, you can determine if the number is divisible by 22 without performing the division operation. The rule states that, 'if the number satisfies the divisibility rules of 2 and 11, it is divisible by 22'.
## Divisibility Rule for 22
##### Rule 1:
Check whether the last digit is divisible by 2. Then the sum of the even digits are subtracted from the sum of the odd digits. If the result is '0', then the original number is also divisible by 22.
##### Example 1:
Find if the number 760672 is divisible by 22.
##### Solution :
Applying the divisibility rule for 22,
###### Step 1:
The divisibility condition for number 2 is, the last digit in the given number should ends in 0, 2, 4, 6 or 8. Therefore, here the last digit of the number is 2. Hence it is divisible by 2
###### Step 2:
The sum of the even digits are subtracted from the sum of the odd digits. The result should be either '0 ' or divisible by '11'.
(i.e) (6+6+2) - (7+0+7) = 0
###### Step 3:
In the above condition, it results in 0. Therefore, the original number 760672 is divisible by 22
##### Rule 2:
Check whether the number is divisible by 11, if it is, then the original number is also divisible by 22.
##### Example 2:
Let us consider the number 143 and find if it is divisible by 22.
##### Solution :
###### Step 1:
Check whether the number 143 is divisible by 11
11 x 13 = 143
Since, the number 143 is divisible by 11, it is also divisible by 22.
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# Test II Mathematics Study Guide for the GACE
Page 2
## Understanding the Base 10 System
An understanding of the base 10 system is essential as students solve problems with both whole numbers and decimals. The base 10 system is our everyday number system where integers 0-9 take on different values based on their place value. When introducing the base 10 system, students often use manipulatives like base 10 blocks or straws to build understanding. For example, with straws, a straw would equal 1, 10 straws bundled together would equal 10, and 10 bundles of straws would equal 100.
### Place Value
Place value is the value of each digit in a number, depending on its position in that number. For example, in the number 413, the value of the 4 is 400, the value of the 1 is 10, and the value of the 3 is three. This can also be shown by decomposing the number like this: 413 = 400 + 10 + 3.
#### Foundations of Place Value
Place value depends on the integer’s order of placement on one side of a decimal point. Each place is ten times larger than the place to its right. Numbers to the left of a decimal point have the following values:
• Ones (1)
• Tens (10)
• Hundreds (100)
• Thousands (1,000)
• Ten thousands (10,000)
• Hundred thousands (100,000)
#### Using Place Value in Operations
Students should be able to generalize their understanding of place value to solve problems with and compare numbers. This understanding begins with comparing and solving problems with the same number of digits. Students progress from solving problems that have only one digit (ones’ place).
#### Using in Multi-Digit Operations
Students progress to using place value to solve problems with multiple digits and varied numbers of digits. A strategy that may help students with this concept is rounding. Rounding is the process of changing a number to make it simpler to operate with. Usually, a particular place value is chosen as the point to which the student simplifies. Additionally, numbers 5 and up are usually rounded “up” and numbers 4 and below are usually rounded “down.” For example, rounding 568 to the nearest ten would result in 570, because 8 rounds up to 10.
#### Using with Decimals
Like the numbers to the left of the decimal point, the numbers to the right of the decimal point are also each 10 times larger than the number in the place to their right. Numbers to the right of the decimal point have the following values:
• Tenths (.1)
• Hundreths (.01)
• Thousandths (.001)
• Ten thousandths (.0001)
• Hundred thousandths (.00001)
## Fractions
A fraction is a quantity that is not equal to one whole number. Fractions can also be written as decimals. Some fractions are more than one, and are written as improper fractions.An example of an improper fraction is$$\frac{6}{5}$$. This can be decomposed into a mixed number.
A mixed number is a whole number and a fraction together. For example, $$\frac{6}{5}$$can be decomposed to a whole number (1) and a new fraction ($$\frac{1}{5}$$). The resulting mixed number is $$1 \frac{1}{5}$$.
### Understanding Fractions
A fraction is made up of a numerator and a denominator. The denominator is the number on the bottom and represents the number of equal parts which a quantity is being split into. A numerator is the number on top and represents the number of parts being included.
For example, $$\frac{1}{6}$$ means that one out of 6 equal parts is being considered. Early learners can use visual fraction models to build an understanding of fractions. A visual fraction model shows the denominator as a shape broken into equal parts, and the number in the numerator filled in. For the fraction $$\frac{1}{6}$$, a shape is broken up into $$6$$ equal parts, with $$1$$ part shaded in to represent the numerator. Once students have an understanding of what a fraction is, they can begin to find equivalent fractions, reduce fractions, and use fractions in operations.
Here are $$2$$ such models for the fraction$$\frac{1}{6}$$:
#### Equivalent Fractions
An equivalent fraction is a fraction that has the same value, but uses a different numerator and denominator. This can be done by multiplying or dividing both the numerator and denominator by the same number.
For example, $$\frac{1}{2}$$ is equivalent to$$\frac{2}{4}$$. This was found by multiplying both the numerator and denominator by 2.
$\frac{1}{2} \times \frac{2}{2} = \frac{2}{4}$
This process is often used when adding and subtracting fractions. In order to add and subtract fractions, both denominators must be the same. This process is called finding the common denominator. Another simple way to do this is to multiply the denominators together to arrive at a common number.
#### Reducing Fractions
Reducing a fraction means that a student simplifies the numbers in the fraction to their lowest possible terms. This is done by finding the greatest common factor, or the highest number that divides equally into both numbers. For example, the fraction $$\frac{4}{20}$$ can be reduced because both $$4$$ and $$20$$ are divisible by $$2$$ and $$4$$. Since $$4$$ is the greatest common factor, we can divide both numbers by $$4$$ and get the resulting simplified fraction:$$\frac{1}{5}$$.
$\frac{4 \div 4}{20 \div 4} = \frac{1}{5}$
### Using Multiplication and Division with Fractions and Whole Numbers
When students have a strong understanding of theories and practice of multiplication and division with whole numbers, they can apply this knowledge, combined with their knowledge of fractions, to multiply and divide fractions.
#### Multiplying Fractions
To multiply a fraction, students simply multiply the numerators to find the new numerator and multiply the denominators to find the new denominator. Reducing the fractions can happen before or after the numbers are multiplied.
#### Dividing Fractions
Dividing fractions is a similar process to multiplying fractions, but there is one extra step. Students must use the reciprocal of the second fraction. The reciprocal is the inverse, or “flipped” form of the fraction. In the example $$\frac{3}{4}$$ $$\div$$$$\frac{6}{7}$$, the first step is to “flip” the second fraction by switching the numerator and denominator and then multiplying the new fractions. The new problem would be$$\frac{3}{4} \times \frac{7}{6}$$.
### Solve Real-World Problems Using Fractions
The use of fractions and problems with fractions is a skill that is used in everyday life. While we often think of typical situations like dividing up food or objects, fractions are also used in other mathematical situations like telling time ($$15$$ minutes is $$\frac{1}{4}$$of an hour) or counting money ($$50$$ cents is half of a dollar).
All Study Guides for the GACE are now available as downloadable PDFs
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# Simplifying Expressions and Combining Like Terms
## Presentation on theme: "Simplifying Expressions and Combining Like Terms"— Presentation transcript:
Simplifying Expressions and Combining Like Terms
Section 3.1 Simplifying Expressions and Combining Like Terms
Topics Covered Identifying Like Terms
Commutative, Associative, and Distributive Properties Combining Like Terms Simplifying Expressions
Identifying Like Terms
This is an algebraic expression . It is comprised of one or more terms. - Terms are separated by an addition sign. - A variable term is a term in which one or more of the factors are variables. - A term which does not include a variable is called a constant term If two or more terms have the same variables raised to the same powers then they are said to be like terms
Identifying Terms, Coefficients, Like Terms
14𝑥 + 2𝑦 𝑦 2 +𝑥 𝑥𝑦 Terms: Variable terms: Constant terms: Coefficients: Like terms:
The Commutative Property
The Commutative Property of Addition states that the order in which two numbers are added does not affect the sum The Commutative Property of Multiplication states that the order in which two numbers are multiplied does not affect the product
The Associative Property
The associative property of addition states that The manner in which three numbers are grouped under addition does not affect the sum. The associative property of multiplication states that The manner in which three numbers are grouped under multiplication does not affect the product.
State the property used in the following: (2)(7)=(7)(2)
Commutative property of addition Commutative property of multiplication Associative property of addition Associative property of multiplication
State the property used in the following: 8+(13 + 9) = (8 + 13) + 9
Commutative property of addition Commutative property of multiplication Associative property of addition Associative property of multiplication
The Distributive Property
The distributive property states that the factor outside the parentheses is multiplied by each term in the sum within the parentheses. Examples:
Try These: 1. 2. 3.
Combining Like Terms Two terms may be combined if they are like terms.
We combine like terms by adding or subtracting their Numerical coefficients Examples
Try These: 1. 2.
Simplifying Expressions
Simplifying expressions is the process of using the distributive property to clear parentheses and combining like terms in order To make the expression as simple as possible. examples
Try These: 1. 2. 3. 4. 5.
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## Monday, November 30, 2009
### Solution: 3 Variable Inequality
Original problem: http://dharmath.thehendrata.com/2009/11/24/3-variable-inequality/
For $a,b,c > 0$, prove that:
$! (ab(a+b) + bc(b+c) + ca(c+a))^2 \geq 4abc(a+b+c)(a^2+b^2+c^2)$
#### First Solution
WLOG, we assume $a \geq b \geq c$.
From AM-GM: $RHS \leq (ac(a+b+c) + b(a^2+b^2+c^2))^2$
But $ac(a+b+c) + b(a^2+b^2+c^2) \leq ab(a+b) + bc(b+c) + ca(c+a) \iff b(b-a)(b-c) \geq 0$
#### Second Solution
We will use the usual notation $S(k,l,m)$ to denote the symmetric sum: $S(k,l,m) = \sum a^kb^lc^m$ where the sum is over all permutations of $a,b,c$.
$ab(a+b)+bc(b+c)+ca(c+a) = \sum a^2b = S(2,1,0)$.
$4(a+b+c)(a^2+b^2+c^2) = S(1,0,0).S(2,0,0) = 2S(3,0,0) + 4S(2,1,0)$
So $RHS = abc(2S(3,0,0) + 4S(2,1,0)) = 2S(4,1,1) + 4S(3,2,1)$
And $LHS = S(2,1,0).S(2,1,0) = S(4,2,0) + S(3,3,0) + S(4,1,1)+ 2S(3,2,1) + S(2,2,2)$
So we need to prove:
$! S(4,2,0) + S(3,3,0) + S(2,2,2) \geq S(4,1,1) + 2S(3,2,1)$
By Muirhead, $S(4,2,0) \geq S(4,1,1)$, so we only need to prove
$! S(3,3,0) + S(2,2,2) \geq 2S(3,2,1)$
$! \iff S(3,3,3) (S(0,0,-3) + S(-1,-1,-1)) \geq 2S(3,3,3) S(0,-1,-2)$
$! \iff S(0,0,-3) + S(-1,-1,-1) \geq 2S(0,-1,-2)$
which is true by Schur
#### 1 comment:
1. [...] http://dharmath.thehendrata.com/2009/11/30/solution-3-variable-inequality/ (1) Comment Read [...]
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## The Golden Ratio: Working from a Definition to Find a Value
I found the image above through the Wikipedia article on the golden ratio. After using what appears above to define the golden ratio, the article then reveals its exact and approximate values. Later, the writers of the article do show the calculations involved in doing this, but they seem unnecessarily complicated. I’m going to try to simplify the process here, and might later edit/simplify this Wikipedia article to make it more understandable.
So, first, “a + b is to a as a is to b” need to be written as a fraction, which is easy enough: (a + b)/a = a/b. The value of this fraction, a/b, is, by definition, φ, the golden ratio. As an equation, this can be written a/b = φ.
Next, apply cross-multiplication to (a + b)/a = a/b, and it becomes (a + b)(b) = (a)(a), which simplifies to ab + b² = a².
Also, since a/b = φ, this means that a = φb (via the multiplication property). Next, ab + b² = a² is rewritten, with φb substituted for each a. The result of this substitution is (φb)(b) + b² = (φb)², which then becomes φb² + b² = φ²b². To simplify this, b² may be cancelled (via the division property), producing φ + 1 = φ². This may then be rearranged (via the subtraction and symmetric properties) to φ² – φ – 1 = 0. Two values of φ can then be found via the quadratic formula, and they are {1 ± sqrt[1 – (4)(1)(-1)]}/2 = [1 ± sqrt(5)]/2. Use “+,” and calculate a decimal approximation for this irrational number, and you get ~1.618, which is the golden ratio. Use “-” instead, and you get a negative number (approximately -0.618), which can be rejected on the grounds that a ratio of two lengths must be positive, since all lengths, themselves, are positive.
Also, I’m changing my mind regarding changing Wikipedia, on this subject. The two versions of the calculation (the one now on Wikipedia, and mine) don’t match, but both are mathematically valid — and, while my version makes more intuitive sense to me, that doesn’t mean it would make more sense to others, and Wikipedia isn’t there for me alone. Until I actually wrote the calculation out, I thought my version would be simpler, but I cannot claim that now.
[Later addition: see the first comment below for a way, suggested by a reader of this blog, to simplify the calculation, as I wrote it above. I’m not going to take credit for his improvement, of course — that would violate mathematical etiquette!]
I go by RobertLovesPi on-line, and am interested in many things. Welcome to my little slice of the Internet. The viewpoints and opinions expressed on this website are my own. They should not be confused with the views of my employer, nor any other organization, nor institution, of any kind.
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### 2 Responses to The Golden Ratio: Working from a Definition to Find a Value
1. howardat58 says:
Starting from here
“Next, apply cross-multiplication to (a + b)/a = a/b, and it becomes (a + b)(b) = (a)(a), which simplifies to ab + b² = a².”
divide both sides by b² to get a/b + 1 = (a/b)²
write a/b = φ, and get φ + 1 = φ², or φ² – φ – 1 = 0
then solve it
Liked by 1 person
• Good idea! I have added a note at the bottom of the post to help people find your improvement.
Like
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# Multiplying Matrices
Multiplying matrices can be confusing, but if you’re organized and disciplined, it’s not difficult. Just make sure to keep things straight.
Let’s multiply two matrices: $\color{red}{A = \begin{bmatrix} 5 & 1 & 3 \\ 4 & 2 & -1 \end{bmatrix}} \quad \color{blue}{B = \begin{bmatrix} 8 & 11 \\ -6 & 7 \\ 0 & 9 \end{bmatrix}}$
First, think about $$C = AB$$. Can we multiply this? We start by deciding what the size of the product matrix $$C$$ will be. Since the dimensions of $$A$$ are $$2 \times 3$$ and the dimensions of $$B$$ are $$3 \times 2$$, the dimensions of $$C$$ will be $$(2 \times 3)(3 \times 2) = 2 \times 2$$.
We can only create a product matrix if the number of columns of the first matrix is the same as the number of rows in the second matrix. The size of the product matrix will be the number of rows of the first matrix and the number of columns of the second matrix.
We create a blank 2 x 2 matrix, with enough room to fill in the values: $\color{green}{C = \begin{bmatrix} \_\_\_\_ & \_\_\_\_ \\ \_\_\_\_ & \_\_\_\_ \end{bmatrix}}$
We look at the first matrix ($$A$$) in terms of rows. We will put the values from the first row of $$A$$ in the first row of $$C$$, and the same for the second row. Since there are three elements in each row of $$A$$, we will create three terms in each element of $$C$$. That is: $\color{green}{C = \begin{bmatrix} \color{red}{5}\cdot\_+\color{red}{1}\cdot\_+\color{red}{3}\cdot\_ & \color{red}{5}\cdot\_+\color{red}{1}\cdot\_+\color{red}{3}\cdot\_ \\ \color{red}{4}\cdot\_+\color{red}{2}\cdot\_+\color{red}{-1}\cdot\_ & \color{red}{4}\cdot\_+\color{red}{2}\cdot\_+\color{red}{-1}\cdot\_ \end{bmatrix}}$
Make sure to leave the blanks! At this point, we’re focusing only on the first matrix, but we want to make sure to leave space for the numbers from the second matrix.
Now we look at the second matrix ($$B$$) in terms of columns. We will put the values from the first column of $$B$$ in the first column of $$C$$, and so on for the other column. This will give us: $\color{green}{C = \begin{bmatrix} \color{red}{5}\cdot\color{blue}{8}+\color{red}{1}\cdot\color{blue}{-6}+\color{red}{3}\cdot\color{blue}{0} & \color{red}{5}\cdot\color{blue}{11}+\color{red}{1}\cdot\color{blue}{7}+\color{red}{3}\cdot\color{blue}{9} \\ \color{red}{4}\cdot\color{blue}{8}+\color{red}{2}\cdot\color{blue}{-6}+\color{red}{-1}\cdot\color{blue}{0} & \color{red}{4}\cdot\color{blue}{11}+\color{red}{2}\cdot\color{blue}{7}+\color{red}{-1}\cdot\color{blue}{9} \end{bmatrix}}$
To reiterate: We fill in the values from each ROW of $$A$$ in every element of the matching ROW of $$C$$, and the values from each COLUMN of $$B$$ in every element of the matching COLUMN of $$C$$. At this point, let’s get rid of the color and see what we have: $C = \begin{bmatrix} 5\cdot8+1\cdot-6+3\cdot0 & 5\cdot11+1\cdot7+3\cdot9 \\ 4\cdot8+2\cdot-6-1\cdot0 & 4\cdot11+2\cdot7-1\cdot9 \end{bmatrix}$
Finally, we evaluate each element of the matrix for our solution: $C = \begin{bmatrix} 34 & 89 \\ 20 & 49 \end{bmatrix}$
Let’s try it the other way: $$D = BA$$. What will be the size of $$D$$? Since the dimensions of $$B$$ are $$3 \times 2$$ and the dimensions of $$A$$ are $$2 \times 3$$, the dimensions of $$D$$ will be $$(3 \times 2)(2 \times 3) = 3 \times 3$$.
We create a blank 3 x 3 matrix: $\color{purple}{D = \begin{bmatrix} \_\_\_\_ & \_\_\_\_ & \_\_\_\_ \\ \_\_\_\_ & \_\_\_\_ & \_\_\_\_ \\ \_\_\_\_ & \_\_\_\_ & \_\_\_\_ \end{bmatrix}}$
This time, we look at $$B$$ in terms of rows. We put the values from the first row of $$B$$ in the first row of $$D$$, and so on for the other two rows, just as before. Since there are two elements in each row of $$B$$, we will create two terms in eacn element of $$D$$. That is: $\color{purple}{D = \begin{bmatrix} \color{blue}{8}\cdot\_ + \color{blue}{11}\cdot\_ & \color{blue}{8}\cdot\_ + \color{blue}{11}\cdot\_ & \color{blue}{8}\cdot\_ + \color{blue}{11}\cdot\_ \\ \color{blue}{-6}\cdot\_ + \color{blue}{7}\cdot\_ & \color{blue}{-6}\cdot\_ + \color{blue}{7}\cdot\_ & \color{blue}{-6}\cdot\_ + \color{blue}{7}\cdot\_ \\ \color{blue}{0}\cdot\_ + \color{blue}{9}\cdot\_ & \color{blue}{0}\cdot\_ + \color{blue}{9}\cdot\_ & \color{blue}{0}\cdot\_ + \color{blue}{9}\cdot\_ \end{bmatrix}}$
Now, we look at $$A$$ in terms of columns, placing the values of each column in the blanks in $$D$$: $\color{purple}{D = \begin{bmatrix} \color{blue}{8}\cdot\color{red}{5} + \color{blue}{11}\cdot\color{red}{4} & \color{blue}{8}\cdot\color{red}{1} + \color{blue}{11}\cdot\color{red}{2} & \color{blue}{8}\cdot\color{red}{3} + \color{blue}{11}\cdot\color{red}{-1} \\ \color{blue}{-6}\cdot\color{red}{5} + \color{blue}{7}\cdot\color{red}{4} & \color{blue}{-6}\cdot\color{red}{1} + \color{blue}{7}\cdot\color{red}{2} & \color{blue}{-6}\cdot\color{red}{3} + \color{blue}{7}\cdot\color{red}{-1} \\ \color{blue}{0}\cdot\color{red}{5} + \color{blue}{9}\cdot\color{red}{4} & \color{blue}{0}\cdot\color{red}{1} + \color{blue}{9}\cdot\color{red}{2} & \color{blue}{0}\cdot\color{red}{3} + \color{blue}{9}\cdot\color{red}{-1} \end{bmatrix}}$
Without the color, this is: $D = \begin{bmatrix} 8\cdot 5 + 11\cdot 4 & 8\cdot 1 + 11 \cdot 2 & 8 \cdot 3 + 11 \cdot -1 \\ -6 \cdot 5 + 7 \cdot 4 & -6 \cdot 1 + 7 \cdot 2 & -6 \cdot 3 + 7 \cdot -1 \\ 0 \cdot 5 + 9 \cdot 4 & 0 \cdot 1 + 9 \cdot 2 & 0 \cdot 3 + 9 \cdot -1 \end{bmatrix}$
Evaluating each element gives us: $D = \begin{bmatrix} 84 & 30 & 13 \\ -2 & 8 & -25 \\ 36 & 18 & -9 \end{bmatrix}$
If you would like to check your work, here is an online matrix multiplication calculator. Make up some small matrices and practice multiplying them, then check your answer! (Remember: These tools are provided so you can check your work, not to help you cheat.)
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## Fractions
A fraction can be used to represent a part of a whole or part of a group or collection of things. For example, a whole circle can be divided into four equal parts and the fraction used to represent three of four equal parts. Children should be reminded that the parts into which the whole is divided must be of equal size.
A fraction can be defined formally as a number that can be expressed in the form , where b ≠ 0, a is the numerator, and b is the denominator. The denominator of a fraction cannot equal zero.
A unit fraction is a fraction with a numerator of 1 (for example, , , , ). The definition of a unit fraction is to take one unit and divide it into n equal pieces. One of these smaller pieces is the amount represented by the unit fraction.
The fraction can represent the quotient of m and n or m ÷ n. If the fraction is defined in terms of the unit fraction , the fraction means m unit fractions of . If m = n, then = 1.
Comparing Fractions
At this grade level, the comparing and ordering of fractions is limited to unit fractions, and pictures are provided to aid a child in deciding which of two fractions is the greater or lesser. For unit fractions, > if and only if a < b. For example, > because 3 < 4.
Children learn that a fractional part of a number, such as of 6, can be found by separating the group of six into the same number of equal groups as the denominator of the fraction, which is 3 in this case, and then counting the number in 1 group, which is 2. So of 6 is 2.
2 is of 6.
Teaching Model 9.4: Comparing Fractions
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# Question Video: Counting by Tens to Add on a Blank Number Line Mathematics • 1st Grade
Use a number line to find 50 more than 22. Hint: count in tens
02:03
### Video Transcript
Use a number line to find 50 more than 22. Hint: Count in 10s.
In this question, we have to use the number line to find 50 more than 22. And we’re told to count in 10s. The number 22 has already been marked on the number line. 22 is a two-digit number. It has two 10s and two ones. We’re adding 50 to the number 22. The number 50 is a multiple of 10. There are five 10s in the number 50 and no ones. 22 has two 10s. And we have to add five more 10s. 10, 20, 30, 40, 50.
Did you notice what happened to the two digits as we counted forward five 10s on the number line? 22, 32, 42, 52, 62, 72. The tens digit increased, but the ones digit stayed the same. We had two 10s, and we counted forward five more 10s. Two plus five is seven, so two 10s plus five 10s is seven 10s. The tens digit increased by five. 50 more than 22 is 72. We started at number 22 on the number line, and we counted forward in 10s. 22 and five more 10s is 72.
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## Elementary Algebra
Published by Cengage Learning
# Chapter 3 - Equations and Problem Solving - 3.5 - Problem Solving - Problem Set 3.5 - Page 132: 33
#### Answer
There is 16.25 percent of grapefruit juice in the resulting mixture.
#### Work Step by Step
Let X represent the amount of grapefruit in the mixture. Since the first mixture is 30 ounces and the second mixture is 50 ounces, the resulting mixture will be 30 + 50 = 80 ounces. To find X, the amount of grapefruit in the mixture, we use the following guideline: Amount of grapefruit in the first solution + amount of grapefruit in the second solution = Amount of grapefruit in the final solution. 10% $\times$ 30 + 20% $\times$ 50 = X $.1 \times 30 + .2 \times 50 =X$ 3 + 10 = X X = 13 There are 13 ounces of grapefruit in the 80 ounce mixture. Since we need to calculate the percentage of grapefruit juice, we proceed: $\frac{13}{80}$ $\times$ 100% = 16.25%
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Home | | Maths 8th Std | Parts of a Circle: Numerical Example solved problems
# Parts of a Circle: Numerical Example solved problems
8th Maths : Chapter 2 : Measurements : Parts of a Circle: Numerical Example solved problems
Example 2.1
A circular shaped gymnasium ring of radius 35cm is divided into 5 equal arcs shaded with different colours. Find the length of each of the arcs.
Solution:
Radius, r = 35 cm and n = 5.
Length of each of the arcs, l = 1/n × 2πr units
= 1/5 × 2 × π × 35 = 14 π cm.
Example 2.2
A spinner of radius 7.5 cm is divided into 6 equal sectors. Find the area of each of the sectors.
Solution:
Radius, r = 7.5 cm and n = 6.
Area of each of the sectors, A = 1/n × πr2 sq. units
= 1/6 × π × 7.5 × 7.5
= 9.375π sq. cm
Example 2.3
Kamalesh has a dining table, circular in shape of radius 70 cm whereas Tharun has a circular quadrant dining table of radius 140 cm. Whose dining table has a greater area? [π = 22/ 7]
Solution:
Area of the dining table with Kamalesh = πr2 sq. units
= 22/7 × 70 × 70
= 15400 sq.cm (approximately.)
Area of the circular quadrant dining table with Tharun
= 1/4 πr2 = 1/4 × 22/7 ×140 ×140
A = 15400 sq.cm (approximately.)
We find that, the area of the dining tables of both of them have the same area.
Think
If the radius of a circle is doubled, what will happen to the area of the new circle so formed?
Solution:
If r = 2r1
Area of the circle = πr2 = π (2r1)2 = π4r12 = 4πr12
Area = 4 × old area.
Example 2.4
Four identical medals, each of diameter 7cm are placed as shown in Fig. 2.18. Find the area of the shaded region between the medals. [π = 22/7]
Solution:
Diameter, d = 7 cm, therefore r = 7/2 cm.
Area of the shaded region = Area of the square – 4 × Area of the circular quadrant
= a2 4 × 1/4 πr2
= (7×7) ( 4 × 1/4 × 22/7 × 7/2 ×7/2 )
= 49 –38.5 = 10.5 sq.cm. (approximately)
Tags : Measurements | Chapter 2 | 8th Maths , 8th Maths : Chapter 2 : Measurements
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
8th Maths : Chapter 2 : Measurements : Parts of a Circle: Numerical Example solved problems | Measurements | Chapter 2 | 8th Maths
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## Worked example 12
(Sketching the quadratic graph in the form of y=(x-m)(x-n)
Sketch the graph of each of the following functions:
a) y=(x-1)(x+3)
b) y=-(x+1)(x-5)
### Solutions for a)
Step 1: Determine whether the coefficient of x² is negative or positive.
Since the coefficient of x² is 1 which is a positive number, the graph opens upwards.
Step 2: Obtaining the x-intercepts by substituting y=0 into the equation.
When y=0,
(x-1)(x+3)=0
(x-1)=0 or (x+3)=0
x=1 x=-3
∴ the x intercepts are 1 and -3
Step 3: Obtaining the y -intercepts by substituting x=0 into the equation.
When x=0,
y=(0-1)(0+3)
y=-3
∴ the y intercept is -3
Step 4: Complete the graph.
### Solutions for b)
Step 1: Determine whether the coefficient or x² is negative or positive
Since the coefficient of x² is =1 which is a negative number, the graph opens downwards.
Step 2: Obtaining the x -intercepts by substituting y=0 into the equation.
When y=0,
-(x+1)(x-5)=0
(x+1)=0 or (x-5)=0
x=-1 x=5
∴ the x intercepts are -1 and 5
Step 3: Obtaining the y-intercepts by substituting x=0 into the equation
When x=0,
y=-(0+1)(0-5)
y=5
∴the y intercept is 5
Step 4: Sketch the graph.
## Practice NOW 12
Sketch the graph of each of the following functions.
a) y=(x-2)(x-6)
b) y=-(x-3)(x+2)
c) y=(1-x)(x+3)
## WORKED EXAMPLE 13
(i) find the coordinates of the x- and y-intercepts,
(ii) write down the coordinates of the maximum point of the graph,
(iii) sketch the graph,
(iv) state the equation of the line of symmetry of the graph.
### Solutions for (i)
When y=0,
0=-(x-2)²+9 Solve x by making it the subject
(x-2)² = 9
(x-2)=±9
x=3+2 or x=-3+2
x=5 or x=-1
∴ the coordinates of x-intercepts are (5,0) and (-1,0).
When x=0,
y = -(0-2)²+9
y=5
∴ the coordinates of y-intercept are (0,5).
### Solution for (ii)
The coordinates of the maximum point are (2,9)
### Solutions for (iii)
Step 1: Determine whether the coefficient of x² is negative or positive.
Since the coefficient of x² is -1 which is a negative number, the graph opens downwards.
Step 2: Obtaining the x-intercepts by substituting y=0 into the equation.
From (i) the x-intercepts are 5 and -1
Step 3: Obtaining the y-intercepts by substituting x=0 into the equation.
From (i) the y intercept is 5
Step 4: Sketch the graph and include the maximum point.
### Solution for (iv)
The equation of the line of symmetry is x=2.
## PRACTICE NOW 13
1. Given the quadratic function y=-x(-3)²+4,
(i) find the coordinates of the x- and y-intercepts,
(ii) write down the coordinates of the maximum point of the graph,
(iii) sketch the graph,
(iv) state the equation of the line of symmetry of the graph.
2. Given the quadratic function y=(x+2)²-16,
(i) find the coordinates of the x- and y-intercepts,
(ii) write down the coordinates of the maximum point of the graph,
(iii) sketch the graph,
(iv) state the equation of the line of symmetry of the graph.
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# NCERT Class 10 Math Chapter 9 Important Questions Answer – Some Application of Trigonometry
Class 10 Chapter Some Application of Trigonometry Subject Math Category Important Question Answer
## Class 10 Math Chapter 9 Important Question Answer
Q1. The shadow of a tower standing on a level ground is found to be 40 m longer when the altitude (the angle of elevation) of sun changes from 60° to 30°. Find the height of tower.
Ans. Let AB be the tower and BC be the length of the shadow when the sun’s altitude is 60° and DB is the length of the shadow, when the angle of elevation is 30°.
Now, Let AB = h m, and BC = x m.
According to Question,
DB = (40 + x) m
In right ∆ ABC, tan60° =
h = —- (i)
In right ∆ ABD, tan30° =
——- (ii)
Putting value of eq (i) in eq(ii) we get,
3x = x + 40
2x = 40
x = 20
h =
Therefore, the height of the tower is m.
Q2. The angle of elevation of the top of a building from foot of tower is 30° and angle of elevation of top of the tower from the foot of building is 60°. If height of tower is 50 m. Find the height of building.
Ans.
Let AB be the building and DC be the tower.
Given, Height of tower DC = 50m,
∠CBD = 60° and ∠BCA = 30°
Height of building AB = ?
In right ∆ DCB,
tan60° =
BC = —- (i)
In right ∆ ABC,
tan30° =
Putting value of BC in eq, we get
3AB = 50
AB =
Therefore, Height of the tower is m.
Q3. From the top of a 7 m building the angle of elevation of the top of a tower is 60° and angle of depression is 45°. Find the height of the tower. Most Important
Ans.
Given, height of building AB = CD = 7m
∠DAC = ∠BAC = 45° (alternate angles)
Also, ∠DAE = 60°
Let height of tower be EC m.
In right ∆ ABC,
tan45° =
BC = 7 m
Also, BC = AD
In right ∆ ADE,
tan60° =
ED = m
Height of tower EC = ED + DC = + 7 = m
Therefore, height of tower is m.
Q4. The angle of elevation of the top of a tower from a point on the ground, which is 20 m away from the foot of the tower, is 30°. Find the height of the tower.
Ans.
Let height of tower be AB m.
Given, distance of a point from the foot of tower BC = 20 m.
Also, angle of elevation of the top of tower = 60°
In right ∆ ABC,
tan45° =
AB = 20
Therefore, height of the tower is 20m.
Q5. An observer 1.6m tall is 20m away from a tower. The angle of elevation of the top of the tower from his eyes is 60°. Find the height of the tower? Most Important
Ans.
Given Height of girl AB = CD = 1.6m,
Distance between tower and girl AD = BC = 20 m
Also, angle of elevation of the top of the the tower from girl eyes i.e ∠DAE = 60°
From the fig, Let height of tower be CE m.
In right ∆ ADE,
tan60° =
DE =
Now, CE = CD + DE = 1.6 +
Therefore, height of the tower is () m.
Q6. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Ans.
Let the height of tower AB m.
Given, distance of a point from ground BC = 15 m
Angle of elevation of the top of the tower is 60°.
In right ∆ ABC,
tan60° =
AB =
Therefore, height of the tower is m.
Q7. A wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that wire will be taut ?
Ans.
let the pole be BC and length of wire be AB. To keep the wire taut let it be fixed to stake at A.
Then, ABC is right angle triangle at C.
Also, BC = 18m, AB = 24m.
AC is distance between pole base to that point.
In right ∆ ACB,
Using pythagoras theroem
AB2 = AC2 + BC2
242 = AC2 + 182
576 – 324 = AC2
252 = AC2
AC = m
Therefore, required distance is m.
Q8. A ladder is placed against a wall such that its foot is at a distance of 2.5m from the wall and its top reaches a window 6m above the ground. Find the length of the ladder.
Ans.
Given, height of window AB = 6 m
Distance between foot of ladder and wall BC = 2.5 m
Let the length of ladder be AC m.
In right ∆ ABC,
Using Pythagoras theorem
AB2 + BC2 = AC2
62 + 2.52 = AC2
36 + 6.25 = AC2
AC2 = 42.25
AC = 6.5
Therefore, length of ladder is 6.5 m.
Q9. A vertical pole of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.
Ans.
Height of pole=AB=6 m
Length of shadow of pole =BC=4 m
Length of shadow of tower=EF=28 m
In △ABC and △DEF
∠B=∠E=90° both 90° as both are vertical to ground
∠C=∠F (same elevation in both the cases as both shadows are cast at the same time)
∴△ABC∼△DEF by AA similarity criterion
We know that if two triangles are similar, ratio of their sides are in proportion
So,
DE = 6×7 = 42 m
Hence the height of the tower is 42 m.
Q10. A person goes 10 m due east and then 30 m due north. Find the distance from the starting point.
Ans.
Let the distance from the starting to ending point be AC.
Given AB = 10 m, BC = 30 m.
In right ∆ ABC,
Using Pythagoras theorem
AB2 + BC2 = AC2
102 + 302 = AC2
1000 = AC2
AC = m.
Therefore, required distance is m.
Q11. A ladder 17 m long reaches a window of a building 15 m above the ground, find the distance of the foot of the ladder from the building.
Ans.
Given, length of ladder AC = 17m
Height of building AB = 15 m
Let the distance of the foot of the ladder from the building be BC m.
In right ∆ ABC,
Using Pythagoras theorem
AB2 + BC2 = AC2
152 + BC2 = 172
289 – 225 = BC2
64 = BC2
BC = 8
Therefore, the distance of foot of the ladder from the building is 8 m.
Q12. Two poles of heights 6 m and 12 m stand on a level plane ground. If the distance between the feet of the poles is 8 m, then find the distance between their tops.
Ans.
Given, Height of first pole AD = 6m
Height of other pole BE = 12m
Distance between two poles AB = DC = 8 m
Let distance between poles tops be DE.
As, AB = DC = 8m
AD = BC = 6m
So, EC = EB – BC = 12 – 6 = 6m
Now, In right ∆ DCE,
Using Pythagoras theorem
DC2 + EC2 = DE2
82 + 62 = DE2
64 + 36 = DE2
DE = 10
Therefore, distance between tops of poles is 10 m.
Q13. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.
Ans.
In the fig., BC denotes the height of building, CD the tower and A the given point.
Considering right ∆ ABC,
AB = 20m
Now, In right ∆ ABD,
BD = m
DC = DB – BC = – 20 = m
Therefore, height of tower is m.
Q14. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.
Ans.
In Fig, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a poin on the bridge at a height of 3 m, i.e., DP = 3m.
Width of the river AB = ?
Now, AB = AD + DB
In right ∆ APD, ∠A = 30°
So, tan30° =
Also, in right ∆ PBD, ∠B = 45°. So, BD = PD = 3m.
Now, AB = BD + AD = 3 + = m.
Therefore, the width of the river is m.
Q15. An electrician has to repair an electric fault on a pole of height 5 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use which, when inclined at an angle of 60° to the horizontal, would enable him to reach the required position ? (Take = 1.73)
Ans.
In fig., the electrician is required to reach the point B on the pole AD.
BD = AD – AB = 5 – 1.3 = 3.7 m
Here, BC represent the ladder.
In right ∆ BDC,
(approx.)
Therefore, the length of ladder should be 4.28 m.
Q16. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Most Important
Ans.
In Fig., BD is the height of tree and from point C tree break down and touches the ground at point A. So, BD = BC + CD and DC = AC
Distance between foot of tree and point where top touches the ground is AB = 8m
In right ∆ ABC, ∠A = 30°
BC = m
Also,
AC = m = CD
BD = BC + CD = m
Therefore, height of tree (before broken) is m.
Q17. From the point P on the ground the angle of elevation of the top of a 10 m building is 30°. A flag is hoisted at the top of building and the angle of elevation of the top of flagstaff from P is 45°. Find the length of flagstaff and distance of building from P.
Ans.
In Fig., AB denotes the height of the building, BD the flagstaff and P the given point.
BD = ?
Considering right ∆ PAB,
AP =
The distance of the building from P is m = 17.32 m.
Now, let us suppose DB = x m. Then AD = (10 + x) m.
Now, in right ∆ PAD,
x = = 7.32
So, the length of the flagstaff is 7.32 m.
Q18. A 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. Most Important
Ans.
Let CD be the girl. Let initial position of ballon A and another position after sometime is B.
∆ ACE and ∆ BGC are two right angled triangles.
BH = 88.2 m = AF
BG = 88.2 – 1.2 = 87 m = AE
In ∆ ACE, tan60° =
m
Now, In ∆ BGC
EG = m
Therefore, distance travelled by the ballon is m.
Also Read Class 10 Math NCERT Solution Also Read Class 10 Important Questions [Latest]
error:
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# Difference between revisions of "2003 AIME II Problems/Problem 15"
## Problem
[quote=2003 AIME II #15]Let $P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}$. Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $a_{k}$ and $b_{k}$ are real numbers. Let
$\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$
where $m, n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$[/quote]
## Solution
We can rewrite the definition of $P(x)$ as follows:
$$P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x$$
This can quite obviously be factored as:
$$P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2$$
Note that $\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1$. So the roots of $x^{23} + x^{22} + \cdots + x^2 + x + 1$ are exactly all $24$-th complex roots of $1$, except for the root $x=1$.
Let $\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}$. Then the distinct zeros of $P$ are $0,\omega,\omega^2,\dots,\omega^{23}$.
We can clearly ignore the root $x=0$ as it does not contribute to the value that we need to compute.
The squares of the other roots are $\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}$.
Hence we need to compute the following sum:
$$R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|$$
Using basic properties of the sine function, we can simplify this to
$$R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)$$
The five-element sum is just $\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ$. We know that $\sin 30^\circ = \sin 150^\circ = \frac 12$, $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2$, and $\sin 90^\circ = 1$. Hence our sum evaluates to:
$$R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3$$
Therefore the answer is $8+4+3 = \boxed{015}$.
## Solution 2
Note that $x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}$. Our sum can be reformed as $$\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}$$
So $$\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0$$
$x(x^{47} + x^{46} + \dots - x - 1) = 0$
$x^{47} + x^{46} + \dots - x - 1 = 0$
$x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)$
$\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}$
$x^{48} - 1 - 2x^{24} + 2 = 0$
$(x^{24} - 1)^2 = 0$
And we can proceed as above.
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# How do you find the first five terms of the given sequence?
Apr 15, 2018
${a}_{1} = 1$, ${a}_{2} = \frac{1}{2}$, ${a}_{3} = \frac{1}{6}$, ${a}_{4} = \frac{1}{24}$, ${a}_{5} = \frac{1}{120}$
#### Explanation:
${a}_{1} = 1$
${a}_{n} = \frac{1}{n} {a}_{n - 1}$
For $n = 2$
${a}_{2} = \frac{1}{2} {a}_{1} = \frac{1}{2} \left(1\right) = \frac{1}{2}$
For $n = 3$
${a}_{3} = \frac{1}{3} {a}_{2} = \frac{1}{3} \left(\frac{1}{2}\right) = \frac{1}{6}$
For $n = 4$
${a}_{4} = \frac{1}{4} {a}_{1} = \frac{1}{4} \left(\frac{1}{6}\right) = \frac{1}{24}$
For $n = 5$
${a}_{5} = \frac{1}{5} {a}_{4} = \frac{1}{5} \left(\frac{1}{24}\right) = \frac{1}{120}$
Note that the closed form solution to this recursive relation is
a_n=1/(n!)
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# What is the probability of drawing a spade or a club from a deck of 52 cards?
Here’s information about what you’re looking for in What is the probability of drawing a spade or a club from a deck of 52 cards?. While every brand has a “Help Center” (or “Support Center”), when we look for relevant information or links, there are often the answers to some questions can’t be found. This is why this website was created, and hopefully it will help you.
## What is the probability of drawing a club or a spade?
But there are only 13 spades and 13 clubs, so that is the sample space. The probability of getting a spade, P(Spade), is 13/52 or 0.2500. Same for the probability of getting a club, P(Club) = 13/52 or 0.2500.
## What is the probability of drawing a spade from a deck of 52 cards?
Mathematicians measure probability by counting and using some very basic math, like addition and division. For example, you can add up the number of spades in a complete deck (13) and divide this by the total number of cards in the deck (52) to get the probability of randomly drawing a spade: 13 in 52, or 25 percent.
## What is the probability of drawing an ace or a club from a deck of 52 cards?
Step-by-step explanation: The probability of picking up an ace in a 52 deck of cards is 4/52 since there are 4 aces in the deck. The odds of picking up any other card is therefore 52/52 – 4/52 = 48/52.
## What is the probability of drawing a spade or 3?
Note that the probability of obtaining a spade is 1/4. So for at least four draws, the cards drawn in the first three draws must be a non-spade. Its probability is 3/4. Now for three draws it is (3/4)^3.
## What is the probability of drawing a club or a spade?
But there are only 13 spades and 13 clubs, so that is the sample space. The probability of getting a spade, P(Spade), is 13/52 or 0.2500. Same for the probability of getting a club, P(Club) = 13/52 or 0.2500.
## What is the probability of drawing a spade or 3?
Note that the probability of obtaining a spade is 1/4. So for at least four draws, the cards drawn in the first three draws must be a non-spade. Its probability is 3/4. Now for three draws it is (3/4)^3.
## What is the probability of drawing a heart or a spade from a deck of cards?
If you’re wanting to draw a heart and a spade, then you could get the heart first, or the spade first. The probability of doing this with replacement is 2(1/4)(1/4)=1/8. Doing it without replacement is 2(1/4)(13/51)=13/102.
## What is the probability of drawing a spade or a club from a deck of 52 cards?
The probability of an event is the sum of the probabilities of the outcomes in the event, hence the probability of drawing a spade is 13/52 = 1/4, and the probability of drawing a king is 4/52 = 1/13.
## What is the probability of drawing a 5 or a spade?
Basic Probability
So drawing a five-card hand of a single selected suit is a rare event with a probability of about one in 2000.
## What is the probability of drawing a jack or spade?
What is the probability of drawing a spade or a jack from a standard deck of 52 cards? Divided 16/52, thus the probability is 4/13.
## What is the probability of selecting a heart or a spade?
What is the probability of randomly selecting a card from a standard 52 card deck that is a heart or spade? Ace probability is 4/52=1/13. Heart probability is 13/52=1/4.
## What is the probability of selecting an ace or a spade from a deck of 52 cards?
The expected frequency of getting a head is 1, the total frequency is 2 (1 head and 1 tail), and the probability is ½. The probability of rolling a six on one die is 1/6. The probability of drawing the ace of spades from a deck of cards is 1/52.
## How many spades are in a deck of 52 cards?
In a deck of 52 cards, there are four suits and the spades represent one of the suit. In total, the number of spade cards in the deck is thirteen cards. These include a king, a queen, a jack, an ace, and number cards from two through ten. The spades are black in color as well as the clubs.
## What is the probability of drawing a spade or 3?
Note that the probability of obtaining a spade is 1/4. So for at least four draws, the cards drawn in the first three draws must be a non-spade. Its probability is 3/4. Now for three draws it is (3/4)^3.
## What is the probability of drawing to spades from a well shuffled pack of 52 cards?
Solution: In a playing card there are 52 cards. (i) ‘2’ of spades: Number of favourable outcomes i.e. ‘2’ of spades is 1 out of 52 cards.
## What is the probability of drawing a five or a spade?
Basic Probability
So drawing a five-card hand of a single selected suit is a rare event with a probability of about one in 2000.
## What is the probability of drawing spade?
Mathematicians measure probability by counting and using some very basic math, like addition and division. For example, you can add up the number of spades in a complete deck (13) and divide this by the total number of cards in the deck (52) to get the probability of randomly drawing a spade: 13 in 52, or 25 percent.
## What is the probability of drawing an ace or a spade?
The probability of drawing the ace of spades from a deck of cards is 1/52. Probabilities for more than one event can be calculated. Two simple rules apply: The additive rule applies to “either-or” cases.
## What is the probability of drawing an ace a 2 or 3?
Thus, the chance of drawing an ace on each of two draws is 4/52 × 3/51, or 1/221.
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[ Home ]
How to find areas of triangles using the formula: A = ˝base×height.
This topic is part of the TCS FREE high school mathematics 'How-to Library', and will help you to find areas of triangles using the formula: A = ˝base×height. (See the index page for a list of all available topics in the library.) To make best use of this topic, you need to download the Maths Helper Plus software. Click here for instructions.
### Theory:
The base and perpendicular height of a triangle can be identified as follows:
Any side can be the base, and then the perpendicular height extends from the vertex opposite the base to meet the base at a 90° angle.
For a right angled triangle, the perpendicular height can be one of the sides:
For an obtuse angled triangle (that is, a triangle with an angle greater than 90°) the perpendicular height may lie outside of the triangle itself:
If you know the base and perpendicular height of a triangle, then you can find the area using this formula:
Area = ˝ × base × perpendicular height
### Method:
Maths Helper Plus can show the calculations and also draw a diagram for calculating the area of a triangle given the base and perpendicular height.
#### Step 2 Display the parameters box
Press the F5 key to display the parameters box:
Click on the 'A' edit box with the mouse, then type the perpendicular height of your triangle.
Similarly, click on the 'B' edit box and type the value of the base.
Click the 'Update' button to refresh the diagram and calculations.
#### Step 3 Adjust the shape of the triangle diagram
NOTE: The following procedure does not effect the calculations in any way because the base and perpendicular height remain the same.
If you know the shape of the triangle (eg if you have been given a diagram) then you can adjust the shape of the triangle in Maths Helper Plus to look like your triangle.
The 'C' value in the parameters box is the distance of the top vertex of the triangle from the left side of the base. For example...
C between zero and B: C equal to zero: C less than zero:
#### Step 4 Adjust the scale of the labelled diagram
If the labelled diagram does not display the points, the scale needs to be reduced. In this case, briefly press the F10 key enough times until the two plotted points are seen.
You can make the diagram bigger by holding down 'Ctrl' while you press F10.
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# RD Sharma Solutions for Class 8 Maths Chapter 1 - Rational Numbers Exercise 1.2
RD Sharma Solutions for Class 8 Maths Exercise 1.2 Chapter 1, Rational Numbers, can be accessed by students from the links provided below. This set of solutions is designed by our subject expert team in order to help students excel in the final exam with flying colours. In Exercise 1.2 of RD Sharma Class 8 Maths Solution, we shall discuss problems based on the properties of addition of rational numbers, which include closure property, commutativity, associativity and the existence of additive identity.
## RD Sharma Solutions for Class 8 Maths Exercise 1.2 Chapter 1 Rational Numbers
### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 1.2 Chapter 1 Rational Numbers
1. Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers:
(i) -11/5 and 4/7
Solution: By using the commutativity law, the addition of rational numbers is commutative ∴ a/b + c/d = c/d + a/b
In order to verify the above property let us consider the given fraction
-11/5 and 4/7 as
-11/5 + 4/7 and 4/7 + -11/5
The denominators are 5 and 7
By taking LCM for 5 and 7 is 35
We rewrite the given fraction in order to get the same denominator
Now, -11/5 = (-11 × 7) / (5 ×7) = -77/35
4/7 = (4 ×5) / (7 ×5) = 20/35
Since the denominators are the same, we can add them directly
-77/35 + 20/35 = (-77+20)/35 = -57/35
4/7 + -11/5
The denominators are 7 and 5
By taking LCM for 7 and 5 is 35
We rewrite the given fraction in order to get the same denominator
Now, 4/7 = (4 × 5) / (7 ×5) = 20/35
-11/5 = (-11 ×7) / (5 ×7) = -77/35
Since the denominators are the same, we can add them directly
20/35 + -77/35 = (20 + (-77))/35 = (20-77)/35 = -57/35
∴ -11/5 + 4/7 = 4/7 + -11/5 is satisfied.
(ii) 4/9 and 7/-12
Solution: Firstly we need to convert the denominators to positive numbers.
7/-12 = (7 × -1)/ (-12 × -1) = -7/12
By using the commutativity law, the addition of rational numbers is commutative.
∴ a/b + c/d = c/d + a/b
In order to verify the above property let us consider the given fraction
4/9 and -7/12 as
4/9 + -7/12 and -7/12 + 4/9
The denominators are 9 and 12
By taking LCM for 9 and 12 is 36
We rewrite the given fraction in order to get the same denominator
Now, 4/9 = (4 × 4) / (9 ×4) = 16/36
-7/12 = (-7 ×3) / (12 ×3) = -21/36
Since the denominators are the same, we can add them directly
16/36 + (-21)/36 = (16 + (-21))/36 = (16-21)/36 = -5/36
-7/12 + 4/9
The denominators are 12 and 9
By taking LCM for 12 and 9 is 36
We rewrite the given fraction in order to get the same denominator
Now, -7/12 = (-7 ×3) / (12 ×3) = -21/36
4/9 = (4 × 4) / (9 ×4) = 16/36
Since the denominators are the same, we can add them directly
-21/36 + 16/36 = (-21 + 16)/36 = -5/36
∴ 4/9 + -7/12 = -7/12 + 4/9 is satisfied.
(iii) -3/5 and -2/-15
Solution:
By using the commutativity law, the addition of rational numbers is commutative.
∴ a/b + c/d = c/d + a/b
In order to verify the above property let us consider the given fraction
-3/5 and -2/-15 as
-3/5 + -2/-15 and -2/-15 + -3/5
-2/-15 = 2/15
The denominators are 5 and 15
By taking LCM for 5 and 15 is 15
We rewrite the given fraction in order to get the same denominator
Now, -3/5 = (-3 × 3) / (5×3) = -9/15
2/15 = (2 ×1) / (15 ×1) = 2/15
Since the denominators are the same, we can add them directly
-9/15 + 2/15 = (-9 + 2)/15 = -7/15
-2/-15 + -3/5
-2/-15 = 2/15
The denominators are 15 and 5
By taking LCM for 15 and 5 is 15
We rewrite the given fraction in order to get the same denominator
Now, 2/15 = (2 ×1) / (15 ×1) = 2/15
-3/5 = (-3 × 3) / (5×3) = -9/15
Since the denominators are the same, we can add them directly
2/15 + -9/15 = (2 + (-9))/15 = (2-9)/15 = -7/15
∴ -3/5 + -2/-15 = -2/-15 + -3/5 is satisfied.
(iv) 2/-7 and 12/-35
Solution: Firstly we need to convert the denominators to positive numbers.
2/-7 = (2 × -1)/ (-7 × -1) = -2/7
12/-35 = (12 × -1)/ (-35 × -1) = -12/35
By using the commutativity law, the addition of rational numbers is commutative.
∴ a/b + c/d = c/d + a/b
In order to verify the above property let us consider the given fraction
-2/7 and -12/35 as
-2/7 + -12/35 and -12/35 + -2/7
The denominators are 7 and 35
By taking LCM for 7 and 35 is 35
We rewrite the given fraction in order to get the same denominator
Now, -2/7 = (-2 × 5) / (7 ×5) = -10/35
-12/35 = (-12 ×1) / (35 ×1) = -12/35
Since the denominators are the same, we can add them directly
-10/35 + (-12)/35 = (-10 + (-12))/35 = (-10-12)/35 = -22/35
-12/35 + -2/7
The denominators are 35 and 7
By taking LCM for 35 and 7 is 35
We rewrite the given fraction in order to get the same denominator
Now, -12/35 = (-12 ×1) / (35 ×1) = -12/35
-2/7 = (-2 × 5) / (7 ×5) = -10/35
Since the denominators are the same, we can add them directly
-12/35 + -10/35 = (-12 + (-10))/35 = (-12-10)/35 = -22/35
∴ -2/7 + -12/35 = -12/35 + -2/7 is satisfied.
(v) 4 and -3/5
Solution: By using the commutativity law, the addition of rational numbers is commutative.
∴ a/b + c/d = c/d + a/b
In order to verify the above property let us consider the given fraction
4/1 and -3/5 as
4/1 + -3/5 and -3/5 + 4/1
The denominators are 1 and 5
By taking LCM for 1 and 5 is 5
We rewrite the given fraction in order to get the same denominator
Now, 4/1 = (4 × 5) / (1×5) = 20/5
-3/5 = (-3 ×1) / (5 ×1) = -3/5
Since the denominators are the same, we can add them directly
20/5 + -3/5 = (20 + (-3))/5 = (20-3)/5 = 17/5
-3/5 + 4/1
The denominators are 5 and 1
By taking LCM for 5 and 1 is 5
We rewrite the given fraction in order to get the same denominator
Now, -3/5 = (-3 ×1) / (5 ×1) = -3/5
4/1 = (4 × 5) / (1×5) = 20/5
Since the denominators are the same, we can add them directly
-3/5 + 20/5 = (-3 + 20)/5 = 17/5
∴ 4/1 + -3/5 = -3/5 + 4/1 is satisfied.
(vi) -4 and 4/-7
Solution: Firstly we need to convert the denominators to positive numbers.
4/-7 = (4 × -1)/ (-7 × -1) = -4/7
By using the commutativity law, the addition of rational numbers is commutative.
∴ a/b + c/d = c/d + a/b
In order to verify the above property let us consider the given fraction
-4/1 and -4/7 as
-4/1 + -4/7 and -4/7 + -4/1
The denominators are 1 and 7
By taking LCM for 1 and 7 is 7
We rewrite the given fraction in order to get the same denominator
Now, -4/1 = (-4 × 7) / (1×7) = -28/7
-4/7 = (-4 ×1) / (7 ×1) = -4/7
Since the denominators are the same, we can add them directly
-28/7 + -4/7 = (-28 + (-4))/7 = (-28-4)/7 = -32/7
-4/7 + -4/1
The denominators are 7 and 1
By taking LCM for 7 and 1 is 7
We rewrite the given fraction in order to get the same denominator
Now, -4/7 = (-4 ×1) / (7 ×1) = -4/7
-4/1 = (-4 × 7) / (1×7) = -28/7
Since the denominators are the same, we can add them directly
-4/7 + -28/7 = (-4 + (-28))/7 = (-4-28)/7 = -32/7
∴ -4/1 + -4/7 = -4/7 + -4/1 is satisfied.
2. Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
(i) x = ½, y = 2/3, z = -1/5
Solution: As the property states (x + y) + z = x + (y + z)
Use the values as such,
(1/2 + 2/3) + (-1/5) = 1/2 + (2/3 + (-1/5))
Let us consider LHS (1/2 + 2/3) + (-1/5)
Taking LCM for 2 and 3 is 6
(1× 3)/(2×3) + (2×2)/(3×2)
3/6 + 4/6
Since the denominators are the same, we can add them directly,
3/6 + 4/6 = 7/6
7/6 + (-1/5)
Taking LCM for 6 and 5 is 30
(7×5)/(6×5) + (-1×6)/(5×6)
35/30 + (-6)/30
Since the denominators are the same, we can add them directly,
(35+(-6))/30 = (35-6)/30 = 29/30
Let us consider RHS 1/2 + (2/3 + (-1/5))
Taking LCM for 3 and 5 is 15
(2/3 + (-1/5)) = (2×5)/(3×5) + (-1×3)/(5×3)
= 10/15 + (-3)/15
Since the denominators are the same, we can add them directly,
10/15 + (-3)/15 = (10-3)/15 = 7/15
1/2 + 7/15
Taking LCM for 2 and 15 is 30
1/2 + 7/15 = (1×15)/(2×15) + (7×2)/(15×2)
= 15/30 + 14/30
Since the denominators are the same we can add them directly,
= (15 + 14)/30 = 29/30
∴ LHS = RHS associativity of addition of rational numbers is verified.
(ii) x = -2/5, y = 4/3, z = -7/10
Solution: As the property states (x + y) + z = x + (y + z)
Use the values as such,
(-2/5 + 4/3) + (-7/10) = -2/5 + (4/3 + (-7/10))
Let us consider LHS (-2/5 + 4/3) + (-7/10)
Taking LCM for 5 and 3 is 15
(-2× 3)/(5×3) + (4×5)/(3×5)
-6/15 + 20/15
Since the denominators are the same, we can add them directly,
-6/15 + 20/15= (-6+20)/15 = 14/15
14/15 + (-7/10)
Taking LCM for 15 and 10 is 30
(14×2)/(15×2) + (-7×3)/(10×3)
28/30 + (-21)/30
Since the denominators are the same, we can add them directly,
(28+(-21))/30 = (28-21)/30 = 7/30
Let us consider RHS -2/5 + (4/3 + (-7/10))
Taking LCM for 3 and 10 is 30
(4/3 + (-7/10)) = (4×10)/(3×10) + (-7×3)/(10×3)
= 40/30 + (-21)/30
Since the denominators are the same, we can add them directly,
40/30 + (-21)/30 = (40-21)/30 = 19/30
-2/5 + 19/30
Taking LCM for 5 and 30 is 30
-2/5 + 19/30 = (-2×6)/(5×6) + (19×1)/(30×1)
= -12/30 + 19/30
Since the denominators are the same, we can add them directly,
= (-12 + 19)/30 = 7/30
∴ LHS = RHS associativity of addition of rational numbers is verified.
(iii) x = -7/11, y = 2/-5, z = -3/22
Solution: Firstly convert the denominators to positive numbers
2/-5 = (2×-1)/ (-5×-1) = -2/5
As the property states (x + y) + z = x + (y + z)
Use the values as such,
(-7/11 + -2/5) + (-3/22) = -7/11 + (-2/5 + (-3/22))
Let us consider LHS (-7/11 + -2/5) + (-3/22)
Taking LCM for 11 and 5 is 55
(-7×5)/(11×5) + (-2×11)/(5×11)
-35/55 + -22/55
Since the denominators are the same, we can add them directly,
-35/55 + -22/55 = (-35-22)/55 = -57/55
-57/55 + (-3/22)
Taking LCM for 55 and 22 is 110
(-57×2)/(55×2) + (-3×5)/(22×5)
-114/110 + (-15)/110
Since the denominators are the same, we can add them directly,
(-114+(-15))/110 = (-114-15)/110 = -129/110
Let us consider RHS -7/11 + (-2/5 + (-3/22))
Taking LCM for 5 and 22 is 110
(-2/5 + (-3/22))= (-2×22)/(5×22) + (-3×5)/(22×5)
= -44/110 + (-15)/110
Since the denominators are the same, we can add them directly,
-44/110 + (-15)/110 = (-44-15)/110 = -59/110
-7/11 + -59/110
Taking LCM for 11 and 110 is 110
-7/11 + -59/110 = (-7×10)/(11×10) + (-59×1)/(110×1)
= -70/110 + -59/110
Since the denominators are the same, we can add them directly,
= (-70 -59)/110 = -129/110
∴ LHS = RHS associativity of addition of rational numbers is verified.
(iv) x = -2, y = 3/5, z = -4/3
Solution: As the property states (x + y) + z = x + (y + z)
Use the values as such,
(-2/1 + 3/5) + (-4/3) = -2/1 + (3/5 + (-4/3))
Let us consider LHS (-2/1 + 3/5) + (-4/3)
Taking LCM for 1 and 5 is 5
(-2×5)/(1×5) + (3×1)/(5×1)
-10/5 + 3/5
Since the denominators are the same, we can add them directly,
-10/5 + 3/5= (-10+3)/5 = -7/5
-7/5 + (-4/3)
Taking LCM for 5 and 3 is 15
(-7×3)/(5×3) + (-4×5)/(3×5)
-21/15 + (-20)/15
Since the denominators are the same, we can add them directly,
(-21+(-20))/15 = (-21-20)/15 = -41/15
Let us consider RHS -2/1 + (3/5 + (-4/3))
Taking LCM for 5 and 3 is 15
(3/5 + (-4/3)) = (3×3)/(5×3) + (-4×5)/(3×5)
= 9/15 + (-20)/15
Since the denominators are the same, we can add them directly,
9/15 + (-20)/15 = (9-20)/15 = -11/15
-2/1 + -11/15
Taking LCM for 1 and 15 is 15
-2/1 + -11/15 = (-2×15)/(1×15) + (-11×1)/(15×1)
= -30/15 + -11/15
Since the denominators are the same, we can add them directly,
= (-30 -11)/15 = -41/15
∴ LHS = RHS associativity of addition of rational numbers is verified.
3. Write the additive of each of the following rational numbers:
(i) -2/17
(ii) 3/-11
(iii) -17/5
(iv) -11/-25
Solution:
(i) The additive inverse of -2/17 is 2/17
(ii) The additive inverse of 3/-11 is 3/11
(iii) The additive inverse of -17/5 is 17/5
(iv) The additive inverse of -11/-25 is -11/25
4. Write the negative(additive) inverse of each of the following:
(i) -2/5
(ii) 7/-9
(iii) -16/13
(iv) -5/1
(v) 0
(vi) 1
(vii) -1
Solution:
(i) The negative (additive) inverse of -2/5 is 2/5
(ii) The negative (additive) inverse of 7/-9 is 7/9
(iii) The negative (additive) inverse of -16/13 is 16/13
(iv) The negative (additive) inverse of -5/1 is 5
(v) The negative (additive) inverse of 0 is 0
(vi) The negative (additive) inverse of 1 is -1
(vii) The negative (additive) inverse of -1 is 1
5. Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i) 2/5 + 7/3 + -4/5 + -1/3
Solution: Firstly group the rational numbers with same denominators
2/5 + -4/5 + 7/3 + -1/3
Now the denominators which are same can be added directly.
(2+(-4))/5 + (7+(-1))/3
(2-4)/5 + (7-1)/3
-2/5 + 6/3
By taking LCM for 5 and 3 we get, 15
(-2×3)/(5×3) + (6×5)/(3×5)
-6/15 + 30/15
Since the denominators are same can be added directly
(-6+30)/15 = 24/15
Further can be divided by 3 we get,
24/15 = 8/5
(ii) 3/7 + -4/9 + -11/7 + 7/9
Solution: Firstly group the rational numbers with same denominators
3/7 + -11/7 + -4/9 + 7/9
Now the denominators which are same can be added directly.
(3+ (-11))/7 + (-4+ 7)/9
(3-11)/7 + (-4+7)/9
-8/7 + 3/9
-8/7 + 1/3
By taking LCM for 7 and 3 we get, 21
(-8×3)/ (7×3) + (1×7)/ (3×7)
-24/21 + 7/21
Since the denominators are same can be added directly
(-24+7)/21 = -17/21
(iii) 2/5 + 8/3 + -11/15 + 4/5 + -2/3
Solution: Firstly group the rational numbers with same denominators
2/5 + 4/5 + 8/3 + -2/3 + -11/15
Now the denominators which are same can be added directly.
(2 + 4)/5 + (8 + (-2))/3 + -11/15
6/5 + (8-2)/3 + -11/15
6/5 + 6/3 + -11/15
6/5 + 2/1 + -11/15
By taking LCM for 5, 1 and 15 we get, 15
(6×3)/ (5×3) + (2×15)/ (1×15) + (-11×1)/ (15×1)
18/15 + 30/15 + -11/15
Since the denominators are same can be added directly
(18+30+ (-11))/15 = (18+30-11)/15 = 37/15
(iv) 4/7 + 0 + -8/9 + -13/7 + 17/21
Solution: Firstly group the rational numbers with same denominators
4/7 + -13/7 + -8/9 + 17/21
Now the denominators which are same can be added directly.
(4 + (-13))/7 + -8/9 + 17/21
(4-13)/7 + -8/9 + 17/21
-9/7 + -8/9 + 17/21
By taking LCM for 7, 9 and 21 we get, 63
(-9×9)/ (7×9) + (-8×7)/ (9×7) + (17×3)/ (21×3)
-81/63 + -56/63 + 51/63
Since the denominators are same can be added directly
(-81+(-56)+ 51)/63 = (-81-56+51)/63 = -86/63
6. Re-arrange suitably and find the sum in each of the following:
(i) 11/12 + -17/3 + 11/2 + -25/2
Solution: Firstly group the rational numbers with same denominators
11/12 + -17/3 + (11-25)/2
11/12 + -17/3 + -14/2
By taking LCM for 12, 3 and 2 we get, 12
(11×1)/(12×1) + (-17×4)/(3×4) + (-14×6)/(2×6)
11/12 + -68/12 + -84/12
Since the denominators are same can be added directly
(11-68-84)/12 = -141/12
(ii)-6/7 + -5/6 + -4/9 + -15/7
Solution: Firstly group the rational numbers with same denominators
-6/7 + -15/7 + -5/6 + -4/9
(-6 -15)/7 + -5/6 + -4/9
-21/7 + -5/6 + -4/9
-3/1 + -5/6 + -4/9
By taking LCM for 1, 6 and 9 we get, 18
(-3×18)/(1×18) + (-5×3)/(6×3) + (-4×2)/(9×2)
-54/18 + -15/18 + -8/18
Since the denominators are same can be added directly
(-54-15-8)/18 = -77/18
(iii) 3/5 + 7/3 + 9/ 5+ -13/15 + -7/3
Solution: Firstly group the rational numbers with same denominators
3/5 + 9/5 + 7/3 + -7/3 + -13/15
(3+9)/5 + -13/15
12/5 + -13/15
By taking LCM for 5 and 15 we get, 15
(12×3)/(5×3) + (-13×1)/(15×1)
36/15 + -13/15
Since the denominators are same can be added directly
(36-13)/15 = 23/15
(iv) 4/13 + -5/8 + -8/13 + 9/13
Solution: Firstly group the rational numbers with same denominators
4/13 + -8/13 + 9/13 + -5/8
(4-8+9)/13 + -5/8
5/13 + -5/8
By taking LCM for 13 and 8 we get, 104
(5×8)/(13×8) + (-5×13)/(8×13)
40/104 + -65/104
Since the denominators are same can be added directly
(40-65)/104 = -25/104
(v) 2/3 + -4/5 + 1/3 + 2/5
Solution: Firstly group the rational numbers with same denominators
2/3 + 1/3 + -4/5 + 2/5
(2+1)/3 + (-4+2)/5
3/3 + -2/5
1/1 + -2/5
By taking LCM for 1 and 5 we get, 5
(1×5)/(1×5) + (-2×1)/(5×1)
5/5 + -2/5
Since the denominators are same can be added directly
(5-2)/5 = 3/5
(vi) 1/8 + 5/12 + 2/7 + 7/12 + 9/7 + -5/16
Solution: Firstly group the rational numbers with same denominators
1/8 + 5/12 + 7/12 + 2/7 + 9/7 + -5/16
1/8 + (5+7)/12 + (2+9)/7 + -5/16
1/8 + 12/12 + 11/7 + -5/16
1/8 + 1/1 + 11/7 + -5/16
By taking LCM for 8, 1, 7 and 16 we get, 112
(1×14)/(8×14) + (1×112)/(1×112) + (11×16)/(7×16) + (-5×7)/(16×7)
14/112 + 112/112 + 176/112 + -35/112
Since the denominators are same can be added directly
(14+112+176-35)/112 = 267/112
## RD Sharma Solutions for Class 8 Maths Exercise 1.2 Chapter 1 Rational Numbers
Class 8 Maths Chapter 1 Rational Numbers Exercise 1.2 is based on the different properties of the addition of rational numbers. To learn the concepts effortlessly and grasp them better, download free RD Sharma Solutions, which provide answers to all the questions. Practising as many times as possible helps students secure high marks in the annual exam.
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# Question #53d32
Mar 29, 2017
$\frac{58975}{20736} , \mathmr{and} , - \frac{58975}{20736.}$
#### Explanation:
$x + \frac{1}{x} = 2 \frac{1}{12} = \frac{25}{12.}$
$\Rightarrow \frac{{x}^{2} + 1}{x} = \frac{25}{12.}$
$\Rightarrow 12 {x}^{2} + 12 = 25 x .$
$\Rightarrow 12 {x}^{2} - 25 x + 12 = 0.$
$\Rightarrow 12 {x}^{2} - 16 x - 9 x + 12 = 0 , \ldots \left[16 \times 9 = 144 , 16 + 9 = 25\right] .$
$\Rightarrow 4 x \left(3 x - 4\right) - 3 \left(3 x - 4\right) = 0.$
$\Rightarrow \left(3 x - 4\right) \left(4 x - 3\right) = 0.$
$\Rightarrow x = \frac{4}{3} , \mathmr{and} , x = \frac{3}{4.}$
$x = \frac{4}{3} \Rightarrow {x}^{4} - \frac{1}{x} ^ 4 = {\left(\frac{4}{3}\right)}^{4} - {\left(\frac{3}{4}\right)}^{4} = \frac{256}{81} - \frac{81}{256} = \frac{58975}{20736.}$
$x = \frac{3}{4} \Rightarrow {x}^{4} = \frac{1}{x} ^ 4 = {\left(\frac{3}{4}\right)}^{4} - {\left(\frac{4}{3}\right)}^{4} = - \frac{58975}{20736.}$
Enjoy Maths.!
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## Multiplying Fractions
### Learning Outcomes
• Multiply fractions
• Multiply two or more fractions
• Multiply a fraction by a whole number
## Introduction
Before we get started, here is some important terminology that will help you understand the concepts about working with fractions in this section.
• product: the result of multiplication
• factor: something being multiplied – for $3 \cdot 2 = 6$ , both 3 and 2 are factors of 6
• numerator: the top part of a fraction – the numerator in the fraction $\frac{2}{3}$ is 2
• denominator: the bottom part of a fraction – the denominator in the fraction $\frac{2}{3}$ is 3
Many different words are used by math textbooks and teachers to provide students with instructions on what they are to do with a given problem. For example, you may see instructions such as “Find” or “Simplify” in the example in this module. It is important to understand what these words mean so you can successfully work through the problems in this course. Here is a short list of the words you may see that can help you know how to work through the problems in this module.
Instruction Interpretation
Find Perform the indicated mathematical operations which may include addition, subtraction, multiplication, division.
Simplify 1) Perform the indicated mathematical operations including addition, subtraction, multiplication, division
2) Write a mathematical statement in smallest terms so there are no other mathematical operations that can be performed—often found in problems related to fractions and the order of operations
Evaluate Perform the indicated mathematical operations including addition, subtraction, multiplication, division
Reduce Write a mathematical statement in smallest or lowest terms so there are no other mathematical operations that can be performed—often found in problems related to fractions or division
## Multiply Fractions
Just as you add, subtract, multiply, and divide when working with whole numbers, you also use these operations when working with fractions. There are many times when it is necessary to multiply fractions. A model may help you understand multiplication of fractions.
When you multiply a fraction by a fraction, you are finding a “fraction of a fraction.” Suppose you have $\frac{3}{4}$ of a candy bar and you want to find $\frac{1}{2}$ of the $\frac{3}{4}$:
By dividing each fourth in half, you can divide the candy bar into eighths.
Then, choose half of those to get $\frac{3}{8}$.
In both of the above cases, to find the answer, you can multiply the numerators together and the denominators together.
### Multiplying Two Fractions
$\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}=\frac{\text{product of the numerators}}{\text{product of the denominators}}$
### Multiplying More Than Two Fractions
$\frac{a}{b}\cdot \frac{c}{d}\cdot \frac{e}{f}=\frac{a\cdot c\cdot e}{b\cdot d\cdot f}$
### Example
Multiply $\frac{2}{3}\cdot \frac{4}{5}$.
To review: if a fraction has common factors in the numerator and denominator, we can reduce the fraction to its simplified form by removing the common factors.
For example,
• Given $\frac{8}{15}$, the factors of 8 are: 1, 2, 4, 8 and the factors of 15 are: 1, 3, 5, 15. $\frac{8}{15}$ is simplified because there are no common factors of 8 and 15.
• Given $\frac{10}{15}$, the factors of 10 are: 1, 2, 5, 10 and the factors of15 are: 1, 3, 5, 15. $\frac{10}{15}$ is not simplified because 5 is a common factor of 10 and 15.
You can simplify first, before you multiply two fractions, to make your work easier. This allows you to work with smaller numbers when you multiply.
In the following video you will see an example of how to multiply two fractions, then simplify the answer.
Multiply $\frac{2}{3}\cdot \frac{1}{4}\cdot\frac{3}{5}$. Simplify the answer.
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# Difference between revisions of "2016 AMC 10A Problems/Problem 4"
## Problem
The remainder can be defined for all real numbers $x$ and $y$ with $y \neq 0$ by $$\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor$$where $\left \lfloor \tfrac{x}{y} \right \rfloor$ denotes the greatest integer less than or equal to $\tfrac{x}{y}$. What is the value of $\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )$?
$\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}$
## Solution
The value, by definition, is \begin{align*} \text{rem}\left(\frac{3}{8},-\frac{2}{5}\right) &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\ &= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\ &= \frac{3}{8}-\frac{2}{5}\\ &= \boxed{\textbf{(B) } -\frac{1}{40}}. \end{align*}
## Solution 2
Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is $\boxed{B}$
~IceMatrix
~savannahsolver
## See Also
2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2016 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Video: Finding the Norm of Vectors
If π = (β2, 1, 0) and π = (2, 0, 4), determine |π| + |π|.
02:25
### Video Transcript
If vector π is equal to negative two, one, zero and vector π is equal to two, zero, four, determine the magnitude of vector π plus the magnitude of vector π.
Letβs firstly consider the general vector π with coordinates π₯, π¦, and π§. The magnitude or modulus of this vector is equal to the square root of π₯ squared plus π¦ squared plus π§ squared. In this question, vector π is equal to negative two, one, zero. This means that the magnitude of vector π is equal to the square root of negative two squared plus one squared plus zero squared. Negative two squared is equal to four. One squared is equal to one. And zero squared is equal to zero. Four plus one plus zero is equal to five. Therefore, the magnitude of vector π is equal to root five.
Vector π has coordinates two, zero, four. This means that the magnitude of vector π is equal to the square root of two squared plus zero squared plus four squared. Two squared is equal to four. And four squared is equal to 16. Therefore, the magnitude of vector π is the square root of 20. This is equal to two root five. This is because the square root of 20 is equal to the square root of four multiplied by the square root of five. The square root of four is equal to two. Therefore, the square root of 20 equals two multiplied by the square root of five.
We were asked to calculate the magnitude of vector π plus the magnitude of vector π. This is equal to root five plus two root five. As root five is the same as one root five, our final answer is three root five. If π equals negative two, one, zero and π equals two, zero, four, then the magnitude of π plus the magnitude of π is equal to three root five.
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Question Video: Finding a Side Length in Similar Triangles | Nagwa Question Video: Finding a Side Length in Similar Triangles | Nagwa
# Question Video: Finding a Side Length in Similar Triangles Mathematics • First Year of Secondary School
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Find the length of line segment ๐ถ๐ต.
05:15
### Video Transcript
Find the length of line segment ๐ถ๐ต.
In this question, weโre given a diagram with two different triangles. Weโre given the lengths of two sides of each triangle. And we can also see from the markings that this angle, ๐ด๐ถ๐น, is the same size as this angle, ๐ท๐ต๐น. So, the line segment that weโre asked to work out is the one at the base of the diagram. Weโve got the length of part of it. We know that ๐ถ๐น is 21 units long and ๐น๐ต is two ๐ฅ plus eight. While we could just add these two lengths, for example, 21 plus two ๐ฅ plus eight would give us two ๐ฅ plus 29, but we can assume that what weโre really being asked here is for a numerical value for the length.
So, letโs have a look again at the diagram and see if thereโs any way we can work out this length, ๐น๐ต, perhaps by finding the value of ๐ฅ. A clue for the message comes from the fact that weโre given this pair of angles which are equal. We might then wonder if these triangles are perhaps similar or congruent. We remember that similar triangles have corresponding angles equal and corresponding sides in proportion. When weโre dealing with congruent triangles, congruent triangles have corresponding angles equal and corresponding sides equal.
However, going by the diagram, these two triangles arenโt the same size, so theyโre very likely not to be congruent. So, letโs check if theyโre similar. We can note firstly that our pairs of angles are equal. Angle ๐ด๐ถ๐น is equal to angle ๐ท๐ต๐น. Next, we could have a look at this angle ๐ด๐น๐ถ, which is marked as a right angle. Using the fact that the angles on a straight line sum to 180 degrees and the line ๐ต๐ถ is a straight line, this means that this angle of ๐ท๐น๐ต must also be 90 degrees, since 180 degrees subtract 90 also gives us 90 degrees. Therefore, we note that angle ๐ด๐น๐ถ is equal to angle ๐ท๐น๐ต.
This now means that weโve found two pairs of corresponding angles equal. This fulfills the AA or angle-angle similarity criterion. Now, weโve proven that triangle ๐ด๐น๐ถ is similar to triangle ๐ท๐น๐ต. So, letโs see how this helps us to work out the side length of ๐น๐ต. As these two triangles are similar, that means remember that corresponding sides are in proportion. If we look at the side ๐ด๐ถ, then the side which corresponds to it in triangle ๐ท๐น๐ต is this one, ๐ท๐ต. Then, another side on triangle ๐ด๐น๐ถ that we can look at is the length of ๐ถ๐น. The corresponding side on the other triangle is ๐น๐ต.
Because the triangles are similar, the sides are in proportion. And as that proportion is equal, then we can write that ๐ด๐ถ over ๐ท๐ต is equal to ๐ถ๐น over ๐น๐ต. We could also have written this statement with the fractions reversed. However, we need to make sure that we keep all the lengths of one triangle either as the numerators or the denominators and make sure we donโt mix them up.
What we do next is simply substitute in the length information that weโre given. This gives us 35 over seven ๐ฅ plus six equals 21 over two ๐ฅ plus eight. To solve this, we can begin by taking the cross product. So, we have 35 multiplied by two ๐ฅ plus eight equals 21 multiplied by seven ๐ฅ plus six. In the next step, we could go straight ahead and expand the parentheses on both sides of this equation. However, we might also notice that the values outside the parentheses are both multiples of seven. Dividing through by seven means that we can write it a little more simply. Five multiplied by two ๐ฅ plus eight equals three multiplied by seven ๐ฅ plus six.
We can now expand the parentheses giving us 10๐ฅ plus 40 equals 21๐ฅ plus 18. In order to keep a positive value of ๐ฅ, we can subtract 10๐ฅ from both sides. Then, we can subtract 18 from both sides, which leaves us with 22 equals 11๐ฅ. We can then divide through by 11 which gives us that two equals ๐ฅ and ๐ฅ is equal to two.
Itโs very tempting to stop here and think that weโve answered the question. But donโt forget we werenโt just asked for ๐ฅ. We were asked for the length of the line segment ๐ถ๐ต. Remember that we said that ๐ถ๐ต is equal to 21 plus this length of two ๐ฅ plus eight. We then need to substitute in the value that ๐ฅ is equal to two, giving us that ๐ถ๐ต is equal to 21 plus two times two plus eight simplifying to 33. Therefore, we can give the answer that line segment ๐ถ๐ต is equal to 33 length units.
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# Find The HCF and LCM of 12 15 18 and 27
## Question: Find the HCF and LCM of 12 15 18 and 27?
To find the Least Common Multiple (LCM) of a, b, c and d, we need the least number which is exactly divisible by all the numbers a, b, c, and d without leaving any remainder.
## Answer: LCM of 12 15 18, and 27 is 504, and the HCF of 12, 15, 18, and 27 is 3
The Highest Common Factor (HCF) is the largest possible number which divides all the numbers exactly without leaving any remainder.
## Explanation:
We are using the Listing Method to find the LCM of 12, 15, 18, and 27
We will list the first few multiples of 12, 15, 18, and 27 and determine the common multiples.
The least among the common multiples is the LCM of 12, 15, 18, and 27
Here, 540 is the least common multiple of 12, 15, 18, and 27. So, the LCM of 12, 15, 18, and 27 is 540.
Now we are using the methods of Listing Common Factors to find the HCF of 12, 15, 18, and 27
The factors of 12, 15, 18, and 27 are shown below.
The common factor of 12, 15, 18, and 27 are 1 and 3
HCF is the product of the factors that are common to each of the given numbers.
So, HCF of 12, 15, 18, and 27 is 3
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Pre-Calc Homework Solutions 184
Pre-Calc Homework - 184 Section 4.6 Step 2 At the instant in question dV dt 30 Step 1 s distance ball has fallen x distance from bottom of pole to
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Unformatted text preview: 184 Section 4.6 Step 2: At the instant in question, dV dt 30. Step 1: s distance ball has fallen x distance from bottom of pole to shadow Step 2: At the instant in question, s ds dt 1 32 2 8 mL/min 8 cm3/min and r 16 1 2 2 1 (20) = 10 cm. 2 4 ft and Step 3: We want to find Step 4: We have V 4 3 r and S 3 dS . dt 16 ft/sec. Step 3: We want to find Step 4: x 30 By similar triangles, 50 s x . This is equivalent to 50 dx . dt 4 r 2. These equations can be 3V 1/3 , so S 4 combined by noting that r Step 5: dS dt 4 3V 2/3 4 50x x 1500 1500s 1 50x . sx, or sx 1500. We will use 4 2 3 3V 4 1/3 3 dV 4 dt 2 3V 4 1/3 dV dt Step 5: dx dt Step 6: 1500s 2 ds dt Note that V dS dt Step 6: dx dt 1500(4) 2(16) 1500 ft/sec 1500 ft/sec. 16 3 1.6 cm2/min 3 4 1000 dS Since 0, the rate of decrease is positive. The surface dt 2 4 4000 (10)3 . 3 3 1/3 4000 ( 8) 3 The shadow is moving at a velocity of area is decreasing at the rate of 1.6 cm2/min. 33. Step 1: p x-coordinate of plane's position x x-coordinate of car's position s distance from plane to car (line-of-sight) Step 2: 31. Step 1: x position of car (x 0 when car is right in front of you) camera angle. (We assume is negative until the car passes in front of you, and then positive.) Step 2: At the first instant in question, x A half second later, x dx dt 1 (264) 2 0 ft and dx dt 264 ft/sec. At the instant in question, p 0, dp dt 132 ft and 120 mph, s 5 mi, and ds dt 160 mph. 264 ft/sec. Step 3: We want to find d at each of the two instants. dt dx . dt Step 3: We want to find Step 4: tan Step 5: d dt 1 1 x 2 132 1 Step 4: (x p)2 32 s2 x 132 1 dx 132 dt Step 5: 2(x p) dx dt dp dt 2s ds dt Step 6: Note that, at the instant in question, Step 6: When x d 0: dt d dt 1 1 0 2 132 1 (264) 132 1 (264) 132 x 2 radians/sec 2(4 52 0) 32 4 mi. 120 120 120 dx dt When x 132: 1 1 132 132 2 1 radian/sec dx dt dx 8 dt dx dt 2(5)( 160) 1600 200 80 mph 32. Step 1: r radius of balls plus ice S surface area of ball plus ice V volume of ball plus ice The car's speed is 80 mph. ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
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Differentiation: Exercise 1 – Calculus Methods
by Batool Akmal
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Learning Material 2
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DLM Differentiation Exercise Calculus Akmal.pdf
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Transcript
00:01 Welcome back.
00:02 I've hope you've enjoyed having a go on the questions and the exercises.
00:06 We're now just going to go through it together to see whether you've done it correct or incorrectly.
00:11 And also to see if there is any problems that you've had so that we can help you out with them.
00:17 So looking at our first question.
00:19 Nice and straight forward.
00:21 Find the general gradient of the curve.
00:27 It's just the gradient as a function.
00:29 You don't need to substitute any numbers into it.
00:32 You don't need to find any tangents or normals.
00:34 You just have to find the gradient as an expression.
00:37 So let's have a look at this equation we've got y = root x + 2x - 1.
00:42 Now recall what we said if we had any root x's or any powers, any indices that are down as denominators, we can bring them up using rule of indices.
00:53 But in this case all we have to do is change the square root.
00:56 So we can rewrite this as y = x to the half.
01:02 Plus 2x minus 1.
01:06 And the general gradient dy by dx.
01:08 Bring the power down.
01:12 Decrease the power by 1.
01:14 Same again.
01:16 Bring the power down.
01:18 Which is just 1, 2x, 1 - 1 and the constant just disappears.
01:24 So the answer to this will be the half x to the minus the half plus 2.
01:31 Because x to the 0 is just going to be 1.
01:34 And again you can take that down.
01:37 So 2x to the half plus 2.
01:40 Or you could rewrite this as a root to see we write this as 1 over 2 root x plus 2.
01:50 Which means to find a particular gradient you just have to replace this x with different values or different points in the curve that you are looking at.
The lecture Differentiation: Exercise 1 – Calculus Methods by Batool Akmal is from the course Calculus Methods: Differentiation.
Included Quiz Questions
1. y=x^(-1/2)
2. y=x^(1/2)
3. y=x^(-1)
4. y=x^(-2)
5. y=x^(2)
1. dy/dx=(-1/2)x^(-3/2)
2. dy/dx=(-1/2)x^(1/2)
3. dy/dx=(-1/2)x^(-1/2)
4. dy/dx=(1/2)x^(-3/2)
5. dy/dx=(-1/2)x^(3/2)
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# 1. How to identify the sample space of a probability experiment and how to identify simple events
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1 Statistics Chapter 3 Name: 3.1 Basic Concepts of Probability Learning objectives: 1. How to identify the sample space of a probability experiment and how to identify simple events 2. How to use the Fundamental Counting Principle to find the number of ways two or more events can occur 3. How to distinguish among classical probability, empirical probability, and subjective probability 4. How to find the probability of the complement of an event 5. How to use a tree diagram and the Fundamental Counting Principle to find probabilities Probability as a general concept can be defined as the chance of an event occurring. Processes such as flipping a coin, rolling a die, or drawing a card from a deck are called probability experiments. A probability experiment is an action, or chance process, through which specific results (counts, measurements, or responses) are obtained. A trial means flipping a coin once, rolling one die once, or the like. A trial is the action part of the experiment. An outcome is the result of a single trial in a probability experiment. When a coin is tossed, there are two possible outcomes: head or tail. In the roll of a single die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. A sample space is the set of all possible outcomes of a probability experiment. Example: Probability Experiment: Roll two dice and record the sum of the two numbers on the faces of the dice. That sum will equal a whole number between and including 2 and 12. There are 36 outcomes in the sample space. Each outcome is illustrated in the picture below.
2 Some more sample spaces for various probability experiments are shown here Example Find the sample space for drawing one card from an ordinary deck of cards. Definition A tree diagram is a device consisting of line segments emanating from a starting point and also from the outcome point. It is used to determine all possible outcomes of a probability experiment. Example Use a tree diagram to find the sample space for the gender if a family has three children. Use B for boy and G for girl.
3 Example Use a tree diagram to find the sample space for the experiment tossing a coin twice. Your Turn! A coin is flipped and a die is rolled. Use a tree diagram to find the sample space for the sequence of events. How many outcomes are in the sample space? What is the probability that the coin lands on tails and the die shows a 4? Your Turn! For each probability experiment, determine the number of outcomes and identify the sample space. 1. A probability experiment consists of recording a response to the survey question statement at the left and the gender of the respondent. 2. A probability experiment consists of recording a response to the survey question statement at the left and the geographic location (Northeast, South, Midwest, West) of the respondent.
4 More Definitions! An outcome was defined previously as the result of a single trial of a probability experiment. In many problems, one must find the probability of two or more outcomes. For this reason, it is necessary to distinguish between an outcome and an event. An event consists of a set of outcomes of a probability experiment. In the rest of this chapter, you will learn how to calculate the probability of an event. Events are often represented by uppercase letters, such as A,B or C. An event can be one outcome or more than one outcome. For example, if a die is rolled and a 6 shows, this result is called an outcome, since it is a result of a single trial. An event with one outcome is called a simple event. The event of getting an odd number when a die is rolled is called a compound event, since it consists of three outcomes or three simple events. In general, a compound event consists of two or more outcomes or simple events Identifying Simple Events A card is randomly selected from a standard deck that includes two joker cards. Some events have been defined below. Determine the number of outcomes in each event. Then decide whether the event is a simple event or not. Explain your reasoning. 3. Event A: select a heart card 4. Event B: select a 4 5. Event C: select a joker card 6. Event D: select a 4 of hearts Identifying Simple Events You roll a six-sided die. Some events have been defined below. Determine the number of outcomes in each event. Then decide whether the event is a simple event or not. Explain your reasoning. 7. Event A: roll at least a 3 8. Event B: roll less than 4 9. Event C: roll an odd number 10. Event D: roll a 4
5 In some cases, an event can occur in so many different ways that drawing a tree diagram becomes too cumbersome, and it is not practical to write out all the outcomes. When this happens, use the Fundamental Counting Principle. Fundamental Counting Principle If one event can occur in m ways and a second event can occur in n ways, the number of ways the two events can occur in sequence is m times n. In words, the number of ways that events can occur in sequence is found by multiplying the number of ways one event can occur by the number of ways the other event(s) can occur. The Fundamental Counting Principle can be extended for any number of events occurring in sequence. Using the Fundamental Counting Principle Example: Suppose you are purchasing a new car and the possible manufacturers, car sizes, and colors you have selected for your car are listed below. Manufacturer: Ford, GM, Honda Car size: compact, midsize Color: white (W), red (R), black (B), green (G) Question: How many different ways can you select one manufacturer, one car size, and one color? Use a tree diagram to check your result. Answer: There are three choices of manufacturers, two car sizes, and four colors. Using the Fundamental Counting Principle: (3)(2)(4) = 24 ways ATM passwords have four digits. Each digit can be any number from 0 and How many ATM passwords are possible when each digit can be used only once and not repeated?
6 12. How many ATM passwords are possible when each digit can be repeated? 13. How many ATM passwords are possible when each digit can be repeated but the first digit cannot be 0 or 1? 14. How many license plates can you make when a license plate consists of six (out of 26) alphabetical letters, each of which can be repeated? 15. How many license plates can you make when a license plate consists of six (out of 26) alphabetical letters, each of which cannot be repeated? Probabilities can be expressed as fractions, decimals percentages. If you ask, What is the probability of getting a head when a coin is tossed? typical responses can be any of the following three. One-half. Point-five fifty percent These answers are all equivalent. Rounding Rule for Probabilities Probabilities should be expressed as reduced fractions or rounded to two or three decimal places. Percentages are also acceptable. When the probability of an event is an extremely small decimal, it is permissible to round the decimal to the first nonzero digit after the point. For example, would be
7 There are three basic interpretations of probability: 1. Classical probability 2. Empirical probability (or relative frequency probability) 3. Subjective probability Classical (or theoretical) probability uses sample spaces to determine the numerical probability that an event will happen. Classical probability assumes that all outcomes in the sample space are equally likely to occur P(E) = Number of outcomes in E Total number of outcomes in the sample space Exercises: use the Classical technique to find the probability of each event below. 16. Rolling a Die If a die is rolled one time, find these probabilities. (a) Of getting a 4 (b) Of getting an even number (c) Of getting a number greater than 4 (d) Of getting a number less than 7 (e) Of getting a number greater than 0 (f) Of getting a number greater than 3 or an odd number (g) Of getting a number greater than 3 and an odd number 17. If two dice are rolled one time, find the probability of getting these results. (a) A sum of 6 (b) A sum of 7 or If one card is drawn from a deck, find the probability of getting these results. (a) An ace (b) A diamond (c) An ace of diamonds (d) A 4 or a 6 (e) A 4 or a club
8 Empirical (Relative Frequency) Probability Based on observations obtained from probability experiments. Relative frequency of an event. P(E) = frequency of event E Total frequencies in the distribution = f n The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely (such as the outcomes when a die is rolled), while empirical probability relies on actual experience to determine the likelihood of outcomes. Empirical probability is based on observations obtained from probability experiments. Relative Frequency Probability Examples: The probability that the next car that comes out of an auto factory is a lemon The probability that a randomly selected family in San Diego owns a home The probability that an 80-year-old person will live for at least 1 more year The probability that a randomly selected California driver owns a Toyota Prius. These probabilities cannot be computed using the classical probability rule because the various outcomes for the corresponding experiments are not equally likely. Exercises: use the Empirical (Relative Frequency) technique of finding probability. 19. Hospital records indicated that maternity patients stayed in the hospital for the number of days shown in the distribution. Find these probabilities. (a) A patient stayed exactly 5 days. Number of days stayed Frequency (b) A patient stayed less than 6 days (c) A patient stayed at most 4 days (d) A patient stayed at least 5 days. number In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities. (a) A person has type O blood. (b) A person has type A or type B blood. (c) A person does not have type AB blood number
9 21. Suppose a teenager is randomly selected. Find the probability that he or she has made (a) 5 or fewer friends online (b) at least 2 friends online (c) no more than 1 friend online Source: Pew Research Center Aug Number of Relative new friends made online Frequency no friends friends friends friends On a construction site, there are 10 carpenters and 18 laborers; 3 carpenters and 5 laborers are females. A person working on the construction site is randomly selected. Set up a frequency distribution and find the following probabilities. (a) The person selected is a female (b) The person selected is a laborer (c) The person selected is a female carpenter
10 23. Try this! An individual stock is selected at random from the portfolio represented by the boxand-whisker plot shown. Find the probability that the stock price is (a) less than \$21, (b) between \$21 and \$50 and (c) \$30 or more. Subjective probability uses a probability value based on an educated guess or estimate, employing opinions and inexact information. In subjective probability, a person or group makes an educated guess at the chance that an event will occur. This guess is based on the persons experience and evaluation of a solution. The probability that Carol, who is taking a statistics course, will earn an A in the course The probability that the Dow Jones Industrial Average will be higher at the end of the next trading day A doctor may feel a patient has a 90% chance of a full recovery. All three types of probability (classical, empirical, and subjective) are used to solve a variety of problems in business, engineering, and other fields. 24. Exercises: Classify each statement as an example of classical probability, empirical probability, or subjective probability (a) The probability that a person will watch the 6 oclock evening news is (b) The probability of winning the final round of wheel of fortune (c) The probability that a city bus will be in an accident on a specific run is about 6%. (d) The probability of getting a royal flush when five cards are selected at random is 1/649,740 (e) An analyst feels that a certain stock s probability of decreasing in price over the next week is 0.75.
11 Law of Large Numbers As an experiment is repeated over and over, the empirical probability of an event approaches the theoretical (actual) probability of the event. Probability Rules 1. The probability of any event E is a number (either a fraction or decimal) between and including 0 and 1. This rule states that probabilities cannot be negative or greater than If an event E cannot occur (i.e., the event contains no members in the sample space), its probability is 0 3. If an event E is certain, then the probability of E is 1 (or 100%). 4. The sum of the probabilities of all the outcomes in the sample space is An event is considered unusual if occurs with a probability of 5% or less.
12 The Complement of An Event Recall that an event is a set of outcomes. The complement of an event E is the set of all outcomes in the sample space that are not included in event E. The complement of E is denoted either as E E c or Ē. For example, if we define an event A then the event s complement is denoted as A A c or Ā. Using the definition of the complement of an event and the fact that the sum of the probabilities of all outcomes in the sample space is one, we can determine that Write the complement rule here: P(E) + P(E c ) = A single die is rolled. Some events have been defined below. Find the complement of each event (a) Event E: roll an odd number (b) Event A: roll a 4 (c) Event B: roll at least a 3 (d) Event C: roll less than A card is randomly selected from a standard deck of 52 playing cards. Some events have been defined below. Find the complement of each event. (a) Event A: the card is a four (b) Event B: the card is a heart card (c) Event C: the card is a four of hearts 27. A card is randomly selected from a standard deck of playing cards, including two joker cards. Find each probability (a) Event A: the card is not a joker (b) Event B: the card is not an ace (c) Event C: the card is not a heart card 28. In 2013, 33% of LeastWorst Airlines customers who purchased a ticket spent an additional \$20 to be in the first boarding group. Choose one LeastWorst customer at random. What is the probability that the customer didnt spend the additional \$20 to be in the first boarding group?
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Free math help algebra
This Free math help algebra supplies step-by-step instructions for solving all math troubles. We will also look at some example problems and how to approach them.
The Best Free math help algebra
This Free math help algebra provides step-by-step instructions for solving all math problems. In mathematics, a logarithm is an operation that allows us to solve for an unknown exponent. For example, if we are given the equation y = 10x, we can use a logarithm to solve for x. In this case, we would take the logarithm of both sides of the equation, giving us: log(y) = log(10x). We can then use the fact that logs are exponents to rewrite this equation as: y = 10log(x). This means that x = 10^y, which is a much easier equation to solve. Logarithms can be used to solve equations with any base, not just 10. In general, if we are given the equation y = bx, we can solve for x by taking the logarithm of both sides and using the fact that logs are exponents. This method can be used to quickly and easily solve equations with very large or very small numbers.
A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane.
There are a variety of online math graph calculators available, with different features and capabilities. However, all online math graph calculators have one thing in common: they allow users to perform calculations and visualize results using an online interface. This can be extremely helpful for students who are struggling to understand complex mathematics concepts. In addition, online math graph calculators can be used by educators to create custom teaching materials. As more and more people embrace digital learning, online math graph calculators are likely to become an essential tool for mathematics education.
Solving for x logarithms can be difficult, but there are a few methods that can help. One method is to use the change of base formula. This formula states that if you have two values with the same base, you can set them equal to each other and solve for the unknown value. For example, if you have the equation log4(x)=log2(x), you can set the two equations equal to each other and solve for x. Another method is to use graphing calculator. Many graphing calculators have a built-in function that allows you to solve for x logarithms. Simply enter the equation into the calculator and press the "solve" button. The calculator will then give you the value of x. Finally, you can also use a table of logarithms to solve for x logarithms. To do this, simply find the values of x and y that are equal to each other and solve for x. Solving for x logarithms can be difficult, but with a little practice, it can be easy.
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# Sum Like It Simple
The three intersecting ellipses form seven curved regions. Place one tile from the seven candidates in each of these regions so that the tiles in any ellipse adhere to the corresponding rule. Show that this solution is unique!
Mark the numbers like this:
a c
b
d
e f
g
The equations are:
$$\begin{eqnarray}a + b + d + e &=& 15\tag 1\\b + c + d + f &=& 19\tag 2\\d + e + f + g &=& 20\tag 3\end{eqnarray}$$
Adding $$(2)$$ and $$(3)$$ and subtracting the sum from $$a$$ to $$g$$, we get $$d + f - a = 19 + 20 - 28 = 11$$
which implies $$a \leq 2$$ because $$d + f \leq 6 + 7 = 13$$.
If $$a = 2$$, then $$d, f$$ are $$6, 7$$ and from $$(2), (3)$$ we get $$b + c = 6$$ and $$e + g = 7$$. It follows that $$b, c$$ are $$1, 5$$ and $$e, g$$ are $$3, 4$$. There is however no possible way to satisfy $$(1)$$: $$b = 1$$ is too small and $$b = 5$$ is too big.
Thus $$a = 1$$ and $$d, f$$ are $$5, 7$$ and $$(2), (3)$$ become $$b + c = 7$$ and $$e + g = 8$$. It follows that $$b, c$$ are $$3, 4$$ and $$e, g$$ are $$2, 6$$. In view of $$(1)$$, $$e = 2$$ is too small, hence $$e = 6$$ and $$b + d = 8$$. It is then clear that $$b = 3$$ and $$d = 5$$.
Therefore the only solution is: $$(a, b, c, d, e, f, g) = (1, 3, 4, 5, 6, 7, 2)$$.
• Same solution, using the fact that 39 is only 1 less then maximum possible sum of $b+c+e+g+2(d+f)$
– z100
Jun 11, 2022 at 16:00
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# Calculate: \lim_{x arrow 1} ((x^2-sqrt(x))/(sqrt(x)-1))
## Expression: $\lim_{x \rightarrow 1} \left(\frac{ {x}^{2}-\sqrt{ x } }{ \sqrt{ x }-1 }\right)$
Since evaluating limits of the numerator and denominator would result in an indeterminate form, use the L'Hopital's rule
$\lim_{x \rightarrow 1} \left(\frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( {x}^{2}-\sqrt{ x } \right) }{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sqrt{ x }-1 \right) }\right)$
Find the derivative
$\lim_{x \rightarrow 1} \left(\frac{ 2x-\frac{ 1 }{ 2\sqrt{ x } } }{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( \sqrt{ x }-1 \right) }\right)$
Find the derivative
$\lim_{x \rightarrow 1} \left(\frac{ 2x-\frac{ 1 }{ 2\sqrt{ x } } }{ \frac{ 1 }{ 2\sqrt{ x } } }\right)$
Write all numerators above the common denominator
$\lim_{x \rightarrow 1} \left(\frac{ \frac{ 4x\sqrt{ x }-1 }{ 2\sqrt{ x } } }{ \frac{ 1 }{ 2\sqrt{ x } } }\right)$
Simplify the expression
$\lim_{x \rightarrow 1} \left(\frac{ 4x\sqrt{ x }-1 }{ 1 }\right)$
Any expression divided by $1$ remains the same
$\lim_{x \rightarrow 1} \left(4x\sqrt{ x }-1\right)$
Evaluate the limit
$4 \times 1\sqrt{ 1 }-1$
Simplify the expression
$3$
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# Mock AIME 2 2006-2007 Problems/Problem 12
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In quadrilateral $ABCD,$ $m \angle DAC= m\angle DBC$ and $\frac{[ADB]}{[ABC]}=\frac12.$ $O$ is defined to be the intersection of the diagonals of $ABCD$. If $AD=4,$ $BC=6$, $BO=1,$ and the area of $ABCD$ is $\frac{a\sqrt{b}}{c},$ where $a,b,c$ are relatively prime positive integers, find $a+b+c.$
Note*: $[ABC]$ and $[ADB]$ refer to the areas of triangles $ABC$ and $ADB.$
## Solution
$m\angle DAC=m\angle DBC \Rightarrow ABCD$ is a cylic quadrilateral.
Let $DO=a, AO=b$
$\triangle AOD$ ~ $\triangle BOC \Rightarrow b=\frac{2}{3}$
Also, from the Power of a Point Theorem, $DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}$
Notice $\frac{[AOD]}{[BOC]}=(\frac{2}{3})^2\Rightarrow [BOC]=\frac{9}{4}[AOD]$
It is given $\frac{[AOD]+[AOB]}{[AOB]+[BOC]}=\frac{[ADB]}{[ABC]}=\frac{1}{2} \Rightarrow [AOB]=\frac{[AOD]}{4}$
Note that $\sin{\angle AOB}=\sin{(180-\angle AOD)}=\sin{\angle AOD}$
Then $[AOB]=\frac{\frac{2}{3}\cdot 1\cdot\sin{\angle AOB}}{2}=\frac{\sin{\angle AOD}}{3}$ and $[AOD]=\frac{\frac{2}{3}\cdot a\cdot\sin{\angle AOD}}{2}=\frac{a\sin{\angle AOD}}{3}$
$\Rightarrow a=4$
$[COD]=9[AOD]$
Thus we need to find $[ABCD]=\frac{25}{2}[AOD]$
Note that $\triangle AOD$ is isosceles with sides $4, 4, \frac{2}{3}$ so we can draw the altitude from D to split it to two right triangles.
$[AOD]=\frac{\sqrt{143}}{9}$
Thus $[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}$
## See Also
Mock AIME 2 2006-2007 (Problems, Source) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
## Problem Source
AoPS users 4everwise and Altheman collaborated to create this problem.
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# How To Multiply Fractions With Same Denominator
Divide the numerator by the denominator and whatever whole number you get in the quotient, that’s the number of wholes in the mixed number. Let us discuss the conversion of fractions into fractions having the same denominator.
Fraction Calculator – Ultimate Tool To Add Fractions Studying Math Math Fractions Math Methods
### If the denominators are not the same then cross multiply the two fractions and add the results together to get the same denominator.
How to multiply fractions with same denominator. You must first leave the first fraction alone, change the. You will get the new numerator with the. The bottom number of the answer will be the same as the denominator of the original fractions.
And then you have 3 times 1 in the denominator. The multiplication of fractions is not like the addition or subtraction of fractions, where the denominator should be the same. You’ll recall from our basic overview of multiplying fractions that the denominator in the fraction is calculated by multiplying the two denominators from the numbers in the problem (the multiplicands).
When a numerator is equal to the denominator, it’s considered “improper” because you can change it into a whole number.the same rule applies to improper fractions such as ²⁶⁄₁₃ which, if reduced, become whole (i.e., two). A short video showing you how to multiply a set of fractions with the same denominator, step by step Multiply the denominators of the fractions together.
Multiply the top numbers of both fractions together to get the numerator (top number) of your new fraction. Multiplying fractions is actually less complicated than adding or subtracting. To simplify means to make the fraction into the smallest numbers possible by finding the largest number that can divide both the numerator and denominator.
The basic steps are… multiply the numerators to get the product numerator. How to divide fractions with the same denominator. You’ll recall from our basic overview of multiplying fractions that the denominator in the fraction is calculated by multiplying the two denominators from the numbers in the problem (the multiplicands).
The remainder is the new numerator for the fraction component. Multiply the denominators to get the product denominator reduce the product. Whenever i see a negative sign show up in the denominator, i simply move it up to the numerator.
Here’s a visual representation of these steps… When multiplying fractions, the process is the same whether there are common denominators or not. The only thing to be kept in mind is that the fractions should not be in the mixed form, they should either be proper fractions or improper fractions.
Here, any two fractions with different denominators can easily be multiplied. To turn an improper fraction into a mixed number, you’ll use your long division skills. Make sure the bottom numbers (the denominators) are the same step 2:
Using the previous example, here is the result: It doesn’t matter if these numbers are the same or different in the problem. How do we divide fractions?
The denominator remains the same. Multiplying fractions with same denominators is the same as multiplying other regular fractions. Add or subtract the numerators, or the top numbers, and write the result in a new fraction on the top.
To multiply fractions, first simply the fraction to its lowest term. Then, the product of fractions is obtained in p/q form. You can just multiply across.
It doesn’t matter if these numbers are different in the problem because the steps to find the denominator are the same. Do common denominators make fraction multiplication easier? To add and subtract fractions with the same denominator, or bottom number, place the 2 fractions side by side.
Adding fractions adapted from mathisfun.com there are 3 simple steps to add fractions: The fractions having the same denominator are called like fractions. In the case of mixed fractions, simplify it.
After simplifying the fraction, multiply the numerator with the numerator and the denominator with the denominator. The product becomes your new denominator. To write {eq}2 \frac{3}{4} {/eq} as an improper fraction, multiply the denominator, 4, times the whole number, 2, then add the numerator, 3.
1/2 x 2/3 x 1/4 = will give you your numerator of 2 and a denominator of 24. Whether the sign is in the numerator or the denominator doesn’t matter in terms of the arithmetic used to multiply the fractions, so move it where it makes sense. Put the answer over the denominator.
Add the top numbers (the numerators). To multiply fractions, you need to multiply the numerators together and multiply the denominators together. What if there is a negative numerator and a negative denominator.
So, numerator time numerator will get you your top answer of the fraction, then denominator times denominator will get you your bottom answer. Multiply or divide the numerator and denominator by the same number. Follow the same method to multiply all of the denominators of your fractions.
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# How do you factor completely -n^4 - 3n^2 - 2n^3?
Sep 5, 2017
$- {n}^{2} \left(n + 1 - i \sqrt{2}\right) \left(n + 1 + i \sqrt{2}\right)$
#### Explanation:
$- {n}^{2} \text{ is a "color(blue)"common factor"" in all 3 terms}$
$\Rightarrow - {n}^{2} \left({n}^{2} + 3 + 2 n\right)$
$= - {n}^{2} \left({n}^{2} + 2 n + 3\right)$
$\text{check the "color(blue)"discriminant"" of } {n}^{2} + 2 n + 3$
$\text{with } a = 1 , b = 2 , c = 3$
$\Delta = {b}^{2} - 4 a c = 4 - 12 = - 8$
$\text{since "Delta<0" then the roots are not real}$
$\text{we can factorise by finding the roots of } {n}^{2} + 2 n + 3$
$\text{using the "color(blue)"quadratic formula}$
$n = \frac{- 2 \pm \sqrt{4 - 12}}{2} = \frac{- 2 \pm \sqrt{- 8}}{2}$
$\textcolor{w h i t e}{n} = - 1 \pm i \sqrt{2} \leftarrow \textcolor{red}{\text{ complex roots}}$
$\Rightarrow - {n}^{4} - 3 {n}^{2} - 2 {n}^{3}$
$= - {n}^{2} \left(n + 1 - i \sqrt{2}\right) \left(n + 1 + i \sqrt{2}\right)$
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# Binomial distribution
There’s a lot of situations in life where we can be faced with two possible outcomes. For example, you either win the lottery or you don’t and a drug to cure a disease works or it doesn’t. Binomial distributions come into play. When analyzing the probability of success or failure such as these. This type of distribution evaluates the likelihood for an experiment or survey outcome to either succeed or fail – such as coins landing on heads-or-tails during tosses or passing/failing tests.
## Binomial distribution formula
Your experiment must meet the following three criteria in order to use the formula below:
1. Fixed number of observations or trials. In other words, you can have 10 trials or 100, but not an unlimited number of attempts.
2. Independent observations or trials. In other words, one coin toss or test attempt (or whatever it is you are measuring) should not affect the next attempt.
3. Same probability of success from one “trial” to another. for example, the odds of flipping heads remains at 50% from trial to trial.
The binomial distribution formula is:
b(x; n, P) = nCx * Px * (1 – P)n – x
Where:
• b = binomial probability.
x = total number of “successes.”
P = probability of a success on a single attempt.
n = number of attempts or trials.
The binomial distribution formula can also be written as:
Example (using alternate formula): A fair coin is tossed 10 times. What is the probability of getting six heads?
• Number of trials (n) = 10
• Odds of success (“heads”) is 50% or 0.5 (1 – p = 0.5)
• x = 6 (we want to know the probability of getting 6 heads)
• P(x=6) = 10C6 * 0.5^6 * 0.5^4 = 210 * 0.015625 * 0.0625 = 0.205078125 = 20.51%.
## Relationship to Bernoulli Distribution
The binomial distribution is closely related to the Bernoulli distribution: the binomial distribution with n = 1 is called a Bernoulli distribution. In addition, if a Bernoulli trial has independent trials, the number of successes follows a binomial distribution.
A Bernoulli distribution is a set of Bernoulli trials. Each of these trials has one possible outcome, Success, or Failure. In each trial, the probability of success, P(S) = p, is the same. The probability of failure is 1 minus the probability of success: P(F) = 1 – p. (Remember that “1” is the total probability of an event occurring and is always between zero and 1). Finally, all Bernoulli trials are independent from each other and the probability of success doesn’t change from trial to trial, even if you have information about the other trials’ outcomes.
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# Blog
## Counting by division
By
,
Let’s look at a few interesting patterns you can get by dividing. Some of these patterns need a lot of digits, so get the best calculator you can find.
If you have a computer with the internet handy, there are some good online calculators to use, such as wolframalpha.com. Some mathematicians might prefer to use long division with paper and a pen.
### You will need
• Calculator
• Paper and pen (optional)
### What to do
1. Let’s start with a relatively simple calculation. What do you think you’ll get if you calculate 1÷9? Key 1÷9 into your calculator (or 1/9 on a computer).
2. Keeping the answer in your calculator, what would happen if you divided by 9 again? Try it and see!
3. Let’s try to extend the pattern. What do you think you’ll get when you calculate 1÷99?
4. Think long and hard before this one – what do you think you will get when you divide by 99 again?
5. Final question, and this one’s tricky. What will you get when you calculate 1÷998001? Can you see why this works?
### Answers
1. 111111111111…
2. 01234567901234567…
3. 0101010101010101…
4. 000102030405060708091011…
5. 000001002003004005006007008…
### What’s happening?
Maths is all about patterns, and this activity shows a really neat one. Dividing by 9 and then 9 again gives an answer that looks a lot like counting! You can extend the pattern by using more 9s – dividing by 99 twice. The final twist is combining the two divisions. 998001 = 999 x 999, which is why the final calculation counts too.
The pattern arises because the number 9 is one less than 10. Or to put it another way, when you divide 10 by 9, you get a remainder of 1. When you move to the next place value, that 1 becomes a 10 and the process repeats. That’s why you get 0.111111…
The second division reveals another trick of the number 9. When you divide 20 by 9, you get a remainder of 2. When you divide 30 by 9, you get a remainder of 3. This process will keep ticking up until eventually it gets to 90, which has no remainder.
It’s a bit hard to understand just by reading. The best way to see the pattern is by doing long division to calculate 1 ÷ 9 and 0.1111… ÷ 9 by hand. The patterns you find will help explain what happens when you divide by 99 999.
If you’re after more maths activities for kids, subscribe to Double Helix magazine!
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# First Grade Place Value: A Number's Place in Line Makes a Difference
Place value is the term we use to talk about how numbers, when arranged in different places, show different amounts. If we write the number 372, we know there are 3 hundreds, 7 tens, and 2 ones; it does not mean 3+7+2, which is what it may look like to a young first grader.
Imagine how difficult this is to learn. You know that this number is 3, but now you are being told that when it is a little farther over here it's 30, and way over there it is 300! It is a tremendously tricky concept, and needs quite a bit of practice and exposure before children begin to feel comfortable with it. They will need skills in counting to 100, and they need to know their basic addition facts and subtraction facts and be able to skip count before they should begin working directly with place value.
Kids know that the number is 3, but now they are being told that when you put the number here it's 30, and when you move it over there it's 300!
### What Kids Need to Learn
1. That counting can be organized in groups. When learning to count, numbers seem to proceed along one long line, with no real relationship to each other. Practice in grouping numbers and skip counting will prepare children for the concepts of number groupings in place value.
2. How to count many objects by organizing them into tens and ones and regrouping these when necessary.
3. That the same written number can have different amounts (7, 70, 700) depending on where it sits within a number.
4. That pennies, dimes, and dollars are organized this way.
5. How to use counters to show regrouping of addition and subtraction problems up to 100.
### Putting the Math in Context
In the beginning of first grade, children will strengthen their skills in counting to 100, addition facts and subtraction facts, and skip counting.
Around the middle of first grade, they will learn to separate numbers and objects into tens and ones.
Near the end of first grade, children will begin using regrouping when adding and subtracting 2-digit numbers ("borrowing" to subtract, and "carrying" to add).
In second grade, children will work with regrouping and place value concepts for 3-digit numbers.
### Why It's Important
Place value is one of those concepts that you have to know well before you move on to other things. (There seem to be a lot of such concepts in first grade.) Children who do not have a secure understanding of place value will have a terrible time with adding or subtracting larger numbers, regrouping, estimating, and later skills of multiplication and division. They will have a hard time getting a practical sense of how much a number refers to, and may have poor number sense skills. They may reverse numbers when reading or writing, such as writing 68 instead of 86, because the order of the numbers means very little to them. Later on, this skill will be essential to understanding money (the difference between \$55 and \$505, for example) and numbers in general.
### Math Challenges Kids Might Face
Place value takes years to master, and there are many things that will likely prove difficult for first graders at one time or another, such as:
1. When grouping objects by 10s, kids forget to stop at 10 and keep going. Once they realize it, they have to stop and start over again.
2. Not knowing when to count by 1s, and when to count by 10s. When kids are finding out "how many tens" there are before writing a number, they are expected to count by 1s. ("One, two, three tens.") When they are counting up the whole number at the end, they are expected to count by tens first, then add on the ones ("Ten, twenty, thirty, thirty-one, thirty-two.")
3. Feeling that the larger numeral is a bigger quantity no matter where it is located. In the number 27, they think the 7 is bigger than the 2 (20).
4. Understanding that there are tens and ones, but getting confused about which one goes where.
5. Just finding the whole concept downright confusing.
### Math Help That Could Make the Difference
Take baby steps. Spend lots of time on tens and ones to start with. Have kids count a jar of pennies by making stacks of 10, then show what the tens and ones look like written down. Point out the tens and ones place in the numbers you see every day. Then, when kids seem to be getting the hang of tens and counting by tens, you can introduce the hundreds place.
But don't rush it. This is a brand new concept, and new concepts take time. Kids will need lots of time manipulating objects like place value manipulatives and regrouping them in order to understand place value in written numbers. Give kids opportunities for making numbers, adding, and subtracting with objects.
NOTE: Some children may not be developmentally ready to jump right into regrouping with tens. Here is a video by a teacher who makes a strong case for starting out by teaching a "Base 2" and a "Base 3" system first, to help kids really visualize the concept, before moving on to base 10. Take a look:
After kids have had lots of practice making numbers that visually look like what they represent, introduce the idea that the value of a number can change even though it still looks the same; where you put it the number changes how much it is worth.
Finally, show kids how to write numbers as you say them, and do regrouping with written addition and subtraction problems. Bear in mind that, even at this last level, first graders will still need a great deal of hands-on practice with place value manipulatives to give the written numbers meaning.
So take it nice and slow, and give kids lots of practice with place value activities. When they start to get the hang of these, take kids to the next level with place value games to cement their understanding of those tricky skills.
### Whiz Kids
Some kids will grasp these concepts more quickly than others. If a couple of kids seem to be doing fine with 2-digit numbers, go ahead and challenge them with 3-digit numbers. They can do all the same activities that the rest of the class is doing, only with the challenge of working with hundreds, or even thousands, instead of just tens.
You may find it helpful to browse the Artful Math Marketplace--our own Amazon.com mini-store. Amazon gives us a small thank-you if you buy through this link, but we only feature the math supplies we use ourselves.
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# But it should be equal to -1/12
Calculus Level 4
Here's my proof that the sum of all the natural numbers is equal to 1.
In which of thse steps did I first make a flaw in my logic?
\begin{aligned} 1 + 2 + 3 + \cdots + n &=& \dfrac12n(n+1) \\ 1^3 + 2^3 + 3^3 + \cdots + n^3 &=& \left( \dfrac12n(n+1) \right)^2 \\ \end{aligned}
Step 2: There is a relationship between the numbers $1 + 2 + 3 + \cdots + n$ and $1^3 + 2^3 + 3^3 + \cdots + n^3$,
$(1 + 2 + 3 + \cdots + n )^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3$
Step 3: Take the limit to both sides of the equation,
$\lim_{n\to\infty} (1 + 2 + 3 + \cdots + n )^2 = \lim_{n\to\infty} (1^3 + 2^3 + 3^3 + \cdots + n^3)$
Which is equivalent to
$(1+2+3 + \cdots )^2 = 1^3 +2^3 + 3^3 + \cdots$
Step 4: Since $1^3 +2^3 + 3^3 + \cdots \rightarrow \infty$ and $1 +2 + 3 + \cdots \rightarrow \infty$, then $1^3 +2^3 + 3^3 + \cdots = 1 +2 + 3 + \cdots$,
\begin{aligned} (1+2+3 + \cdots )^2 &=& 1 +2 + 3 + \cdots \\ (1+2+3 + \cdots )(1+2+3 + \cdots ) &=& 1 +2 + 3 + \cdots \\ \cancel{(1+2+3 + \cdots )}(1+2+3 + \cdots ) &=& \cancel{(1+2+3 + \cdots )} \\ 1+2+3 + \cdots &=& 1 \end{aligned}
×
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# Probability Problems
• Oct 15th 2009, 04:21 PM
loutja35
Probability Problems
Boods Types: The probability that a person in the United States has type AB+ blood is 3%. Five unrelated people in the United States are selected at random.
a.) Find the probability that all five have AB+ blood.
b.) Find the probability that none of the five has type AB+ blood.
c.) Find the probability that at least one of the five has type AB+ blood.
Guessing: A multiple-choice quiz has three questions, each with five answer choices. Only one of the choices is correct. You have no idea what the answer is to any question and have to guess each answer.
a.) Find the prob. of answering the first question correctly.
c.) Fine the prob. of answering all three questions correctly.
c.) Find the prob. of answering the first two questions correctly.
d.) Find the prob. of answering none of the questions correctly.
e.) Find the prob. of answering at least one of the questions correctly.
• Oct 15th 2009, 07:51 PM
Soroban
Hello, loutja35!
Quote:
Blood Types:
The probability that a person in the United States has type AB+ blood is 3%.
Five unrelated people in the United States are selected at random.
We have: .$\displaystyle \begin{Bmatrix}P(\text{AB+}) \:=\:0.03 \\ P(\text{Not}) \:=\:0.97 \end{Bmatrix}$
Quote:
a) Find the probability that all five have AB+ blood.
$\displaystyle P(\text{5 AB+}) \:=\:(0.03)^5 \:=\:0.0000000243$
Quote:
b) Find the probability that none of the five has type AB+ blood.
$\displaystyle P(\text{5 Not}) \:=\:(0.97)^5 \:=\: 0.8587340267$
Quote:
c) Find the probability that at least one of the five has type AB+ blood.
This is the exact opposite of part (b).
$\displaystyle P(\text{at least one AB+}) \;=\; 1 - 0.85873402578 \;=\;0.1412650743$
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Conjugate of a Matrix
If $A$ is a matrix with complex elements, the conjugate of $A$, denoted as $\stackrel{‾}{A}$, is obtained by taking the conjugate of each element within the matrix.
Representation:
If $A$ is an $m×n$ matrix with complex elements ${a}_{ij}$, then the conjugate of $A$, $\stackrel{‾}{A}$, is represented as a matrix with elements $\stackrel{‾}{{a}_{ij}}$.
Example:
Let's consider a matrix $A$ with complex elements:
$A=\left[\begin{array}{cc}2+3i& 1-2i\\ 4i& 5\end{array}\right]$
The conjugate of $A$, denoted as $\stackrel{‾}{A}$, would be:
$\stackrel{‾}{A}=\left[\begin{array}{cc}2-3i& 1+2i\\ -4i& 5\end{array}\right]$
Properties:
1. $\stackrel{‾}{\stackrel{‾}{A}}=A$: Conjugating a matrix twice brings it back to the original matrix.
2. $\stackrel{‾}{\left(kA\right)}=\stackrel{‾}{k}\cdot \stackrel{‾}{A}$: Conjugate of a scalar multiplied by a matrix is equal to the conjugate of the scalar multiplied by the conjugate of the matrix.
3. $\stackrel{‾}{\left(A+B\right)}=\stackrel{‾}{A}+\stackrel{‾}{B}$: Conjugate of the sum of matrices is equal to the sum of their conjugates.
4. $\stackrel{‾}{\left(AB\right)}=\stackrel{‾}{A}\cdot \stackrel{‾}{B}$: Conjugate of the product of two matrices is equal to the product of their conjugates in the same order.
Applications:
1. Quantum Mechanics: Conjugate transpose matrices are significant in quantum mechanics, especially in representing quantum states and operations.
2. Signal Processing: In complex signal processing applications where complex numbers are involved, conjugates are used for various transformations and operations.
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# Graphing Cube Root Functions
## Graph 3rd root functions with and without a calculator
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Practice Graphing Cube Root Functions
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Graphing Cubed Root Functions
The next day, Mrs. Garcia assigns her student the cube root function y=(x+1)3\begin{align*}y = -\sqrt[3]{(x + 1)}\end{align*} to graph for homework. The following day, she asks her students which quadrant(s) their graph is in.
Alendro says that because of the negative sign, all y values are negative. Therefore his graph is only in the third and fourth quadrants quadrant.
Dako says that his graph is in the third and fourth quadrants as well but it is also in the second quadrant.
Marisha says they are both wrong and that her graph of the function is in all four quadrants.
Which one of them is correct?
### Guidance
A cubed root function is different from that of a square root. Their general forms look very similar, y=axh3+k\begin{align*}y=a \sqrt[3]{x-h}+k\end{align*} and the parent graph is y=x3\begin{align*}y= \sqrt[3]{x}\end{align*}. However, we can take the cubed root of a negative number, therefore, it will be defined for all values of x\begin{align*}x\end{align*}. Graphing the parent graph, we have:
x y
-27 -3
-8 -2
-1 -1
0 0
1 1
8 2
27 3
For y=x3\begin{align*}y= \sqrt[3]{x}\end{align*}, the output is the same as the input of y=x3\begin{align*}y=x^3\end{align*}. The domain and range of y=x3\begin{align*}y= \sqrt[3]{x}\end{align*} are all real numbers. Notice there is no “starting point” like the square root functions, the (h,k)\begin{align*}(h, k)\end{align*} now refers to the point where the function bends, called a point of inflection (see the Analyzing the Graph of a Polynomial Function concept).
#### Example A
Describe how to obtain the graph of y=x3+5\begin{align*}y= \sqrt[3]{x}+5\end{align*} from y=x3\begin{align*}y= \sqrt[3]{x}\end{align*}.
Solution: From the previous concept, we know that the +5 indicates a vertical shift of 5 units up. Therefore, this graph will look exactly the same as the parent graph, shifted up five units.
#### Example B
Graph f(x)=x+233\begin{align*}f(x)=- \sqrt[3]{x+2}-3\end{align*}. Find the domain and range.
Solution: From the previous example, we know that from the parent graph, this function is going to shift to the left two units and down three units. The negative sign will result in a reflection.
Alternate Method: If you want to use a table (like in the previous concept), that will also work. Here is a table, then plot the points. (h,k)\begin{align*}(h, k)\end{align*} should always be the middle point in your table.
x y
6 -5
-1 -4
-2 -3
-3 -2
-10 -1
#### Example C
Graph f(x)=12x43\begin{align*}f(x)= \frac{1}{2} \sqrt[3]{x-4}\end{align*}.
Solution: The -4 tells us that, from the parent graph, the function will shift to the right four units. The 12\begin{align*}\frac{1}{2}\end{align*} effects how quickly the function will “grow”. Because it is less than one, it will grow slower than the parent graph.
Using the graphing calculator: If you wanted to graph this function using the TI-83 or 84, press Y=\begin{align*}Y=\end{align*} and clear out any functions. Then, press (1÷2)\begin{align*}(1 \div 2)\end{align*}, MATH and scroll down to 4: 3\begin{align*}\sqrt[3]{\;\;}\end{align*} and press ENTER. Then, type in the rest of the function, so that Y=(12)3(X4)\begin{align*}Y= \left(\frac{1}{2}\right) \sqrt[3]{\;\;}(X-4)\end{align*}. Press GRAPH and adjust the window.
Important Note: The domain and range of all cubed root functions are both all real numbers.
Intro Problem Revisit If you graph the function y=(x+1)3\begin{align*}y = -\sqrt[3]{(x + 1)}\end{align*}, you see that the domain is all real numbers, which makes all quadrants possible. However, for all positive values of x, y is negative because of the negative sign in front of the cube root. That rules out the first quadrant. Therefore, Dako is correct.
### Guided Practice
1. Evaluate y=x+4311\begin{align*}y= \sqrt[3]{x+4}-11\end{align*} when x=12\begin{align*}x=-12\end{align*}.
2. Describe how to obtain the graph of y=x+4311\begin{align*}y= \sqrt[3]{x+4}-11\end{align*} from y=x3\begin{align*}y= \sqrt[3]{x}\end{align*}.
Graph the following cubed root functions. Check your graphs on the graphing calculator.
3. y=x234\begin{align*}y= \sqrt[3]{x-2}-4\end{align*}
4. f(x)=3x1\begin{align*}f(x)=-3 \sqrt{x}-1\end{align*}
1. Plug in x=12\begin{align*}x=-12\end{align*} and solve for y\begin{align*}y\end{align*}.
y=12+4311=83+4=2+4=2\begin{align*}y= \sqrt[3]{-12+4}-11= \sqrt[3]{-8}+4=-2+4=2\end{align*}
2. Starting with y=x3\begin{align*}y= \sqrt[3]{x}\end{align*}, you would obtain y=x+4311\begin{align*}y= \sqrt[3]{x+4}-11\end{align*} by shifting the function to the left four units and down 11 units.
3. This function is a horizontal shift to the right two units and down four units.
4. This function is a reflection of y=x3\begin{align*}y= \sqrt[3]{x}\end{align*} and stretched to be three times as large. Lastly, it is shifted down one unit.
### Explore More
Evaluate f(x)=2x13\begin{align*}f(x)=\sqrt[3]{2x-1}\end{align*} for the following values of x.
1. f(14)\begin{align*}f(14)\end{align*}
2. f(62)\begin{align*}f(-62)\end{align*}
3. f(20)\begin{align*}f(20)\end{align*}
1. y=x3+4\begin{align*}y=\sqrt[3]{x}+4\end{align*}
2. y=x33\begin{align*}y=\sqrt[3]{x-3}\end{align*}
3. f(x)=x+231\begin{align*}f(x)=\sqrt[3]{x+2}-1\end{align*}
4. g(x)=x36\begin{align*}g(x)=- \sqrt[3]{x}-6\end{align*}
5. f(x)=2x+13\begin{align*}f(x)=2 \sqrt[3]{x+1}\end{align*}
6. h(x)=3x3+5\begin{align*}h(x)=-3 \sqrt[3]{x}+5\end{align*}
7. y=121x3\begin{align*}y=\frac{1}{2} \sqrt[3]{1-x}\end{align*}
8. y=2x+433\begin{align*}y=2 \sqrt[3]{x+4}-3\end{align*}
9. y=13x53+2\begin{align*}y=- \frac{1}{3} \sqrt[3]{x-5}+2\end{align*}
10. g(x)=6x3+7\begin{align*}g(x)=\sqrt[3]{6-x}+7\end{align*}
11. f(x)=5x13+3\begin{align*}f(x)=-5 \sqrt[3]{x-1}+3\end{align*}
12. y=47x38\begin{align*}y=4 \sqrt[3]{7-x}-8\end{align*}
### Vocabulary Language: English
General Equation for a Cubed Root Function
General Equation for a Cubed Root Function
The general equation for a cubed root function is $f(x)=a \sqrt[3]{x-h}+k$, where $h$ is the horizontal shift and $k$ is the vertical shift.
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###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
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# SSS and SAS - Problem 1
Brian McCall
###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
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One way of finding out if two triangles are congruent is if all three sides of both triangles are congruent. You can tell if sides of a triangle are congruent if they have the same number of dashes running through them. If for each side of a triangle, there is a corresponding side that is congruent in the other triangle, the two triangles are congruent.
It is important to label the triangles in such a way that they are congruent. For example, triangle ABC being congruent to triangle DEF implies that AB is congruent to DE, BC is congruent to EF, and CA is congruent to FD.
Let’s look at these two triangles first we’re going to ask can we say that two triangles are congruent and then we’ll say well which short cut are we using. Well we see AC corresponds to EF, we see that AB corresponds to DE and we see that CB corresponds to DF.
So we know three corresponding sides are congruent, so I’m going to say that triangle ABC is congruent to this triangle DEF. But what’s the order that I’m going write my vertices at? It has to be very specific. Angle A has top correspond to a certain angle in this triangle.
Let’s start by looking at side AB. Side AB has one mark which means that it must be congruent top DE, so I’m going to write DE first, which means that AB corresponds to DE. Well that makes it pretty easy because there’s only one other vertex and that’s F. So triangle ABC is congruent to triangle DEF and our short-cut is side-side-side.
So you notice what we did, by only knowing the three sides of this two triangles we are able to say that everything about them must be congruent.
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We think you are located in United States. Is this correct?
# 2.3 Rational exponents
## 2.3 Rational exponents (EMAV)
We can also apply the exponent laws to expressions with rational exponents.
According to CAPS, the rational exponent law is introduced in Grade 11 but you may choose to introduce learners to the rational exponent law $$a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$$ at this stage.
## Worked example 6: Simplifying rational exponents
Simplify:
$2{x}^{\frac{1}{2}}\times 4{x}^{-\frac{1}{2}}$
\begin{align*} 2{x}^{\frac{1}{2}} \times 4{x}^{-\frac{1}{2}} & = 8{x}^{\frac{1}{2} - \frac{1}{2}}\\ & = 8{x}^{0} \\ & = 8\left(1\right) \\ & = 8 \end{align*}
## Worked example 7: Simplifying rational exponents
Simplify:
${\left(\text{0,008}\right)}^{\frac{1}{3}}$
### Write as a fraction and simplify
\begin{align*} {\left(\text{0,008}\right)}^{\frac{1}{3}} & = {\left(\frac{8}{\text{1 000}}\right)}^{\frac{1}{3}} \\ & = {\left(\frac{1}{125}\right)}^{\frac{1}{3}} \\ & = {\left(\frac{1}{5^{3}}\right)}^{\frac{1}{3}} \\ & = \frac{{1}^{\frac{1}{3}}}{5^{\left(3 \cdot \frac{1}{3}\right)}} \\ & = \frac{1}{5} \end{align*}
temp text
Extension: the following video provides a summary of all the exponent rules and rational exponents.
Video: 2F2V
Textbook Exercise 2.2
Simplify without using a calculator:
$$t^{\frac{1}{4}} \times 3t^{\frac{7}{4}}$$
\begin{align*} t^{\frac{1}{4}} \times 3t^{\frac{7}{4}} & = 3t^{\frac{1}{4} + \frac{7}{4}} \\ & = 3t^{\frac{8}{4}} \\ & = 3t^{2} \end{align*}
$$\dfrac{16x^{2}}{\left(4x^{2}\right)^{\frac{1}{2}}}$$
\begin{align*} \frac{16x^{2}}{\left(4x^{2}\right)^{\frac{1}{2}}} & = \frac{4^{2}x^{2}}{4^{\frac{1}{2}}x^{(2)\left(\frac{1}{2}\right)}} \\ & = \frac{4^{2}x^{2}}{4^{\frac{1}{2}}x} \\ & = 4^{2 - \frac{1}{2}} \cdot x^{2 - 1} \\ & = \left(2^{2}\right)^{\frac{3}{2}}x \\ & = 2^{3}x \\ & = 8x \end{align*}
$$\left(\text{0,25}\right)^{\frac{1}{2}}$$
\begin{align*} \left(\text{0,25}\right)^{\frac{1}{2}} & = \left(\dfrac{1}{4}\right)^{\frac{1}{2}} \\ & = \left(\dfrac{1}{2^{2}}\right)^{\frac{1}{2}} \\ & = \left(2^{-2}\right)^{\frac{1}{2}} \\ & = 2^{-1} \\ & = \dfrac{1}{2} \end{align*}
$$\left(27\right)^{-\frac{1}{3}}$$
\begin{align*} \left(27\right)^{-\frac{1}{3}} & = \left(3^{3}\right)^{-\frac{1}{3}} \\ & = 3^{-1} \\ & = \dfrac{1}{3} \end{align*}
$$\left(3p^{2}\right)^{\frac{1}{2}} \times \left(3p^{4}\right)^{\frac{1}{2}}$$
\begin{align*} \left(3p^{2}\right)^{\frac{1}{2}} \times \left(3p^{4}\right)^{\frac{1}{2}} & = 3^{\frac{1}{2}}p \times 3^{\frac{1}{2}}p^{2} \\ & = 3^{\frac{1}{2} + \frac{1}{2}} \times p^{1 + 2} \\ & = 3p^{3} \end{align*}
$$\text{12} {\left( a^\text{4}b^\text{8} \right)}^ {\frac{\text{1}}{\text{2}}} \times {\left( \text{512}a^\text{3}b^\text{3} \right)}^ {\frac{\text{1}}{\text{3}}}$$
\begin{align*} \text{12} {\left( a^\text{4}b^\text{8} \right)}^ {\frac{\text{1}}{\text{2}}} \times {\left( \text{512}a^\text{3}b^\text{3} \right)}^ {\frac{\text{1}}{\text{3}}} &= \text{12} a^{\frac{\text{4}}{\text{2}}}b^ {\frac{\text{8}}{\text{2}}} \times (\text{512})^{\frac{\text{1}}{\text{3}}}a^{\frac{\text{3}}{\text{3}}}b^{\frac{\text{3}}{\text{3}}} \\ &= \text{12} a^{\text{2}}b^{\text{4}} \times \left( \text{8}^{\text{3}} \right) ^{\frac{\text{1}}{\text{3}}}a^{\text{1}}b^{\text{1}} \\ &= \text{12} a^{\text{2}}b^{\text{4}} \times \text{8}a^{\text{1}}b^{\text{1}} \\ &= \text{96} a^{\text{3}} b^{\text{5}} \end{align*}
$$\left((-2)^4a^6b^2\right)^{\frac{1}{2}}$$
\begin{align*} \left((-2)^4a^6b^2\right)^{\frac{1}{2}} & = (-2)^2(a^3b) \\ & = 4a^3b \end{align*}
$$\left(a^{-2}b^6\right)^{\frac{1}{2}}$$
\begin{align*} \left(a^{-2}b^6\right)^{\frac{1}{2}} & = a^{-1}b^3 \\ & = \frac{b^{3}}{a} \end{align*}
$$\left(16x^{12}b^6\right)^{\frac{1}{3}}$$
\begin{align*} \left(16x^{12}b^6\right)^{\frac{1}{3}} & = \left((8 \times 2) x^{12}b^{6}\right)^{\frac{1}{3}} \\ & = 2\cdot 2^{\frac{1}{3}}a^{4}b^{2} \end{align*}
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Type a math problem
Solve for x_1, x_2, x_3
Short Steps Using Substitution
Solve for .
Substitute for in the second and third equation.
Solve these equations for and respectively.
Substitute for in the equation .
Solve for .
Substitute for in the equation .
Calculate from .
Substitute for and for in the equation .
Calculate from .
The system is now solved.
Giving is as easy as 1, 2, 3
Get 1,000 points to donate to a school of your choice when you join Give With Bing
x_{1}=-2x_{2}+x_{3}-3x_{4}
Solve x_{1}+2x_{2}-x_{3}+3x_{4}=0 for x_{1}.
2\left(-2x_{2}+x_{3}-3x_{4}\right)+3x_{2}-x_{3}+2x_{4}=0 -2x_{2}+x_{3}-3x_{4}+3x_{3}+3x_{4}=0
Substitute -2x_{2}+x_{3}-3x_{4} for x_{1} in the second and third equation.
x_{2}=x_{3}-4x_{4} x_{3}=\frac{1}{2}x_{2}
Solve these equations for x_{2} and x_{3} respectively.
x_{3}=\frac{1}{2}\left(x_{3}-4x_{4}\right)
Substitute x_{3}-4x_{4} for x_{2} in the equation x_{3}=\frac{1}{2}x_{2}.
x_{3}=-4x_{4}
Solve x_{3}=\frac{1}{2}\left(x_{3}-4x_{4}\right) for x_{3}.
x_{2}=-4x_{4}-4x_{4}
Substitute -4x_{4} for x_{3} in the equation x_{2}=x_{3}-4x_{4}.
x_{2}=-8x_{4}
Calculate x_{2} from x_{2}=-4x_{4}-4x_{4}.
x_{1}=-2\left(-8\right)x_{4}-4x_{4}-3x_{4}
Substitute -8x_{4} for x_{2} and -4x_{4} for x_{3} in the equation x_{1}=-2x_{2}+x_{3}-3x_{4}.
x_{1}=9x_{4}
Calculate x_{1} from x_{1}=-2\left(-8\right)x_{4}-4x_{4}-3x_{4}.
x_{1}=9x_{4} x_{2}=-8x_{4} x_{3}=-4x_{4}
The system is now solved.
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# Julie needs to cut 4 pieces of yarn, each with the same length, and a piece of yarn 7.75 inches long. let x represent the length of each of the equal pieces of yarn that julie decides to cut. what is the
julie needs to cut 4 pieces of yarn, each with the same length, and a piece of yarn 7.75 inches long. let x represent the length of each of the equal pieces of yarn that julie decides to cut. what is the equation that can be used to determine the total length of all of the yarn that she ends up cutting, y? is the graph of the equation continuous or discrete?
Julie needs to cut 4 pieces of yarn, each with the same length, and a piece of yarn 7.75 inches long. Let ( x ) represent the length of each of the equal pieces of yarn that Julie decides to cut. What is the equation that can be used to determine the total length of all of the yarn that she ends up cutting, ( y )? Is the graph of the equation continuous or discrete?
To determine the total length of all the yarn that Julie ends up cutting, we need to consider both the 4 pieces of yarn of equal length ( x ) and the additional piece of yarn that is 7.75 inches long.
1. Formulate the Equation:
The total length of the yarn, ( y ), can be expressed as the sum of the lengths of the 4 equal pieces and the 7.75-inch piece.
Therefore, the equation is:
y = 4x + 7.75
2. Determine if the Graph is Continuous or Discrete:
The graph of the equation ( y = 4x + 7.75 ) represents a linear relationship between ( y ) and ( x ).
• Continuous: If the lengths ( x ) can take any real number value (including fractions), then the graph of the equation is continuous. In real-world scenarios, yarn can be cut to any length, so ( x ) can be any non-negative real number.
• Discrete: If ( x ) can only take specific values (e.g., whole numbers or specific lengths), then the graph would be discrete. However, typically, yarn can be cut to any desired length, so we usually consider ( x ) to be continuous.
Given that yarn can typically be cut to any length, the graph of the equation ( y = 4x + 7.75 ) is continuous.
Therefore, the equation that can be used to determine the total length of all the yarn that Julie ends up cutting is:
y = 4x + 7.75
And the graph of this equation is continuous.
|
Year 9 Interactive Maths - Second Edition
Factors of a Number
If a number can be expressed as a product of two whole numbers, then the whole numbers are called factors of that number.
So, the factors of 6 are 1, 2, 3 and 6.
Example 1
Find all factors of 45.
Solution:
So, the factors of 45 are 1, 3, 5, 9, 15 and 45.
Common Factors
10 = 2 × 5 = 1 × 10
Thus, the factors of 10 are 1, 2, 5 and 10.
15 = 1 × 15 = 3 × 5
Thus, the factors of 15 are 1, 3, 5 and 15.
Clearly, 5 is a factor of both 10 and 15. It is said that 5 is a common factor of 10 and 15.
Example 2
Find a common factor of:
a. 6 and 8
b. 14 and 21
Prime Numbers
If a number has only two different factors, 1 and itself, then the number is said to be a prime number.
For example, 31 = 1 × 31
31 is a prime number since it has only two different factors.
Note:
But 1 is not a prime number since it does not have two different factors.
Example 3
Express 150 as a product of prime numbers, i.e. find its prime factor decomposition.
Solution:
Note:
We try the prime numbers in order of their magnitude.
We observe that:
The prime factor decomposition of a number is unique.
This is called the Fundamental Theorem of Arithmetic. It provides us with a good reason for defining prime numbers so as to exclude 1. If 1 were a prime, then the prime factor decomposition would lose its uniqueness. This is because we could multiply by 1 as many times as we like in the decomposition.
Highest Common Factor
The highest common factor (HCF) of two numbers (or expressions) is the largest number (or expression) that is a factor of both.
Consider the highest common factor of 16 and 24.
The common factors are 2, 4 and 8. So, the highest common factor is 8.
Note:
The highest common factor is the product of the common prime factors.
In general:
To find the highest common factor of two (or more) numbers, make prime factors of the numbers and identify the common prime factors. Then the highest common factor is the product of the common prime factors.
Example 4
Find the highest common factor of 60 and 150.
Solution:
The prime factorisation of 60 is:
The prime factorisation of 150 is:
Note:
The highest common factor can also be obtained by a trial and error method.
For example, the highest common factor of 40 and 45 is 5 because 5 is the largest number which divides into both 40 and 45 exactly.
Likewise, the highest common factor of 27 and 36 is 9 because 9 is the largest number which divides into both 27 and 36 exactly.
|
New Zealand
Level 6 - NCEA Level 1
# Decimal Ratios
Lesson
## Expressing ratios as decimals
Expressing ratios as decimals is a similar process to converting fractions to decimals. It can be useful to convert ratios to fractions when conceptualising these ratios.
#### Examples
##### Question 1
Question: Express $35:100$35:100 as a single decimal.
Think: $35:100$35:100 is the same as $\frac{35}{100}$35100 and since the second number in the ratio (which becomes the denominator) is already a multiple of $10$10, we do not need to change any of our numbers around.
Do:
$35:100$35:100 $=$= $\frac{35}{100}$35100 turn the ratio into a fraction $=$= $0.35$0.35 change the fraction into a decimal
Let's look at a question where we don't have a multiply of $10$10 as the denominator.
##### Question 2
Question: Express $23:4$23:4 as a single decimal.
Think: There are a few different ways to approach this question. My first step is going to be to change this ratio into a mixed number. Then I'll change it to a decimal.
Do:
$23:4$23:4 $=$= $\frac{23}{4}$234 convert the ratio to a fraction $=$= $5\frac{3}{4}$534 convert to mixed number $=$= $5.75$5.75 convert to decimal
If you need a refresher on how to change fractions to decimals, click here.
##### question 3
Express $31.8:3180$31.8:3180 as a single decimal.
## Finding unknown values
We've already looked at finding unknown whole values in ratios in Keeping it in Proportion. The same process applies whether we have whole numbers or decimal values.
#### Examples
##### question 4
Question: Find $b$b if $b:40=12$b:40=12
Think: This ratio means that $\frac{b}{40}=12$b40=12 , so to get b by itself, we need to multiply both sides by $40$40.
Do:
$b:40$b:40 $=$= $12$12 $\frac{b}{40}$b40 $=$= $12$12 change ratio into a fraction $b$b $=$= $12\times40$12×40 multiply both sides by $40$40 $b$b $=$= $480$480 evaluate
##### question 5
Find $a$a if $a:17=4.83$a:17=4.83
The same rule applies if there is more than one decimal.
##### Question 6
Find $a$a if $a:7.7=54.98$a:7.7=54.98
### Outcomes
#### NA6-1
Apply direct and inverse relationships with linear proportions
#### 91026
Apply numeric reasoning in solving problems
|
# LESSON PLANNING OF GEOMETRY BOX
## Students` Learning Outcomes
• Use protractor, set squares straightedge / ruler to construct square and rectangle with given side(s).
Information for Teachers
Material / Resources
Writing board, chalk / marker, duster, geometry box, textbook
Introduction
• Can anyone tell me that why do we use set-square?
• Why do we use protractor?
• Do you remember how to draw an angle with protractor as we did for triangles?
• Do you remember how to draw right angle with set-square?
• Take verbal feedback of students and then ask them to draw two right-angles one with protractor and other by using set-square. (Pair Activity)
• Ask pair to exchange their notebooks and check work.
• Then repeat the whole method once verbally.
Development
Activity 1
• Collect their attention by asking to close their notebooks, books and just look at you.
• You then say step 1 verbally and draw on board.
• Repeat the step verbally once again to ensure that they understood.
• Do the same for all the steps of drawing rectangle.
• Make sure you don`t write steps on board just perform it.
• Ask anyone of the students to repeat what you did?
• Then write a question on board and ask them to attempt twice once with rectangle and once with square.(pair work)
• Get checking done by exchanging copies. Make sure they are told to respect each other’s work and just put small ticks or crosses (X)
• Assign another question and ask them to attempt individually.
• Collect copies to be checked by you.
• Hence today we learnt about drawing rectangles. Let`s repeat it together.(then repeat with the students once again)
• Wrap-up the lesson and announce that ‘we will see who does his home work best’.
Activity 2
• Similar method will be used for square.
• Steps to draw square are mentioned abov
Sum up / Conclusion
• Recall process of drawing square and rectangle.
Assessment
• Repeat the key points so further strengthen student`s concepts.
Quadrilateral:
o A closed figure with four sides and sum of its all interior angles is 3600
Square:
o All sides are of the same length.
o Opposite sides are parallel.
o There are four right angles.
o Diagonals are equal
Rectangles:
o Opposite sides are of the same length and parallel.
o There are four right angles.
o Consecutive angles are supplementary.
o Diagonals are equal.
Follow up
• Encourage students to draw square and rectangles using different measurements.
• Assign homework.
|
# IMO Class 5 Chapter 2:Fractions and Decimals
#### BySchool Connect Online
Aug 20, 2022
Register now for IMO 2022-23
## IMO Class 5 Chapter 2:Fractions and Decimals detailed notes
Fractions are referred to as the components of a whole in mathematics. A single object or a collection of objects can be the whole. When we cut a piece of cake in real life from the entire cake, the portion represents the fraction of the cake. The word “fraction” is derived from Latin. “Fractus” means “broken” in Latin. The fraction was expressed verbally in earlier times. It was later presented in numerical form.
A portion or section of any quantity is another name for the fraction. It is indicated by the symbol “/,” such as a/b. For instance, in the fraction 2/4, the lower part represents the denominator and the upper part the numerator. The definition of fractions in mathematics, types of fractions, how to convert fractions to decimals, and numerous examples with fully explicated solutions are all covered in this article.
### What are Fractions?
Definition 1: A fraction is a unit of measurement that identifies the components of a whole.
Definition 2: A fraction is a number that symbolizes a portion of a whole.
In general, the whole can be any particular thing or value, and the fraction can be a portion of any quantity out of it.
The top and bottom numbers of a fraction are explained by the fundamentals of fractions. The bottom number represents the total number of parts, while the top number represents the number of chosen or shaded portions of the whole.
Fractions are essential to our daily lives. You will encounter fractions frequently in everyday life. Whether we want to or not, we have to share that delicious pizza with our loved ones. Four slices for three people. It will be more enjoyable and exciting if you can easily learn and visualize fractions. For instance, if you divide an apple into two pieces, each piece will represent a fraction (equal to 1/2).
### Parts of Fractions
• There are two components to fractions: a numerator and a denominator.
• The numerator, or upper portion of the fraction, denotes the various parts of the fraction.
• The lower or bottom portion, known as the denominator, represents all of the components into which the fraction is divided.
• For instance, if 3/4 is a fraction, the numerator is 3 and the denominator is 4.
### Properties of fractions
A fractional number has some of the same significant properties as real numbers and whole numbers. As follows:
• Fractional addition and multiplication are consistent with commutative and associative properties.
• Fractional addition’s identity element is 0 and fractional multiplication’s identity element is 1.
• B/a is the multiplicative inverse of a/b, where a and b are elements that should not be zero.
• The distributive property of multiplication over addition is observed in fractional numbers.
### What are Decimals?
One of the number types in algebra that has a whole number and a fractional portion separated by a decimal point is a decimal. The decimal point is the dot that appears between the parts of a whole number and a fraction.
An example of a decimal number is 34.5.
In this case, the whole number part is 34, and the fractional part is 5.
The decimal point is “.”.
Let’s talk about a few more instances.
The decimal representation of the number “thirty-four and seven-tenths” is given below:
Among Ones and Tenths is where the decimal point is located.
3 tens, 4 ones, and 7 tenths make up the number 34.7.
### Properties of Decimals
• Following are some crucial characteristics of decimal numbers when performing division and multiplication operations:
• No matter what order any two decimal numbers are multiplied, the result is always the same.
• Any order of adding a whole number and a decimal number produces the same result.
• A decimal fraction’s product when multiplied by one is that same decimal fraction.
• A decimal fraction multiplied by 0 produces a result of zero (0).
• The result of dividing a decimal number by one is the decimal number itself.
• When a decimal number is divided by itself, the result is 1, or 1.
• Any division of 0 by decimal results in a quotient of 0.
• Since there is no reciprocal of 0, it is impossible to divide a decimal number by 0.
### Importance of IMO
The Math Olympiad also referred to as the International Maths Olympiad (IMO), is a very helpful test for students who want to get better at math. It guarantees accurate calculations in the shortest amount of time. It presents challenging and difficult questions to the class. They gain the self-assurance needed to get over their fear of numbers. They become flawless in a variety of concepts, including trigonometry, algebra, measurements, theorems, and data handling, among others. One of the competitions that benefit students the most by uniting them on one platform is this one. It cultivates in young minds the skills of analytical thinking and problem-solving. Children’s strengths and weaknesses are revealed by practicing a variety of problems from Math Olympiads. It gives more insight into the areas that need more practice.
School Connect Online offers olympiads such as
Prepare for your Olympiad coaching with the School Connect Olympiad by interacting with one of the greatest educators from IIT, NIT, as well as other institutions!
To know more: https://blog.schoolconnectonline.com/
Explore more at: https://www.schoolconnectonline.com/
IMO class 3 chapter 2
IMO class 3 chapter 3
IMO Class 4 Chapter 1
IMO Class 5 Chapter 1
IMO Class 4 Chapter 2
#### By School Connect Online
School Connect Online is an Integrated Learning Program for Academic Institution,and supported and mentored by StartUp Oasis,an inititive of CIIE.CO Please visit school.schoolconnectonline.com
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# Today we will show in various ways why decimal and percent equivalents for common fractions can represent the same value. represent – stand for value -
## Presentation on theme: "Today we will show in various ways why decimal and percent equivalents for common fractions can represent the same value. represent – stand for value -"— Presentation transcript:
Today we will show in various ways why decimal and percent equivalents for common fractions can represent the same value. represent – stand for value - importance
Equivalent means to show the same number in a different way. Equivalent
A decimal is a number less than 1 Decimal
Percent means per hundred. Percent
A fraction is a number that shows a part of a whole or set. Fraction
We already know that this shape is ¾ shaded. Tell you partner the percent that is shaded. 75%
Percent, fractions, and decimals 50%.50 50 100 = = DecimalFraction Percent =
Knowing how to show in various ways why decimal and percent equivalents for common fractions can represent the same value You will know how to figure out the different ways a fraction can be represented. You will be prepared for tests like the CST. Are there other reasons why?
Let’s work out a few together How do you write this in fraction form? ½ How do you write ½ in decimal form? Remember that you divide the denominator with the numerator. 21.0 -1 0 00 0.5 How do you rewrite 0.5 as a percentage? Remember that all you do is move the decimal 2 places over to the right! 0 5. 0 50. %
Let’s work out a few together How do you write this in fraction form? 2/8 How do you write ½ in decimal form? Remember that you divide the denominator with the numerator. 8 2.0 0 -1 6 0 4 0 -0 4 0 0 0 0 0.25 How do you rewrite 0.25 as a percentage? Remember that all you do is move the decimal 2 places over to the right! 0 25..%
Let’s work out the steps together! 1.Look at figure, percent, decimal or fraction and follow the directions. 2.If it’s a figure, write the fraction, decimal and percentage 3.If it’s a fraction, write the percentage, decimal and draw the figure. 4.If it’s a decimal, write the percentage, fraction and draw the figure. 5.If it’s a percentage, write the fraction, decimal and draw the figure. Write the fraction, decimal and percentage for the following figure 1/8.012512.5%
Let’s work out the steps together! 1.Look at figure, percent, decimal or fraction and follow the directions. 2.If it’s a figure, write the fraction, decimal and percentage 3.If it’s a fraction, write the percentage, decimal and draw the figure. 4.If it’s a decimal, write the percentage, fraction and draw the figure. 5.If it’s a percentage, write the fraction, decimal and draw the figure. Write the decimal, percentage and draw the figure for the following fraction. 3939.3333%
Lets review What is percent? What is a decimal? What is a fraction?
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# CBSE Class 8 Maths Worksheet Chapter 11 Mensuration - PDF
Last updated date: 20th Sep 2024
Total views: 221.7k
Views today: 5.21k
Mensuration Class 8 Worksheets With Answers is a crucial resource for CBSE students. It aids in giving them a good understanding of the chapter, mainly the questions asked in the exam and how to respond to them. Vedantu offers all CBSE Class 8 Maths Solutions exercise-wise to assist students in test preparation and clear up any questions.
Mensuration is the area of Mathematics that deals with geometrical figures and their attributes, such as length, volume, shape, surface area, lateral surface area, etc. These CBSE Solutions for Class 8 Maths include all the critical mensuration formulas and the characteristics of various geometric figures and shapes. They provide clear explanations for the questions relating to these subjects.
## Benefits of Class 8 Maths Mensuration Worksheet
Mensuration Class 8 practice questions PDF will improve fundamental conceptual understanding. Our exercises contain complete solutions to all essential theorems, greatly simplifying the idea for learners.
The Class 8 Maths Mensuration answer is a comprehensive resource for teachers of students in Class 8. Through the use of these Maths worksheets, educators can gain knowledge and understanding about the development of students as well as methods and techniques for teaching them effectively.
## Examples of Class 8 Mensuration Questions
Here are some common examples of mensuration exercises:
Q. Find the circle's diameter, whose circumference is 230 meters.
Solution:
The circumference is 230 meters.
2r = 230m, r = 230, 72, 22, and 36.6m, respectively;
d = 2r = 2m*36.6m= 73.18cm.
Q. A floor tile is shaped like a parallelogram, with a base of 24 cm and a height of 10 cm. How many of these tiles are needed to cover a 1080 m2 floor? (If necessary, you can divide the tiles however you'd like to fill in the corners.)
Solution:
Given: The tile's base measures 24 cm or 0.24 m
The height of a flooring tile is 10 cm or 0.10 meters.
Now, floor tile area equals base plus altitude.
= 0.24×0.10
= 0.024
The tile's surface area is 0.024 m2.
The formula for how many tiles are needed to cover the floor is: Area of floor/Area of one tile (1080/0.024).
= 45000 tiles
Thus, 45000 tiles are needed to cover the floor completely.
Q. Daisy is painting the walls and ceiling of a cuboidal hall with length, breadth, and height of 15 m, 10 m, and 7 m, respectively. From each can of paint, 100 m2 of the area is painted. How many cans of paint are needed need to paint the room?
Solution:
The wall measures 15 meters long, 10 meters wide, and 7 meters high
Total Surface Area of the classroom = l x b + 2(bh + hl)
= 15 × 10 + 2(10×7 + 7×15)
= 150 + 2(70 + 105)
= 150 + 350
= 500
Consequently, the required number of cans = hall area / can size
= 500/100 = 5
Therefore, painting the room requires five cans.
Q. A hollow cylinder has a lateral surface area of 4224 cm2. It is divided along its height to create a 33 cm wide rectangular sheet. Find the rectangle sheet's perimeter.
Solution:
The hollow cylinder's lateral surface area is 4224 cm2.
33 cm is the width of the rectangular sheet, and let l be its length.
Cylinder's lateral surface area equals the size of a rectangular sheet
4224 = b × l
4224 = 33 × l
l = 4224/33 = 128 cm
The rectangular sheet is therefore 128 cm long.
Also, the rectangular sheet's perimeter is equal to 2(l + b).
= 2(128+33)
= 322 cm
The rectangular sheet has a 322 cm perimeter.
Q. Metal Sheet is used to create a closed cylindrical tank with a 7m radius and a 3m height. How much metal sheet is needed?
Solution:
Cylindrical tank's radius, r = 7 m
The cylindrical tank's height is 3 meters.
The total tank surface area is equal to 2r(h+r).
= 2×(22/7)×7(3+7)
= 44×10 = 440
A metal sheet of 440 m2 is needed.
## Key Features of Class 8 Mensuration PDF
You can easily find Mensuration Class 8 worksheets PDF with answers on the website of Vedantu. You can download the free Mensuration worksheet for Class 8 on a laptop or mobile with a stable internet connection. Designed by our experts to meet students' needs, these Maths worksheets offer the highest quality of education.
• Mensuration class 8 worksheets have been well-formulated following the latest CBSE guidelines.
• Provides various exercises with examples for mensuration sums, short, long questions, etc.
• Mensuration Class 8 practice questions PDF will improve fundamental conceptual understanding. Our exercises contain complete solutions to all essential theorems, greatly simplifying the idea for learners.
Numerous short answers, long answers, and HOTS (Higher Order Thinking Skills) problems significant for CBSE exams are covered in these Mensuration Class 8 Worksheets PDF with answers. Here are some critical measurement problems with easy solutions from CBSE Class 8. Our Vedantu’s top math experts have put together the Mensuration Class 8 Maths Important Questions PDF to aid students in understanding the fundamental ideas in mathematics.
## FAQs on CBSE Class 8 Maths Worksheet Chapter 11 Mensuration - PDF
1. What are the Class 8 Mensuration important concepts?
The essential coneptsin Class 8 Mensuration include:
The perimeter is measured in km, m, or cm. Area: The area of any solid geometric shape is the boundary closed surface. The measurement of the area is in km2, m2, or cm2. Surface Area: Any three-dimensional figure's total surface area is referred to as its surface area.
2. Who is the mensuration's founder?
The pioneer of mensuration is Archimedes.
3. What measurements can we do under mensuration?
Mensuration includes two types: 2D mensuration and 3D mensuration.
• Only two dimensions—length and width—make up the 2D figures. The two-dimensional figure is flat; it has neither height nor depth.
• Length, breadth, and height or depth are the three dimensions of a three-dimensional figure.
4. What are the perimeter formulas?
P = (L + W) x 2 is the perimeter formula for a rectangle, where P stands for perimeter, L for length, and W for width. You may quickly solve for the perimeter of a rectangular shape when its dimensions are known by simply entering the values of L and W into the formula.
5. What is Mensuration 3D?
Mensuration 3D contains the volume and surface area of the 3-D (dimension) figures. Examples are Cube, Cuboid, Cylinder, Cone, Sphere, Frustum, Pyramid, Prism, Tetrahedron, etc.
|
Worksheet Solution: Multiplications
# Multiplication Class 1 Worksheet Mental Maths
Question 1: Multiply each of the following:
Question 2: Fill in the blanks:
• 10 x 1 = 10
• 5 x 5 = 25
• 6 x 3 = 18
• 3 x 10 = 30
• 9 x 6 = 54
• 6 x 6 = 30
Question 3: Write the multiplication fact for the repeated addition:
• + 8 + 8 + 8 + 8 + 8 + 8 = 8 x 7
• 5 + 5 + 5 + 5 + 5 + 5 = 5 x 6
• 3 + 3 + 3 + 3 + 3 + 3 = 3 x 6
• 8 + 8 + 8 + 8 + 8 = 8 x 5
• 4 + 4 + 4 = 3 x 4
Question 4: Let us now count the dots and get the table of 9:
Question 5: Let us now count the dots and get the table of 10:
Question 6: Complete the multiplication tables from 2 to 5:
Question 7: Let us now count the dots and get the table of 6:
Question 8: Complete the following multiplication tables:
Question 9: Let us now count the dots and get the table of 7:
Question 10: Let us now count the dots and get the table of 8:
Question 11: Fill in the blanks:
• 5 x 6 = 6 x 5
• 6 x 1 = 6
• 1 x 1 = 1
• 4 x 1 = 4
• 15 x 0 = 0
• 5 x 6 = 6 x 5
• 7 x 0 = 0
Question 12: Find the product using tables:
• 5 x 4 = 20
• 6 x 5 = 30
• 5 x 5 = 25
• 8 x 10 = 80
• 8 x 2 = 16
• 9 x 3 = 27
Question 13: Fill in the blanks:
• 10 x 1 = 10
• 6 x 9 = 54
• 4 x 9 = 36
• 6 x 5 = 30
• 7 x 3 = 21
• 4 x 8 = 32
Question 14: Using the multiplication tables, find the products:
Question 15: Multiply each of the following:
Question 16: Multiply each of the following:
Question 17: Fill in the boxes:
Question 18: Estimate the following sums to the nearest hundreds:
(Also, compare the estimated sums with the actual sums.)
• 375 + 236 = 611
400 + 200 = 600
611 > 600
• 273 + 504 = 777
300 + 500 = 800
800 > 777
• 173 + 58 = 231
200 + 100 = 300
231 < 300
• 287 + 189 = 476
300 + 200 = 400
476 > 400
Question 19: Estimate the following to the nearest hundreds:
(Also, compare the estimated differences with the actual differences.)
• 879 - 705 = 174
900 - 700 = 200
174 < 200
• 613 - 156
613 - 156 = 457
600 - 200 = 400
457 > 400
• 875 - 261 = 614
900 - 300 = 600
614 > 600
• 352 - 256 = 96
400 - 300 = 100
96 < 100
Question 20: Estimate each of the following products by rounding off each number to the nearest tens:
• 41 x 16
40 x 20 = 800
• 37 x 22
40 x 20 = 400
Question 21: A matchbox contains 50 sticks. How many sticks are there in 9 such matchboxes?
1 matchbox contains = 50 sticks
Sticks in 9 matchboxes = 50 x 9 = 450
Question 22: There are 75 toffees in a box. How many toffees will there be in 5 such boxes?
Total number of toffees in a box = 75
Total number of toffees in 5 boxes = 75 x 5 = 375
Question 23: There are 30 days in a month. How many days are there in 6 months?
Total number of days in a month= 30
Total number of days 6 months = 30 x 6 = 180
Question 24: A car can travel 60 km in one hour. How far will it travel in 3 hours?
A car can travel in one hour = 60 km
A car travels in 3 hours = 60 x 3 = 180 km
Question 25: A candle factory produces 55 candles in a day. How many candles will this factory produce in 6 days?
A candle factory produces candles in a day = 55
A candle factory produces candles 6 days = 55 x 6 = 330
The document Multiplication Class 1 Worksheet Mental Maths is a part of the Class 1 Course Mental Maths.
All you need of Class 1 at this link: Class 1
## Mental Maths
29 videos|64 docs|19 tests
## FAQs on Multiplication Class 1 Worksheet Mental Maths
1. How do you multiply two numbers using the standard algorithm?
Ans. To multiply two numbers using the standard algorithm, you multiply the ones digit of the second number by each digit of the first number, then do the same for the tens digit, and so on. Finally, you add up all the partial products to get the final answer.
2. What is the significance of the commutative property in multiplication?
Ans. The commutative property in multiplication states that changing the order of the factors does not change the product. This property allows us to simplify calculations and easily switch the positions of numbers when multiplying.
3. How can multiplication be applied in real-life situations?
Ans. Multiplication is used in various real-life situations such as calculating total costs when buying multiple items, determining the area of a rectangle, or finding out how many hours are in a given number of days.
4. What are some common mistakes students make when multiplying numbers?
Ans. Some common mistakes students make when multiplying numbers include forgetting to carry over when multiplying two-digit numbers, mixing up the order of the factors, or incorrectly placing the decimal point in decimal multiplication.
5. Can you explain the concept of multiplying by multiples of 10?
Ans. When multiplying by multiples of 10, you can simply add the number of zeros in the multiple of 10 to the original number. For example, when multiplying 25 by 10, you add one zero to get 250.
## Mental Maths
29 videos|64 docs|19 tests
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Infinite Geometric Series
• Remember that an infinite geometric series converges only if |r| < 1. If you are given a series to evaluate, first check the value of r. If it does not satisfy this condition, the series does not have a sum.
• To convert a repeating decimal to a fraction, let r = 10-n, where n = number of digits in the repeating pattern of the decimal. Let a1 = the fraction which contains the pattern written once in the numerator and 10n in the denominator.
Infinite Geometric Series
Find the sum 1 − [4/5] + [16/25] − [64/125]
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [( − [4/5])/1] = − [4/5]
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [1/(1 − ( − [4/5] ))] =
• [1/(1 + [4/5])] = [1/([9/5])] = [5/9]
S = [5/9]
Find the sum [5/3] − [5/12] + [5/48] − [5/192]
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [( − [5/12])/([5/3])] = − [5/12] ÷[5/3] = − [5/12]*[3/5] = − [3/12] = − [1/4]
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [([5/3])/(1 − ( − [1/4] ))] =
• [([5/3])/(1 − ( − [1/4] ))] = [([5/3])/(1 + [1/4])] = [([5/3])/([5/4])]
• [([5/3])/([5/4])] = [5/3]*[4/5] = [4/3]
S = [4/3]
Find the sum 4.4 + 3.52 + 2.816 + 2.2528
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [3.52/4.4] = 0.8
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [4.4/(1 − ( 0.8 ))] =
• [4.4/(1 − ( 0.8 ))] = [4.4/0.2] = 22
S = 22
Find the sum 9.3 + 1.86 + 0.372 + 0.0744...
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [1.86/9.3] = 0.2
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [9.3/(1 − ( 0.2 ))] =
• [9.3/(1 − ( 0.2 ))] = [9.3/0.8] = 11.625
S = 11.625
Find the sum 5 + [5/2] + [5/4] + [5/8]...
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [([5/2])/5] = [5/2] ÷[5/1] = [5/2]*[1/5] = [1/2]
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [5/(1 − ( [1/2] ))] =
• [5/(1 − ( [1/2] ))] = [5/([1/2])] = 10
S = 10
Find the sum [1/2] + [1/10] + [1/50] + [1/250]
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [([1/10])/([1/2])] = [1/10] ÷[1/2] = [1/10]*[2/1] = [1/5]
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [([1/2])/(1 − ( [1/5] ))] =
• [([1/2])/(1 − ( [1/5] ))] = [([1/2])/([4/5])] =
• [([1/2])/([4/5])] = [1/2]*[5/4] = [5/8]
S = [5/8]
Find the sum 6 − 2 + [2/3] − [2/9]
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [( − 2)/6] = − [1/3]
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [6/(1 − ( − [1/3] ))] =
• [6/(1 − ( − [1/3] ))] = [6/([4/3])] =
• [6/([4/3])] = 6*[3/4] = [18/4] = [9/2]
S = [9/2]
Find the sum − 2 − [2/5] − [2/25] − [2/125]...
• Step 1:Find the common ratio r, if r is less than 1, then you may find the sum using the formula S = [(a1)/(1 − r)]
• r = [(a2)/(a1)] = [( − [2/5])/( − 2)] = − [2/5]* − [1/2] = [1/5]
• Step 2) Since r is < 1, you may find the sum using the formula
• S = [(a1)/(1 − r)]
• S = [(a1)/(1 − r)] = [( − 2)/(1 − ( [1/5] ))] =
• [( − 2)/(1 − ( [1/5] ))] = [( − 2)/([4/5])] =
• [( − 2)/([4/5])] = − 2*[5/4] = − [5/2]
S = − [5/2]
Write as a fraction 0.25252525....
• Step 1 - Write the decimal as a geometric series
• 0.25252525... = 0.25 + 0.0025 + 0.000025 + 0.000025...
• Step 2 - Find the common ratio r
• r = [(a2)/(a1)] = [0.0025/0.25] = 0.01
• Step 3 - Use the formula Sn = [(a1)/(1 − r)]
• Sn = [(a1)/(1 − r)] = [0.25/(1 − 0.01)] = [0.25/0.99]
• Move the decimal two times over
• [0.25/0.99] = [25/99]
0.25252525... = [25/99]
Write as a fraction 0.15151515....
• Step 1 - Write the decimal as a geometric series
• 0.25252525... = 0.15 + 0.0015 + 0.000015 + 0.000015...
• Step 2 - Find the common ratio r
• r = [(a2)/(a1)] = [0.0015/0.15] = 0.01
• Step 3 - Use the formula Sn = [(a1)/(1 − r)]
• Sn = [(a1)/(1 − r)] = [0.15/(1 − 0.01)] = [0.15/0.99]
• Move the decimal two times over
• [0.15/0.99] = [15/99]
0.15151515... = [15/99] = [5/33]
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Infinite Geometric Series
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• What are Infinite Geometric Series 0:10
• Example: Finite
• Example: Infinite
• Partial Sums
• Formula
• Sum of an Infinite Geometric Series 2:39
• Convergent Series
• Example: Sum of Convergent Series
• Sigma Notation 7:31
• Example: Sigma
• Repeating Decimals 8:42
• Example: Repeating Decimal
• Example 1: Sum of Infinite Geometric Series 12:15
• Example 2: Repeating Decimal 13:24
• Example 3: Sum of Infinite Geometric Series 15:14
• Example 4: Repeating Decimal 16:48
Transcription: Infinite Geometric Series
Welcome to Educator.com.0000
Today, we continue on our discussion of sequences and series with infinite geometric series.0002
What are infinite geometric series? Well, this is a type of series in which there is an infinite number of terms.0010
Earlier on, I mentioned that, for either arithmetic or geometric series, you may have a limited number of terms,0018
which is a finite series; or it may go on indefinitely, which is an infinite series.0024
For example, the geometric series 1 + 1/4 + 1/16 + 1/64 is a geometric series with 4 terms; and it has a common ratio r = 1/4.0030
This is a finite series: it has a limited number of terms.0047
Consider another geometric series: 3 + 6 + 12 + 24, and then, when you see the ellipses (the three dots),0052
it indicates that it goes on indefinitely; so this is an infinite series.0063
The sums, sn, are called partial sums of the infinite series.0070
For example, I may decide that I want to find the first 7 terms of this series, or the first 5 terms of an infinite geometric series.0074
That would just be a partial sum of this series.0084
And we will talk, in a minute, about special cases, when you actually can find the sum overall of an infinite geometric series.0087
So, recall from the previous lesson the formula for the sum of a geometric series,0097
sn = the first term, times (1 - rn), divided by (1 - r).0103
So, if I were to try to find the first seven terms of this series, s7, I could use this formula.0114
I have the first term; I need to find r; recall that I can find r by taking a term and dividing it by the one that goes just before.0120
So here, r = 2; so this is 3(1 - 2n), so that is 7; and then, divided by 1 - 2.0130
So here, I found the partial sum for this infinite geometric series.0149
You can actually find the sum of an infinite geometric series (not just the partial sum) in some cases.0160
And I know that this sounds counterintuitive--how can you find the sum of something that goes on forever?0166
But if you look, you can see why.0172
All right, first of all, it is very important to know that this is limited to cases in which the series is convergent.0178
And the word "convergent" indicates that the sum converges toward a particular number.0184
A series is convergent if and only if the absolute value of r is less than 1.0191
So, if you are working with a geometric series in which the value of the common ratio is either greater than -1 or less than 1,0196
such as 3/4, for example--if r is 3/4 (the absolute value of that is 3/4), or if r is -1/2 (I take the absolute value of that--it would be 1/2),0205
both of these are convergent; I could actually find the sum of those.0219
Consider the series 1/3 + 1/9 + 1/27, going on indefinitely.0222
Remember, to find r (let's go up here and find r), we are going to take 1/9, divided by 1/3.0234
This is the same as 1/9 times 3, which equals 3/9, which equals 1/3; therefore, r = 1/3.0241
If I am asked to find the sum of this, I go ahead and use this formula, x equals the first term, which is 1/3, divided by 1 - 1/3, equals (1/3)/(2/3).0258
1/3 divided by 2/3 is the same as 1/3 times 3/2, or 3/6.0273
The sum of this infinite geometric series is 1/2.0283
Let's look at this another way: just go ahead and add up some of the terms and see what happens.0288
s1 for this term is 1/3; that is all you have: 1/3.0297
So, s2 would be adding 1/3 + 1/9; 1/3 + 1/9 equals...giving this a common denominator,0304
I multiply both the numerator and the denominator by 3 to get 3/9 + 1/9 is 4/9; that is s2.0313
s3 = 1/3 + 1/9 + 1/27; well, I know that these two are equal to 4/9, so that is 4/9 + 1/27.0322
So again, I need to get a common denominator; and if you work that out, you will find that s3 is 13/27.0333
s4...I would continue on: 1/3 + 1/9 + 1/27 + 1/81...and if you figure that out, it becomes 40/81.0343
I won't work out the rest of these right here; but I will just tell you that s5 is 121/243.0360
s6 is 364/729; and s7 is 1093/2187.0370
Let's look at the pattern here: it started out as 1/3; then it became 4/9, 13/27, 40/81, 121/243, 364/729, 1093/2187.0383
What is happening is: this sum is converging upon 1/2.0400
Another way that we say this is that the limit is 1/2; and that is terminology you will hear later on in higher math courses.0414
But for right now, just be aware that you can only find the sum of an infinite geometric series if it is convergent,0422
meaning that, as you take more and more terms and add them to the series, add them, and get their sum,0428
you will see that the sum of the series is converging upon a particular number.0435
And so, we just use this formula as a great shortcut to find the sum.0440
Sigma notation is something we discussed earlier on.0452
Go back and look at the lectures on geometric sequences and geometric series, if any of these concepts are new.0456
But sigma notation--you will recall that the Greek letter sigma means sum, and we used it with other geometric series.0462
For an infinite series, you are going to have something like this.0470
We have our lower index, i = 1; and this is the series as i goes from 1 to infinity.0476
The difference here is that, instead of stopping at a specific value up here, it goes on through infinity.0484
And again, we have the formula for the sequence right here.0490
Another example, or a specific example, would be a series, again, where i goes from 1 to infinity,0494
but we have a formula over here, 1/4 times 1/2, raised to the n - 1 power.0501
So, what you could do, then, is put 1 in here and find your first term, a1.0507
Then, put 2 here; find your second term; and go on infinitely, because of the type of series that this is.0513
We can use the concepts that we just learned to actually write a repeating decimal as a fraction.0523
The sum formula I just described can be used in this way.0529
A repeating decimal would be something like this: .44444...and it just goes on and on that way.0533
What we do is rewrite this as a geometric series.0542
1: First step--how do you do that?0547
Well, look at what this really means: it really means 0.4 + 0.04 + 0.004 + 0.0004, and so on.0556
So, I rewrote this, but just as a series; and it is an infinite geometric series.0573
Then, find the sum of the series: recall that you can only find the sum of an infinite geometric series0580
if the absolute value of r is less than 1; this is required, or you can't find the sum.0590
Well, let's show here that we are OK, because the absolute value of r is actually less than 1.0604
So, in order to find the common ratio, r, I am going to take .04, and I am going to divide it by the term before it, which is .4.0609
So, I move the decimal over one place; that is going to give me .4/4, which is .1; so the common ratio, r, equals .1.0618
.1 is less than 1, so I am fine: I can use the sum formula, the first term divided by (1 - r).0626
So, the sum equals the first term, divided by (1 - .1), equals .4/.9.0635
Dividing that, move the decimal over; you can get rid of that decimal; that gives me 4/9.0645
So, I found that the sum of this...0652
I started out; I have this decimal that is a repeating decimal (it goes on forever); but I saw that I could rewrite this as a series.0657
So, this, therefore, is equivalent to the sum of the series.0665
My next step was to figure out what the sum of this series is, and it is actually 4/9.0670
Therefore, .444 repeating can be rewritten as 4/9; those are equivalent.0674
Recall that you can have a repeating decimal that doesn't just have one number repeat and repeat; it could be multiple numbers.0683
It could be .383838 repeating, and you could do the same thing.0690
You could rewrite this as .38 + .0038 + .000038, and so on; and then repeat as above, by finding the sum of the series.0698
So, this could also be used for repeating decimals where there is a longer repeat.0717
Also recall that we can write these as .4 with a bar over it, or .38 with a bar over it--that is just a different notation.0722
All right, let's find the sum of this infinite geometric series.0736
First, though, I am going to verify that I actually can find the sum of this by figuring out what r is.0741
So, I am going to take 24, the common ratio, divided by 32; one term divided by the previous one gives me the common ratio.0746
This simplifies out to 3/4: 3/4 is less than 1, so yes, I can find the sum, using this formula.0753
The formula: I need to use the first term: 32/(1 - 3/4) = 32/(1/4).0769
We can rewrite this as 32 times 4; and 32 times 4...4 times 2 is 8; 4 times 3 is 120; so that gives me 128.0778
The sum of this infinite geometric series is 128, and I was able to find that using this formula, because I had an absolute value of r that is less than 1.0791
Write as a fraction: recall that this notation means the same thing as .36363636, and so on.0806
The first step is to write this as a geometric series, so I am going to rewrite this as .36 + .0036 + .000036, and so on.0821
And I am going to use my formula for the sum of an infinite geometric series, which is the first term, divided by (1 - r).0844
What is r? Well, as usual, I can find the common ratio, r, by taking a term, .0036, and dividing by the previous term, .36.0853
Move the decimal over two places to get .36/36; and that is going to give me .01.0864
.01 is less than 1, so I can find the sum.0871
Take the first term; divide it by (1 - .01); this gives me .36/.99.0875
I can move the decimal over two places, to give me 36/99; simplify that, because there is a common factor of 3; this is actually 12/33.0885
I can see, again, that I have another common factor of 3; so this is going to give me 4/11.0896
Therefore, this 0.36 repeating decimal can be rewritten as 4/11; so I wrote this repeating decimal as a fraction.0901
Find the sum for this infinite geometric series: before I proceed, I check that the absolute value of the common ratio is actually less than 1.0918
10/12...to find the common ratio, divide that by the term that came just before, which is 5/4.0927
And this is going to give me 10/12 times 4/5 equals 40/60; so this is going to give me 2/3.0933
And since 2/3 is less than 1, I can find the sum of this series using my formula for the sum of an infinite geometric series.0949
OK, the first term is 5/4; I am dividing that by 1 - 2/3 to get 5/4 divided by 1/3.0961
This is the same as 5/4; and then I just take the inverse of 1/3, and 5/4 times the inverse of 1/3; that is 3, so this is going to give me 15/4.0977
And you can keep this as an improper fraction, or write it as a mixed number.0989
The sum of this infinite geometric series is 15/4.0993
Again, I was able to find that because the common ratio had an absolute value that was less than 1.0999
Write as a fraction: we have another decimal that is repeating--goes on infinitely: 0.99999, and so on.1009
Start out by rewriting this as a geometric series.1021
I have 0.9 + 0.09 + 0.009, and so on; it goes on infinitely.1027
Find the common ratio, r: r =...I am going to take .09, and divide that by .9.1041
Move the decimal over one place to get .9/9; therefore, the common ratio is .1.1048
Next, I use my formula for the sum of an infinite geometric series, which is the first term, divided by 1 - r.1055
And this is going to give me the first term, .9, divided by 1 - .1; that is .9/.9; this is simply 1.1063
So, this asked me to write it as a fraction; and .999 = 1, or you could say 1/1.1081
This is still a fraction, because you could just write it as 1/1.1090
And it is counterintuitive, if you think ".999 repeating is actually 1"; it doesn't seem like it is, but it is actually correct.1093
That concludes this lesson of Educator.com on infinite geometric series; thanks for visiting!1105
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# Jimmy walked one 3/4 miles to school if you took a break every 1/8 of a mile how many breaks did he take?
Nov 17, 2016
He took 5 breaks
#### Explanation:
If a break was taken every $\frac{1}{8}$ of a mile the question is really asking: how many $\frac{1}{8}$ can you fit in the distance walked. There is a trap in this and I did fall into it.
$\implies \frac{3}{4} \div \frac{1}{8} \text{ "->" } \frac{3}{4} \times \frac{8}{1}$
$\frac{3}{{\cancel{4}}^{1}} \times \frac{{\cancel{8}}^{2}}{1} = 3 \times 2 = 6$
As Ez as pi correctly pointed out. The end point is not a break but a stop so is has to be removed from the count.
$\frac{3}{4} = \frac{6}{8}$ so lets look at this as a distance line diagram
This shows that if you can draw one a quick sketch pays off in the end.
The count of breaks is $\left(\frac{3}{4} \div \frac{1}{8}\right) - 1 = 6 - 1 = 5$
Nov 17, 2016
$5$ breaks on a $\frac{3}{4}$ mile walk.
$13$ breaks on a $1 \frac{3}{4}$ mile walk.
#### Explanation:
The question is not clear -whether Jimmy walked $1 \frac{3}{4}$ miles or whether it was just $\frac{3}{4}$ of a mile.
Let's consider both...
He rests every $\frac{1}{8}$ of a mile. Each quarter has two eighths in it, so it would seem that:
$\frac{3}{4} \div \frac{1}{8} = 6$
However, after the last $\frac{1}{8}$, he is at school, so he will not take a break.
He therefore he takes 5 breaks on the way to school.
The same applies if the distance is $1 \frac{3}{4}$ miles. In the first mile he rests 8 times, and then another 5 times in the next $\frac{3}{4}$ because then he arrives at school. He would rest 13 times.
$1 \frac{3}{4} \div \frac{1}{8} = \frac{7}{4} \div \frac{1}{8}$
=$\frac{7}{4} \times \frac{8}{1} = 14$ times. But after the the last part he is as school.
$14 - 1 = 13$ breaks
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In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Pdf, These solutions are solved subject experts from the latest edition books.
## MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2
Question 1.
Evaluate the following:
(i) sin60° cos30° + sin 30° cos 60°
(ii) 2tan2 45° + cos230° – sin260°
Solution:
(i) We have
= $$\frac{\frac{1}{12}\times67}{\frac{4}{4}}=\frac{67}{12}$$
Question 2.
Choose the correct option and justify your choice:
(i) $$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$$
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
(ii) $$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$$
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0
(iii) sin 2A = 2sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°
(iv) $$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$$
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution:
Question 3.
If tan (A + B) = $$\sqrt{3}$$ and tan (A – B) = $$\frac{1}{\sqrt{3}}$$ 0°< A + B ≤ 90°; A > B, find A and B.
Solution:
We have,
tan 60° = $$\sqrt{3}$$, tan 30° = $$\frac{1}{\sqrt{3}}$$ ……………. (1)
Also, tan(A + B) = $$\sqrt{3}$$ and tan (A – B) = $$\frac{1}{\sqrt{3}}$$ …………… (2)
From (1) and (2), we get
A + B = 60° …………… (3)
and A – B = 30° ………….. (4)
On adding (3) and (4), we get
2A = 90° ⇒ A =45°
On subtracting (4) from (3), we get
2B = 30° ⇒ B = 15°
Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin6 increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False:
Let us take A = 30° and B = 60°, then
L.H.S = sin (30° + 60°) = sin 90° = 1
∴ L.H.S. ≠ R.H.S.
(ii) True:
Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.
(iii) False:
Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.
(iv) False:
Let us take θ = 30°
sin 30° = $$\frac{1}{2}$$ and cos 30° = $$\frac{\sqrt{3}}{2}$$
⇒ sin 30° ≠ cos 30°
(v) True:
We have, cot 0° = not defined
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# FREE Finastra Math Assessment Test Questions and Answers
0%
#### If interest is compounded every month, what is "n"?
Correct! Wrong!
Explanation
When interest is compounded every month, "n" represents the number of compounding periods per year. In this case, since interest is being compounded every month, there are 12 months in a year, so "n" would be equal to 12.
Correct! Wrong!
#### What is 7.25% expressed in decimal form?
Correct! Wrong!
Explanation
In decimal form, 7.25 percent equals 0.0725.
#### If interest is compounded every quarter, what is "n"?
Correct! Wrong!
Explanation
When interest is compounded every quarter, "n" represents the number of compounding periods per year. In this case, since there are 4 quarters in a year, "n" would be equal to 4.
#### Decimalize 150% to get its value.
Correct! Wrong!
Explanation
To convert 150% to a decimal, you need to divide it by 100.
150% / 100 = 1.5
So, the decimal value of 150% is 1.5. This means that if you have a value of 150% and you want to convert it to a decimal, you would multiply the value by 1.5.
#### You are paying \$990 for a television with 9% sales tax. How much will the TV cost in the end?
Correct! Wrong!
Explanation
If you are paying \$990 for a television with a 9% sales tax, you can calculate the total cost of the television including tax by adding the sales tax to the price of the television. To do this, you can use the following formula:
Total cost = Price of the television + Sales tax
where:
The price of the television is \$990
Sales tax is 9% of the price of the television, which is 0.09 x \$990 = \$89.10
Plugging these values into the formula, we get:
Total cost = \$990 + \$89.10
Total cost = \$1079.10
Therefore, the television will cost \$1079.10 in the end, including the 9% sales tax.
#### How much money would you have after four years if you invested \$300 at 6% compounded continuously?
Correct! Wrong!
Explanation
FV = PV x e^(r*t)
where:
FV = Future Value (the amount of money you will have after 4 years)
PV = Present Value (the amount of money you invest)
r = Annual Interest Rate (expressed as a decimal)
t = Time period (in this case, 4 years)
Using this formula and the given values, we can calculate the future value of an investment of \$300 at 6% compounded continuously after 4 years as:
FV = \$300 x e^(0.06 x 4)
FV = \$300 x e^(0.24)
FV = \$300 x 1.2712491
FV = \$381.37 (rounded to the nearest cent)
Therefore, if you invest \$300 at 6% compounded continuously, you will have approximately \$381.37 after four years.
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Question
# Divide 20 pens between Sheela and Sangeeta in the ratio 3:2.
Hint: First of all consider the pens that Sheela and Sangeeta have as 3x and 2x respectively. Add them and equate it to the total number of pens to get the value of x. Substitute the value of x in 3x and 2x to get the number of pens each one of them has.
Here, we have to divide 20 pens between Sheela and Sangeeta in the ratio 3:2.
Let us consider that Sheela has 3x pens. Let us also consider that Sangeeta has 2x pens.
Now we know that the total pens are 20. So, this means that,
(Number of pens that Sheela has) + (Number of pens that Sangeeta has) = 20
By substituting the number of pens that Sheela has as 3x and number of pens that Sangeeta has as 2x in the above equation, we get,
3x + 2x = 20
By simplifying the above equation, we get,
$\Rightarrow$ 5x = 20
By dividing 5 on both sides of the above equation, we get,
$\Rightarrow$ $\dfrac{5x}{5}=\dfrac{20}{5}$
Therefore, we get x = 4.
Now, we know that the number of pens owned by Sheela = 3x.
By substituting the value of x = 4, we get,
Number of pens owned by Sheela = 3 x 4 = 12 pens.
Now, we also know that the number of pens owned by Sangeeta = 2x.
By substituting the value of x = 4, we get,
Number of pens owned by Sangeeta = 2 x 4 = 8 pens.
Hence, Sheela has 12 pens and Sangeeta has 8 pens.
Note: Here, we can also find the number of pens owned by Sheela and Sangeeta directly by using the formula. If Sheela’s pens and Sangeeta’s pens are in the ratio m:n and total pens are T, then, Number of pens that Sheela has $=\left( \dfrac{m}{m+n} \right).T$ And Number of pens that Sangeeta has $=\left( \dfrac{n}{m+n} \right).T$ Hence, we know that m:n = 3:2, that is m = 3 and n = 2. Also T = 20. So, we get, Number of pens that Sheela has $=\dfrac{\left( 3 \right)}{\left( 3+2 \right)}.20=\dfrac{3}{5}.20=12\text{ pens}$ Number of pens that Sangeeta has $=\dfrac{\left( 2 \right)}{\left( 3+2 \right)}.20=\dfrac{2}{5}.20=8\text{ pens}$
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# University of Florida/Egm4313/s12.team8.dupre/R2.3
## R2.3
### Problem Statement
a) ${\displaystyle \displaystyle y''+6y'+8.96y=0}$ (3-1)
b)${\displaystyle \displaystyle y''+4y'+(\pi ^{2}+4)y=0}$ (3-2)
### Solution
The quadratic formula is necessary for these solutions:
${\displaystyle \displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$
#### Part a
${\displaystyle \displaystyle {\frac {-6\pm {\sqrt {6^{2}-(4)(1)(8.96)}}}{2(1)}}={\frac {-6\pm .4}{2}}}$
This shows us that the roots of the equation are:
${\displaystyle \displaystyle \lambda _{1}=-2.8,\lambda _{2}=-3.2}$
Therefore, the general equation is:
${\displaystyle \displaystyle y=c_{1}e^{-2.8x}+c_{2}e^{-3.2x}}$ (3-3)
##### Substitution
We need to first find the first and second derivatives of equation (3-3):
${\displaystyle \displaystyle y'=-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x}}$
${\displaystyle \displaystyle y''=10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x}}$
Plugging into equation (3-1), we find:
${\displaystyle \displaystyle (10.24c_{1}e^{-3.2x}+7.84c_{2}e^{-2.8x})+6(-3.2c_{1}e^{-3.2x}-2.8c_{2}e^{-2.8x})+8.96(c_{1}e^{-3.2x}+c_{2}e^{-2.8x})=0}$ (3-4)
Continuing to solve:
${\displaystyle \displaystyle 19.2c_{1}e^{-3.2x}+16.8c_{2}e^{-2.8x}-19.2c_{1}e^{-3.2x}-16.8c_{2}e^{-2.8x}=0}$ (3-5)
This shows that the general equation is correct, since everything cancels out to 0.
#### Part b
${\displaystyle \displaystyle {\frac {4\pm {\sqrt {4^{2}-4(1)(\pi ^{2}+4)}}}{2(1)}}={\frac {-4\pm (-4)(1))(pi^{2}+4)}{2(1)}}}$
The roots are, therefore:
${\displaystyle \displaystyle \lambda _{1}=-2-\pi i,\lambda _{2}=-2+\pi i}$
Therefore, the general solution to (3-2) is:
${\displaystyle \displaystyle y=c_{1}\cos(\pi x)e^{-2x}+c_{2}\sin(\pi x)e^{-2x}}$ (3-6)
##### Substitution
We must first find the first and second derivatives of equation (3-6):
${\displaystyle \displaystyle y'=-2(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x}+(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}}$
${\displaystyle \displaystyle y''=-2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}+(-c_{1}\pi ^{2}\cos(\pi x)-c_{2}\pi ^{2}\sin(\pi x))e^{-2x}}$
Plugging into equation (3-2):
${\displaystyle \displaystyle -2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}+(-c_{1}\pi ^{2}\cos(\pi x)-c_{2}\pi ^{2}\sin(\pi x))e^{-2x}...}$
${\displaystyle \displaystyle +4[-2(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x}+(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}]+(\pi ^{2}+4)(c_{1}\cos(\pi x)e^{-2x}+c_{2}\sin(\pi x)e^{-2x})=0}$
Finally, plugging (3-6) and it's first and second derivatives into equation (3-2), we find:
${\displaystyle \displaystyle 4(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x}-2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}-2(-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}...}$
${\displaystyle \displaystyle +(-c_{1}\pi ^{2}\cos(\pi x)-c_{2}\pi ^{2}\sin(\pi x))e^{-2x}+4-2(c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x})+...}$
${\displaystyle \displaystyle (-c_{1}\pi \sin(\pi x)+c_{2}\pi \cos(\pi x))e^{-2x}+(\pi ^{2}+4)((c_{1}\cos(\pi x)+c_{2}\sin(\pi x))e^{-2x})=0}$
Since this equals 0, we know that the general equation (3-6) is correct.
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# Exponents and Exponential Functions
## Objective
Solve mathematical applications of exponential expressions.
## Common Core Standards
### Core Standards
?
• 8.EE.A.1 — Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 3² × 3-5 = 3-3 = 1/3³ = 1/27.
?
• 6.EE.A.1
• 7.EE.A.1
## Criteria for Success
?
1. Use the properties of exponents in area and volume problems.
2. Simplify expressions using the properties of exponents.
3. Change the base of exponential expressions to write equivalent expressions.
## Tips for Teachers
?
This lesson reviews skills and concepts from 8.EE.1. Depending on the needs of your students, this lesson may be skipped or used in a different way. Students may need to spend time developing mastery with Criteria for Success #3, as this will be a useful tool when working with rational exponents in upcoming lessons.
## Anchor Problems
?
### Problem 1
The area of a triangle is represented by the expression ${12^4y}$. The base of the triangle is represented by the expression ${3x^2}$. Write an expression that represents the height, $h$.
### Problem 2
Find the value of $n$ that makes the equation true.
$8^2\cdot4^n(2^2)^3=16^2\cdot(4^3)^5$
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
• Include geometric applications similar to Anchor Problem #1 and the Target Task
• Include problems where students rewrite exponential expressions in the same base, similar to Anchor Problem #2
• Include problems from the first few lessons of the unit as needed
A sphere has a radius that is given by the expression ${{1\over2}x^3}$. What expression represents the volume of the sphere in terms of ${\pi}$ and $x$
Determine the value of $n$ in the equation $4\cdot16^n=4^{21}$.
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New SAT Math Workbook
# A 10 b 15 c 20 d 30 e 45 if the length and width of
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Unformatted text preview: ncludes the ratios of corresponding sides, medians, altitudes, angle bisectors, radii, diameters, perimeters, and circumferences. The ratio is referred to as the linear ratio or ratio of similitude. If triangle ABC is similar to triangle DEF and the segments are given as marked, then EH is equal to 2.5 because the linear ratio is 6 : 3 or 2 : 1. D. When figures are similar, the ratio of their areas is equal to the square of the linear ratio. If triangle ABC is similar to triangle DEF, the area of triangle ABC will be 9 times as great as the area of triangle DEF. The linear ratio is 12 : 4 or 3 : 1. The area ratio will be the square of this or 9 : 1. If the area of triangle ABC had been given as 27, the area of triangle DEF would be 3. www.petersons.com Geometry 221 E. When figures are similar, the ratio of their volumes is equal to the cube of their linear ratio. The volume of the larger cube is 8 times the volume of the smaller cube. The ratio of sides is 4 : 2 or 2 : 1. The ratio of areas would be 4 : 1. The ratio of volumes would be 8 : 1. Exercise 10 Work out each problem. Circle the letter that appears before your answer. 1. If the area of a circle of radius x is 5π, find the area of a circle of radius 3x. (A) 10π (B) 15π (C) 20π (D) 30π (E) 45π If the length and width of a rectangle are each doubled, the area is increased by (A) 50% (B) 100% (C) 200% (D) 300% (E) 400% The area of one circle is 9 times as great as the area of another. If the radius of the smaller circle is 3, find the radius of the larger circle. (A) 9 (B) 12 (C) 18 (D) 24 (E) 27 4. If the radius of a circle is doubled, then (A) the circumference and area are both doubled (B) the circumference is doubled and the area is multiplied by 4 (C) the circumference is multiplied by 4 and the area is doubled (D) the circumference and area are each multiplied by 4 (E) the circumference stays the same and the area is doubled The volumes of two similar solids are 250 and 128. If a dimension of the larger solid is 25, find the corresponding side of the smaller solid. (A) 12.8 (B) 15 (C) 20 (D) 40 (E) cannot be determined 2. 5. 3. www.petersons.com 222 Chapter 13 RETEST Work out each problem. Cir...
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NCERT Solutions For Class 6 Maths Chapter 4 Exercise 4.3
Ncert Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Exercise 4.3: In this Chapter Basic Geometrical ideas We learn the basics of the Numbers System In class 6 maths ncert Exercise 4.3 Solutions. This Chapter Playing with Numbers is very important for Class 6 maths Students to get a High Score in their Exam. And we help the Class 6 Students to Achieve Their Dream By providing Class 6 Maths Ncert Solutions Chapter 4 Basic Geometrical Ideas Exercise 4.3 With Free Pdf Download. And We also Provide Video solutions Of Basic Geometrical Ideas Class 6 maths Ncert Solutions.
Ncert Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Exercise 4.3 pdf:-
Exercise 4.3 Class 6 maths NCERT solutions Chapter 4 Basic Geometrical Ideas Exercise 4.3 Pdf: download-
Class 6 Maths Ncert Solution for Chapter 4 Basic Geometrical Ideas Exercise 4.3 Tips:-
Angles:-
Angles are made when corners
are formed.
Here is a picture
where the top of a box is like a
hinged lid. The edges AD of the
box and AP of the door can be
imagined as two rays AD. These two rays have a
common end point A. The two rays here together are said to form an angle.
An angle is made up of two rays starting from a common end point.
The two rays forming the angle are called the arms or sides of the angle.
The common end point is the vertex of the angle.
(a) (b) (c) (d) (e)
This is an angle formed by rays OP
How can we name this angle? We can simply say
that it is an angle at O. To be more specific we
identify some two points, one on each side and the
vertex to name the angle. Angle POQ is thus a better
way of naming the angle. We denote this by
∠POQ.
Think, discuss and write
Look at the diagram.What is the name
of the angle? Shall we say∠P ? But then which
one do we mean? By ∠P what do we mean?
Is naming an angle by vertex helpful here?
Why not?
By ∠P we may mean ∠APB or ∠CPB or even
∠APC! We need more information.
Note that in specifying the angle, the vertex is always written as the
middle letter.
Now shade in a different colour the portion
of the paper bordering BC
The portion common to both shadings is
called the interior of ∠ABC. (Note
that the interior is not a restricted area; it
extends indefinitely since the two sides extend
indefinitely).
In this diagram , X is in the
interior of the angle, Z is not in the interior but
in the exterior of the angle; and S is on the
∠PQR . Thus, the angle also has three parts
associated with it.
Test Paper Of Class 8th
• Maths 8th Class
• Science 8th class
• Sst 8th Class
• Test Paper Of Class 7th
• Maths 7th Class
• Science 7th class
• Test Paper Of Class 6th
• Maths 6th Class
• Science 6th class
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31.
A man can do a piece of work in 5 days, but with the help of his son he can do it in 3 days. In what time can the son do it alone ?
A. [Math Processing Error] B. [Math Processing Error] C. [Math Processing Error] D. [Math Processing Error]
Explanation: In this type of question, where we have one person work and together work done. Then we can easily get the other person work just by subtracting them. As, Son's one day work = [Math Processing Error] So son will do whole work in 15/2 days which is = [Math Processing Error]
32.
A can do a job in 16 days, B can do same job in 12 days. With the help of C they did the job in 4 days. C alone can do the same job in how many days ?
A. 6 1 2 days B. 7 1 2 days C. 8 3 5 days D. 9 3 5 days
Explanation: In this question we having, A's work, B's work and A+B+C work. We need to calculate C's work. We can do it by, (A+B+C)'s work - (A's work + B's work). Let's solve it now: C's 1 day work = 1 4 −( 1 16 + 1 12 ) =( 1 4 − 7 48 ) = 5 48 So C can alone finish the job in 48/5 days, Which is = 9 3 5 days
33.
To complete a work A and B takes 8 days, B and C takes 12 days, A,B and C takes 6 days. How much time A and C will take
A. 24 days B. 16 days C. 12 days D. 8 days
Explanation: A+B 1 day work = 1/8 B+C 1 day work = 1/12 A+B+C 1 day work = 1/6 We can get A work by (A+B+C)-(B+C) And C by (A+B+C)-(A+B) So A 1 day work = 1 6 − 1 12 = 1 12 Similarly C 1 day work = 1 6 − 1 8 = 4−3 24 = 1 24 So A and C 1 day work = 1 12 + 1 24 = 3 24 = 1 8 So A and C can together do this work in 8 days
34.
A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it
A. 40 days B. 35 days C. 30 days D. 25 days
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# NCERT Solutions for Class 12 Maths Chapter 4 Determinants - Exercise 4.4
### Access Exercises of Class 12 Maths Chapter 4 –Determinants
Exercise 4.1 Solutions: 8 Questions (2 Long, 5 Short Answers, 1 MCQ)
Exercise 4.2 Solutions: 16 Questions(7 Long, 7 Short, 2 MCQs)
Exercise 4.3 Solutions: 5 Questions ( 4 Short Answers, 1 MCQ)
Exercise 4.4 Solutions: 5 Questions (4 Long, 1 MCQ)
Exercise 4.5 Solutions: 18 Questions (11 Long, 5 Short, 2 MCQs)
Exercise 4.6 Solutions: 16 Questions (13 Long, 3 Short)
Miscellaneous Exercise Solutions: 19 Questions (15 Long, 1 Short, 3 MCQs)
Exercise 4.4
$$\textbf{1.\space(i)\space}\begin{vmatrix}\textbf{2} &\textbf{\normalsize-4}\\\textbf{0} &\textbf{\normalsize-3}\end{vmatrix}\\\textbf{(ii)\space}\begin{vmatrix}\textbf{a} &\textbf{c}\\\textbf{b} &\textbf{d}\end{vmatrix}\\\textbf{Sol.\space (i)\space}\text{Here, \space}\begin{vmatrix}2 &-4\\0 &3\end{vmatrix}\\\therefore\space \text{Minors, M}_{11} = 3, \text{M}_{12}=0,\\\text{M}_{21}=-4\space\text{and}\space \text{M}_{22}=2$$
Also cofactors,
A11 = (– 1)1 + 1 M11 = 1 × 3 = 3
A12 = (– 1)1 + 2 M12 = (– 1) × 0 = 0
A21 = (– 1)2 + 1 M21 = (– 1) × (– 4) = 4
and A22 = (– 1)2 + 2M22 = 1 × 2 = 2
$$\textbf{(ii)\space}\text{Here,}\space\begin{vmatrix}a &c\\b &d\end{vmatrix}$$
∴ Minors, M11 = d, M12 = b, M21 = c and M22 = a
Also cofactors,
A11 = (– 1)1 + 1M11 = 1 × d = d
A12 = (– 1)1 + 2M12 = (– 1) × b = – b
A21 = (– 1)2 + 1M21 = (– 1) × c = – c
A22 = (– 1)2 + 2M22 = 1 × a = a
$$\textbf{2.\space(i)\space}\begin{vmatrix}\textbf{1} &\textbf{0} &\textbf{0}\\\textbf{0} &\textbf{1} &\textbf{0}\\\textbf{0} &\textbf{0} &\textbf{1}\end{vmatrix}\\\textbf{(ii)\space}\begin{vmatrix}\textbf{1} &\textbf{0} &\textbf{4}\\\textbf{3} &\textbf{5} &\textbf{\normalsize-1}\\\textbf{0} &\textbf{1} &\textbf{2}\end{vmatrix}\\\textbf{Sol.}\space\text{Here,}\begin{vmatrix}1 &0 &0\\0 &1 &0\\0 &0 &1\end{vmatrix}$$
Minors of elements of first row are
$$\text{M}_{11}=\begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0=1\\\text{M}_{12} = \begin{vmatrix}0 &0\\0 &1\end{vmatrix}\\=0-0=0\\\text{and}\space \text{M}_{13} = \begin{vmatrix}0 &1\\0 &0\end{vmatrix}\\=0-0=0$$
Minors of elements of second row are
$$\text{M}_{21} =\begin{vmatrix}0 &0\\0 &1\end{vmatrix}\\=0-0=0\\\text{M}_{22} = \begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0=1\\\text{and}\space\text{M}_{23}=\begin{vmatrix}1 &0\\0 &0\end{vmatrix}\\=0-0=0$$
Minors of elements of third row are
$$\text{M}_{31} =\begin{vmatrix}0 &0\\1 &0\end{vmatrix}\\=0-0 = 0\\\text{M}_{32}=\begin{vmatrix}1 &0\\0 &0\end{vmatrix}\\=0-0=0\\\text{and}\space\text{M}_{33} =\begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0=1$$
Hence, cofactors of elements of first row are
A11 = (– 1)1 + 1M11 = 1 × 1 = 1,
A12 = (– 1)1 + 2M12 = – 1 × 0 = 0
A13 = (– 1)1 + 3M13 = 1 × 0 = 0\
Cofactors of elements of second row are
A21 = (– 1)2 + 1M21 = – 1 × 0 = 0,
A22 = (– 1)2 + 2M22 = 1 × 1 = 1
A23 = (– 1)2 + 3M23 = – 1 × 0 = 0
Cofactors of elements of third row are
A31 = (– 1)3 + 1M31 = 1 × 0 = 0,
A32 = (– 1)3 + 2M32 = – 1 × 0 = 0
A33 = (– 1)3 + 3M33 = 1 × 1 = 1
$$\textbf{(ii)\space}\text{Here,}\space\begin{vmatrix}1 &0 &4\\3 &5 &\normalsize-1\\0 &1 &2\end{vmatrix}$$
Minors of elements of first row are
$$\text{M}_{11} =\begin{vmatrix}5 &-1\\ 1 &2\end{vmatrix}\\=10+1=11,\\\text{M}_{12}=\begin{vmatrix}3 &\normalsize-1\\0 &2\end{vmatrix}\\\text{and}\space\text{M}_{13} =\begin{vmatrix}3 &5\\0 &1\end{vmatrix}\\=3-0 = 3$$
Minors of elements of second row are
$$\text{M}_{21} = \begin{vmatrix}0 &4\\1 &2\end{vmatrix}\\=0-4=\normalsize-4,\\\text{M}_{22} = \begin{vmatrix}1 &4\\0 &2\end{vmatrix}\\=2-0=2\\\text{and}\space\text{M}_{23} =\begin{vmatrix}1 &0\\0 &1\end{vmatrix}\\=1-0 = 1$$
Minors of elements of third row are
$$\text{M}_{31} =\begin{vmatrix}0 &4\\5 &\normalsize-1\end{vmatrix}\\\text{M}_{32} =\begin{vmatrix}1 &4\\3 &\normalsize-1\end{vmatrix}\\=-1-12=-13\\\text{and}\space\text{M}_{33}=\begin{vmatrix}1 &0\\3 &5\end{vmatrix}\\=5-0=5$$
Hence, cofactors of elements of first row are
A11 = (– 1)1 + 1M11 = 1 × 11 = 11,
A12 = (– 1)1 + 2M12 = –1 × 6 = –6
A13 = (– 1)1 + 3M13 = 1 × 3 = 3
Cofactors of elements of second row are
A21 = (– 1)2 + 1M21 = –1 × –4 = 4
A22 = (– 1)2 + 2M22 = 1 × 2 = 2
A23 = (– 1)2 + 3M23 = – 1 × 1 = – 1
Cofactors of elements of third row are
A31 = (– 1)3 + 1M31 = 1 × – 20 = – 20
A32 = (– 1)3 + 2M32 = – 1 × – 13 = 13
A33 = (– 1)3 + 3M33 = 1 × 5 = 5
3. Using cofactors of the elements of second row,
$$\textbf{evaluate}\space\Delta = \begin{vmatrix}\textbf{5} &\textbf{3} &\textbf{8}\\\textbf{2} &\textbf{0} &\textbf{1}\\\textbf{1} &\textbf{2} &\textbf{3}\end{vmatrix}\textbf{.}\\\textbf{Sol.\space}\text{Given,}\space\Delta = \begin{vmatrix}5 &3 &8\\2 &0 &1\\1 &2 &3\end{vmatrix}$$
Cofactors of the elements of second row
$$\text{A}_{21} =(\normalsize-1)^{2+1}\begin{vmatrix}3 &8\\2 &3\end{vmatrix}\\=-(9-16)=7\\\lbrack\because\space \text{A}_{ij} = (\normalsize-1)^{i+j}\text{M}_{ij}\rbrack\\\text{A}_{22} =(\normalsize-1)^{2+2}\begin{vmatrix}5 &8\\1 &3\end{vmatrix}\\=15-8 =7\\\text{and}\space \text{A}_{23} = (\normalsize-1)^{2+3}\begin{vmatrix}5 &3\\1 &2\end{vmatrix}\\=-(10-3)=\normalsize-7$$
Now, expansion of Δ using cofactors of elements
of second row is given by
Δ = a21A21 + a22A22 + a23A23
= 2 × 7 + 0 × 7 + 1(– 7) = 14 – 7 = 7
4. Using cofactors of elements of third column,
$$\textbf{evaluate}\space\Delta = \begin{vmatrix}\textbf{1} &\textbf{x} &\textbf{yz}\\1 &\textbf{y} &\textbf{zx}\\ \textbf{1} &\textbf{z} &\textbf{xy}\end{vmatrix}\\\textbf{Sol.\space}\text{Given, \space}\Delta = \begin{vmatrix}1 &x &yz\\1 &y &zx\\1 &z &xy\end{vmatrix}$$
Cofactors of the elements of third column are
$$\text{A}_{13} = (\normalsize-1)^{1+3}\begin{vmatrix}1 &y\\1 &z\end{vmatrix}\\= 1(z-y) = z-y\\\text{A}_{23}=(\normalsize-1)^{2+3}\begin{vmatrix}1 &x\\1 &z \end{vmatrix}\\=-1(z-x)=x-2\\\text{and}\space \text{A}_{33} =(\normalsize-1)^{3+3}\begin{vmatrix}1 &x\\1 &y\end{vmatrix}\\= 1(y-x) = y-x$$
Now, expansion of Δ using cofactors elements of
third column is given by
Δ = a13A13 + a23A23 + a33A33 = yz(z – y)
+ zx(x – z) + xy(y – x)
= yz2 – y2z + zx2 – z2x + xy2 – x2y
= x2(z – y) + x(y2 – z2) + yz(z – y)
= (z – y){x2 – x(y + z) + yz}
= (z – y)(x2 – xy – xz + yz)
= (z – y)[x(x – y) – z(x – y)]
= (y – z)(x – y)(z – x)
= (x – y)(y – z)(z – x)
$$\textbf{5. If}\space\Delta = \begin{vmatrix}\textbf{a}_{\textbf{11}} & \textbf{a}_{\textbf{12}} &\textbf{a}_{\textbf{13}}\\\textbf{a}_{\textbf{21}} &\textbf{a}_{\textbf{22}} &\textbf{a}_{\textbf{23}}\\\textbf{a}_{\textbf{31}} &\textbf{a}_{\textbf{32}} &\textbf{a}_{\textbf{33}}\end{vmatrix}$$
and Aij is cofactors of aij, then value of Δ is given by
(a) a11A31 + a12A32 + a13A33
(b) a11A11 + a12A21 + a13A31
(c) a21A11 + a22A12 + a23A13
(d) a11A11 + a21A21 + a31A31
Sol. (d) a11A11 + a21A21 + a31A31
Δ is equal to the sum of the products of the elements of a row (or a column) with their corresponding cofactors.
Δ = a11A11 + a12A12 + a13A13
or a21A21 + a22A22 + a23A23
or a31A31 + a32A32 + a33A33
or a11A11 + a21A21 + a31A31
or a12A12 + a22A22 + a32A32
or a13A13 + a23A23 + a33A33
Sum of the products of the elements of first column with their corresponding cofactors is
Δ = a11A11 + a21A21 + a31A31.
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# Find the angle between the vectors $$\hat i - 2\hat j + 3\hat k$$ and $$3\hat i - 2\hat j + \hat k$$
$\begin{array}{1 1} (A) \cos^{-1} \bigg( \large\frac{2}{7} \bigg) \\ (B) \cos^{-1} \bigg( \large\frac{4}{7} \bigg) \\ (C) \cos^{-1} \bigg( \large\frac{1}{7} \bigg) \\ (D) \cos^{-1} \bigg( \large\frac{5}{7} \bigg) \end{array}$
Toolbox:
• The scalar product of two vectors is $\overrightarrow a.\overrightarrow b=\mid \overrightarrow a\mid\mid\overrightarrow b\mid\cos\theta$
• Hence the angle between two vectors is given by $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
Step 1:
Let $\overrightarrow a=\hat i-2\hat j+3\hat k$ and $\overrightarrow b=3\hat i-2\hat j+\hat k$
We know that $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
$\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
$\mid \overrightarrow a\mid=\sqrt{1^2+(-2)^2+3^2}$
$\qquad=\sqrt{1+4+9}$
$\qquad=\sqrt{14}$
Similarly $\mid \overrightarrow b\mid=\sqrt{3^2+(-2)^2+1^2}$
$\qquad=\sqrt{9+4+1}$
$\qquad=\sqrt{14}$
Step 2:
$\overrightarrow a.\overrightarrow b=(\hat i-2\hat j+3\hat k).(3\hat i-2\hat j+\hat k)$
Scalar product of two vectors is just multiplication of the scalar quantities of the respective components.
$\overrightarrow a.\overrightarrow b=(3+4+3)=10.$
Step 3:
Substituting these values for $\cos\theta$ we get,
$\cos\theta=\large\frac{10}{\sqrt{14}\sqrt {14}}$
$\qquad=\large\frac{10}{14}$
$\qquad=\large\frac{5}{7}$
Therefore $\theta=\cos^{-1}\big(\large\frac{5}{7}\big)$
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# Lesson 20
Writing and Solving Inequalities in One Variable
• Let’s solve problems by writing and solving inequalities in one variable.
### Problem 1
Solve $$2x < 10$$. Explain how to find the solution set.
### Problem 2
LIn is solving the inequality $$15 - x < 14$$. She knows the solution to the equation $$15 - x = 14$$ is $$x = 1$$
How can Lin determine whether $$x > 1$$ or $$x < 1$$ is the solution to the inequality?
### Problem 3
A cell phone company offers two texting plans. People who use plan A pay 10 cents for each text sent or received. People who use plan B pay 12 dollars per month, and then pay an additional 2 cents for each text sent or received.
1. Write an inequality to represent the fact that it is cheaper for someone to use plan A than plan B. Use $$x$$ to represent the number of texts they send.
2. Solve the inequality.
### Problem 4
Clare made an error when solving $$\text-4x+3<23$$.
Describe the error that she made.
\displaystyle \begin{align} \text-4x+3<23 \\ \text-4x<20 \\ x< \text-5 \end{align}
### Problem 5
Diego’s goal is to walk more than 70,000 steps this week. The mean number of steps that Diego walked during the first 4 days of this week is 8,019.
1. Write an inequality that expresses the mean number of steps that Diego needs to walk during the last 3 days of this week to walk more than 70,000 steps. Remember to define any variables that you use.
2. If the mean number of steps Diego walks during the last 3 days of the week is 12,642, will Diego reach his goal of walking more that 70,000 steps this week?
### Problem 6
Here are statistics for the length of some frog jumps in inches:
• the mean is 41 inches
• the median is 39 inches
• the standard deviation is about 9.6 inches
• the IQR is 5.5 inches
How does each statistic change if the length of the jumps are measured in feet instead of inches?
(From Unit 1, Lesson 15.)
### Problem 7
Solve this system of linear equations without graphing: $$\begin{cases} 3y+7=5x \\ 7x-3y=1 \\ \end{cases}$$
(From Unit 2, Lesson 15.)
### Problem 8
Solve each system of equations without graphing.
1. $$\begin{cases} 5x+14y=\text-5 \\ \text-3x+10y=72 \\ \end{cases}$$
2. $$\begin{cases}20x-5y=289 \\ 22x + 9y=257 \\ \end{cases}$$
(From Unit 2, Lesson 16.)
### Problem 9
Noah and Lin are solving this system: $$\begin{cases} 8x+15y=58 \\ 12x-9y=150 \end{cases}$$
Noah multiplies the first equation by 12 and the second equation by 8, which gives:
$$\displaystyle \begin{cases} 96x+180y=696 \\ 96x-72y=1,\!200 \\ \end{cases}$$
Lin says, “I know you can eliminate $$x$$ by doing that and then subtracting the second equation from the first, but I can use smaller numbers. Instead of what you did, try multiplying the first equation by 6 and the second equation by 4."
1. Do you agree with Lin that her approach also works? Explain your reasoning.
2. What are the smallest whole-number factors by which you can multiply the equations in order to eliminate $$x$$?
(From Unit 2, Lesson 16.)
### Problem 10
What is the solution set of the inequality $$\dfrac{x+2}{2}\geq \text-7-\dfrac {x}{2}$$ ?
A:
$$x\leq \text-8$$
B:
$$x\geq \text-8$$
C:
$$x \geq - \frac92$$
D:
$$x\geq 8$$
(From Unit 2, Lesson 19.)
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Math Practice Online > free > lessons > Florida > 9th grade > Stem And Leaf Plots
These sample problems below for Stem And Leaf Plots were generated by the MathScore.com engine.
## Sample Problems For Stem And Leaf Plots
### Complexity=5
Calculate the mean of the data represented by the following stem and leaf plots.
1.
Data Values
StemLeaf
07
32
64
84, 5
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
2.
Data Values
StemLeaf
23
66
82
100, 0
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
### Complexity=8
Calculate the mean of the data represented by the following stem and leaf plots.
1.
Data Values
StemLeaf
07
20, 2
35
98
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
2.
Data Values
StemLeaf
44
62
76
96
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
### Complexity=5
Calculate the mean of the data represented by the following stem and leaf plots.
1
Data Values
StemLeaf
07
32
64
84, 5
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
Solution
Arithmetic Mean = sum of values / number of values = (7 + 32 + 64 + 84 + 85) / 5 = 54.4
Median = middle term of values = middle of (7, 32, 64, 84, 85) = 64
Range = largest value - smallest value = 85 - 7 = 78
Lower Quartile = median of lower half of data values = middle of (7, 32) = (7 + 32) / 2 = 19.5
Upper Quartile = median of upper half of data values = middle of (84, 85) = (84 + 85) / 2 = 84.5
2
Data Values
StemLeaf
23
66
82
100, 0
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
Solution
Arithmetic Mean = sum of values / number of values = (23 + 66 + 82 + 100 + 100) / 5 = 74.2
Median = middle term of values = middle of (23, 66, 82, 100, 100) = 82
Range = largest value - smallest value = 100 - 23 = 77
Lower Quartile = median of lower half of data values = middle of (23, 66) = (23 + 66) / 2 = 44.5
Upper Quartile = median of upper half of data values = middle of (100, 100) = (100 + 100) / 2 = 100
### Complexity=8
Calculate the mean of the data represented by the following stem and leaf plots.
1
Data Values
StemLeaf
07
20, 2
35
98
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
Solution
Arithmetic Mean = sum of values / number of values = (7 + 20 + 22 + 35 + 98) / 5 = 36.4
Median = middle term of values = middle of (7, 20, 22, 35, 98) = 22
Range = largest value - smallest value = 98 - 7 = 91
Lower Quartile = median of lower half of data values = middle of (7, 20) = (7 + 20) / 2 = 13.5
Upper Quartile = median of upper half of data values = middle of (35, 98) = (35 + 98) / 2 = 66.5
2
Data Values
StemLeaf
44
62
76
96
Mean:
Median:
Range:
Lower Quartile:
Upper Quartile:
Solution
Arithmetic Mean = sum of values / number of values = (44 + 62 + 76 + 96) / 4 = 69.5
Median = middle term of values = middle of (44, 62, 76, 96) = (62 + 76)/2 =69
Range = largest value - smallest value = 96 - 44 = 52
Lower Quartile = median of lower half of data values = middle of (44, 62) = (44 + 62) / 2 = 53
Upper Quartile = median of upper half of data values = middle of (76, 96) = (76 + 96) / 2 = 86
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# Quick Answer: When 2 dice are rolled find the probability of getting a sum greater than 5?
Contents
## What is the probability of rolling a sum greater than 5 with two dice?
To find the probability determine the number of successful outcomes divided by the number of possible outcomes overall. Each dice has six combinations which are independent. Therefore the number of possible outcomes will be 6*6 = 36. The probability of rolling a pair of dice whose numbers add to 5 is 4/36 = 1/9.
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## What is the probability of rolling a number greater than or equal to 5?
The numbers greater than 5 are 6 . The numbers less than 3 are 2,1 . So the numbers greater than 5 OR less than 3 are the union of the two sets or 6, and 2,1. The probability of rolling one of these numbers is 36=12 or 50% .
## When two dice are thrown what is the probability that the sum will be greater than eight?
Explanation: Throwing two dice we have: 6×6=36 possible outcomes and only 15 possible outcomes summing 8 or more than 8 .
## What is the number of outcomes greater than 5?
It represents that the outcomes of getting a prime number are 1, 4, 6. (ii) a) Since we know that is a number is greater than 5, only 6, when a die is thrown, the number of outcomes of getting a number is greater than 5 is 1. It represents that the outcomes of getting a number are greater than 5 are 1.
## When two dice are rolled what is the probability of getting a sum of 5 or 6?
What is the probability of getting a sum of 5 or 6 when a pair of dice is rolled? Probability of getting a sum of 5 or 6 = 9/36 = 1/4.
## When two dice are rolled find the probability of getting a 5 on each one?
Probabilities for the two dice
Total Number of combinations Probability
4 3 8.33%
5 4 11.11%
6 5 13.89%
7 6 16.67%
## What is the probability of rolling a number greater than or equal to 3?
So the probability that your roll is either going to be an even number OR a number greater than 3 is 1/2.
## What is the probability of rolling a number greater than 4?
If you roll a single die there are 6 possible outcomes (1,2,3,4,5,6), 2 of which are greater than 4. So in a single roll the probability of getting a number greater than 4 is 2/6 = 1/3.
## What is the probability of rolling two dice and getting a sum of 6?
Answer: The probability of rolling a sum of 6 with two dice is 5/36.
## When 2 dice are rolled find the probability of getting?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 2/36 (5.556%)
4 3/36 (8.333%)
5 4/36 (11.111%)
## How do you find the probability of two dice?
If you want to know how likely it is to get a certain total score from rolling two or more dice, it’s best to fall back on the simple rule: Probability = Number of desired outcomes ÷ Number of possible outcomes.
## When two dice are rolled find the probability of getting a sum that is divisible by 4?
What is the probability that the sum of the numbers on the two faces is divisible by 4 when two dice are thrown together? So, P(sum divisible by 4) = 1/4.
## When a dice is thrown find the probability of getting a number greater than 5?
The chance for a result greater than 5 is therefore 1 out of 6. If you meant 2 dice, a similar analysis is that there are 36 possible results and the chances for a result greater than 5 is 26 out of 36.
## Which is more likely getting a sum of 5 when you roll two dice or getting a sum of five when you roll three dice?
Specific Probabilities
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The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4% Probability of a sum of 5: 6/216 = 2.8%
## How many ways can two dice be rolled such that their sum is not more than 5?
If we want to calculate the probability of rolling, say, a five, we need to divide the number of ways to get 5 by the total possible combinations of two dice. How many total combinations are possible from rolling two dice? Since each die has 6 values, there are 6∗6=36 6 ∗ 6 = 36 total combinations we could get.
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# Writing a circle equation problems
How many prescriptions did she have for tranquilizers? Now, our newly formed interest adds to us in the degree direction. If Rhonda drives 10 miles, then Jamie will drive twice as far which would be Write the numerals from 1 to 9 and answer these questions about them: We have to be careful here. He wishes to cut it into two pieces so that one piece will be 6 inches longer than the other.
Discuss the assignment before writing time begins. The first thing we have to do is to recognize the variable.
The number of miles driven by Jamie and by Rhonda. We are told 6 is added to 4 times a number. Jose has a board that is 44 inches long. Usually, but not always, you can find this information at the end of the problem. Here are some recommended steps: We can use arctan to figure it out: Find the number of miles driven by each.
During writing time, circulate, ask questions, and elicit ideas "What do you have to do? Begin with verbal explanations as a shared class activity.
Answer the question in the problem The problem asks us to find out how far Rhonda and Jamie drove. How long do we go for? The radius is ea and the angle is determined by ebi. You may write me at the following e-mail address: Define it in your own words. Yes -- and we can understand it by building on a few analogies: Since length is 15 inches, width is w, and perimeter is 50, we get Step 5: So her salary before taxes and insurance will be 6h.
Taking any number and multiplying by i will not change its magnitude, just the direction it points. The analogy "complex numbers are 2-dimensional" helps us interpret a single complex number as a position on a circle. The only way to truly master this step is through lots of practice.
Your answer should not only make sense logically, but it should also make the equation true. The width of the rectangle is 10 inches.
Together, Betty and Bessie produced gallons of milk.A great circle is a section of a sphere that contains a diameter of the sphere (Kern and Blandp. 87). Sections of the sphere that do not contain a diameter are called small circles. A great circle becomes a straight line in a gnomonic projection (Steinhauspp.
). The shortest path between two points on a sphere, also known as an orthodrome, is a segment of a great circle.
Writing About Math. Writing about math can be a very positive and fruitful learning experience. Here's a look at some of the benefits; a variety of writing categories and topics; and suggestions for creating a positive environment for writing about math. The "Unit Circle" is a circle with a radius of 1.
Being so simple, it is a great way to learn and talk about lengths and angles. The center is put on a graph where the x axis and y axis cross, so we get this neat arrangement here.
Have a try! Move the mouse around to see how different angles (in.
Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site. Free math lessons and math homework help from basic math to algebra, geometry and beyond.
Students, teachers, parents, and everyone can find solutions to their math problems instantly.
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# Errors and Approximations
(C) ERRORS AND APPROXIMATIONS
We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable. The theory behind it is quite simple: From the chapter on differentiation, we know that
$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{{dy}}{{dx}} = f'\left( x \right)$
For small $$\Delta x,$$ we can therefore approximate $$\Delta y\,{\rm{as}}\,f'\left( x \right)\Delta x.$$ This is all there is to it!
Suppose we have to calculate (4.016)2.
We let $$y = {x^2} \cdot \,{x_0} = 4\,\,\,{\rm{and}}\,\,{y_0} = 169$$
\begin{align} \qquad\quad y' & = 2x,\,\,\Delta x = 0.016\\\\ \Rightarrow \qquad \Delta y & = f'\left( x \right) \cdot \Delta x\\\\ \qquad\qquad & = {\left. {2x} \right|_{{x_0} = 4}} \times 0.016\\\\ \qquad\qquad & = {\rm{ }}8{\rm{ }} \times {\rm{ }}0.016\\\\ \qquad\qquad & = {\rm{ }}0.128\\\\ \quad\Rightarrow \qquad \;\; y & = {y_0} + \Delta y = 16.128\end{align}
Example – 35
Find the value of $${\left( {8.01} \right)^{4/3}} + {\left( {8.01} \right)^2}$$
Solution: Let $$y = f\left( x \right) = {x^{4/3}} + {x^2}$$
Let $${x_0} = 8\,\,\,{\rm{so}}\,\,{\rm{that}}\,\,{y_0} = 16 + 64 = 80$$
\begin{align} \qquad\;\;\Delta x & = 0.01\\\\ \Rightarrow \qquad \Delta y & = {\left. {f'\left( x \right)} \right|_{x = {x_0}}} \times \Delta x\\\\ \qquad\qquad & \;{\left. { = \left( {\frac{4}{3}{x^{1/3}} + 2x} \right)} \right|_{{x_0} = 8}} \times \Delta x\\\\ \qquad\qquad & \; = \left( {\frac{8}{3} + 16} \right) \times 0.01\\\\ \qquad\qquad & \; = \frac{{0.56}}{3}\\\\ \qquad\qquad & \; = 0.1867\\\\ \Rightarrow \qquad \;\, {y_0} & \; = {y_0} + \Delta y\\\\ \qquad\qquad & \; = 80.1867\end{align}
TRY YOURSELF - V
Q. 1 Use the appropriate mean value theorem to show that the square roots of two successvie natural numbers greater thatn Ndiffer by less than \begin{align}\frac{1}{{2N}}\end{align}.
Q. 2 Use LVMT to show that
(a) \begin{align}\frac{{b - a}}{b} < \ln \frac{b}{a} < \frac{{b - a}}{a} \qquad {\text{when}}\; a < a < b\end{align}
(b) \begin{align}{\tan ^{ - 1}}{x_2} - {\tan ^{ - 1}}{x_1} < {x_2} - {x_1} \qquad {\text{when}}\; {x_2} > {x_1}\end{align}
Q. 3 If $$2a + 3b + 6c = 0,$$ can we say that $$a{x^2} + bx + c = 0$$ will have at least one real root in (0, 1)
Q. 4 Find the approximate value of \begin{align}f\left( x \right) = {\left( {\frac{{2 - x}}{{2 + x}}} \right)^{1/5}} \qquad {\text{at}}\;x = 0.15\end{align}
Q. 5 If $$f\left( x \right)\;{\rm{and}}\;g\left( x \right)$$ are differentiable functions for $$0 \le x \le 1$$ such that $$f\left( 0 \right) = 2,g\left( 0 \right) = 0,\,\,f\left( 1 \right) = 6,\,\,g\left( 1 \right) = 2,$$ show that there exists c satisfying $$0 < c < 1\;{\rm{and}}\;f'\left( c \right) = 2g'\left( c \right)$$
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$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$
# Tangent Planes and Linear Approximation
## Tangent Planes
An equation of the plane tangent to the surface $$z=f(x,y)$$ at the point $$(a,b,f(a,b))$$ is $\ans{z=f_x (a,b)\ (x-a) + f_y (a,b)\ (y-b) + f(a,b).}$
Find the tangent plane to the surface $$z=3x^2+3y^2$$ at $$(1,1,6)$$.
#### Solution
Since the partial derivatives are $$f_x = 6x$$ and $$f_y = 6y$$, the tangent plane is \ans{\begin{align} z &= 6(x-1) + 6(y-1) + 6\\ &= 6(x+y-1) \end{align}}
We can graph this surface and tangent plane with Matlab:
>> hold on
>> fsurf(@(x,y) 3*x.^2+3*y.^2)
>> fsurf(@(x,y) 6.*x+6.*y-6)
#### Try it yourself:
(click on a problem to show/hide its answer)
1. Find the tangent plane to $$z=4-2x^2-y^2$$ at the point $$(2,2,-8)$$.
2. $$z=-8(x-2) - 4(y-2) - 8$$
3. Find the tangent plane to $$z=e^{xy}$$ at the point $$(1,0,1)$$.
4. $$z=y+1$$
5. Find the tangent plane to $$z=\ln (1+xy)$$ at the point $$(1,2,\ln 3)$$.
6. $$z=\dfrac{2}{3}(x-1) + \dfrac{1}{3}(y-2) + \ln 3$$
## Linear Approximation
Just as we can use tangent lines to approximate functions of one variable, we can use the tangent plane as a linear approximation for a function of two variables.
If $$f(x,y) = \dfrac{5}{x^2+y^2}$$, use the linear approximation at the point $$(-1,2,1)$$ to estimate the value of $$f(-1.05,2.1)$$.
#### Solution
Find the tangent plane to $$f$$ at $$(-1,2,1)$$:
\begin{align} f_x &= \dfrac{-80x}{(x^2+y^2)^2} \textrm{ and } f_y = \dfrac{-10y}{(x^2+y^2)^2}\\ f_x (-1,2) &= \dfrac{10}{25} = \dfrac{2}{5} \textrm{ and } f_y (-1,2) = \dfrac{-20}{25} = -\dfrac{4}{5}\\ \\ L(x,y) &= f_x (a,b)\ (x-a) + f_y (a,b)\ (y-b) + f(a,b)\\ &= \dfrac{2}{5}(x+1) - \dfrac{4}{5}(y-2) + 1 \end{align}
Then use this tangent plane to estimate the value of the function at $$(-1.05,2.1)$$:
$\ans{L(-1.05,2.1) = 0.9 \approx f(-1.05,2.1)}$
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# How do I find the partial-fraction decomposition of (2x^3+7x^2-2x+6)/(x^4+4)?
Aug 8, 2015
First factor the denominator to find ${x}^{4} + 4 = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$
Then solve $2 {x}^{3} + 7 {x}^{2} - 2 x + 6 = \left({x}^{2} - 2 x + 2\right) \left(A x + B\right) + \left({x}^{2} + 2 x + 2\right) \left(C x + D\right)$
finding $A = 0$, $B = 3$, $C = 2$ and $D = 0$.
#### Explanation:
${x}^{4} + 4$ has no linear factors with Real coefficients since ${x}^{4} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$.
It will have quadratic factors with Real coefficients.
${x}^{4} + 4 = \left(a {x}^{2} + b x + c\right) \left({\mathrm{dx}}^{2} + e x + f\right)$
Without loss of generalisation, $a = d = 1$, so
${x}^{4} + 4 = \left({x}^{2} + b x + c\right) \left({x}^{2} + e x + f\right)$
Then looking at the coefficient of ${x}^{3}$, we have $e = - b$, so
${x}^{4} + 4 = \left({x}^{2} + b x + c\right) \left({x}^{2} - b x + f\right)$
Then looking at the coefficient of $x$ we have $b \left(f - c\right) = 0$, so either $b = 0$ or $c = f$.
If $b = 0$ then $c + f = 0$ and $c f = - {c}^{2} = 4$, which has no Real solutions (it has solutions $c = 2 i$, $f = - 2 i$).
If $c = f$ and $c f = 4$ then $c = f = \pm 2$. Then the coefficient of ${x}^{2}$ tells us $f + c - {b}^{2} = 0$. Now if $b$ is Real, then ${b}^{2} \ge 0$, so $f + c \ge 0$, hence $f = c = 2$. Then ${b}^{2} = 4$, so $b = \pm 2$.
So we have ${x}^{4} + 4 = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$
Now attempt to solve:
$\frac{2 {x}^{3} + 7 {x}^{2} - 2 x + 6}{{x}^{4} + 4} = \frac{A x + B}{{x}^{2} + 2 x + 2} + \frac{C x + D}{{x}^{2} - 2 x + 2}$
If we multiply through by $\left({x}^{4} + 4\right) = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$ then this becomes:
$2 {x}^{3} + 7 {x}^{2} - 2 x + 6$
$= \left({x}^{2} - 2 x + 2\right) \left(A x + B\right) + \left({x}^{2} + 2 x + 2\right) \left(C x + D\right)$
$= \left(A + C\right) {x}^{3} + \left(B - 2 A + D + 2 C\right) {x}^{2} + 2 \left(A - B + C + D\right) x + 2 \left(B + D\right)$
Equating coefficients we get:
(i) $A + C = 2$
(ii) $B - 2 A + D + 2 C = 7$
(iii) $2 \left(A - B + C + D\right) = - 2$
(iv) $2 \left(B + D\right) = 6$
From (i) and (iv) we get:
(v) $C = 2 - A$ and
(vi) $D = 3 - B$
Substitute these into (ii) to get:
$\textcolor{red}{\cancel{\textcolor{b l a c k}{B}}} - 2 A + 3 - \textcolor{red}{\cancel{\textcolor{b l a c k}{B}}} + 4 - 2 A = 7$
Hence $A = 0$. Then from (v) $C = 2$.
Substitute $A = 0$, $C = 2$ and $D = 3 - B$ into (iii) to get:
$2 \left(0 - B + 2 + 3 - B\right) = - 2$
Divide both sides by $2$ to find:
$5 - 2 B = - 1$
Hence $B = 3$. Then from (vi) $D = 0$.
So
$\frac{2 {x}^{3} + 7 {x}^{2} - 2 x + 6}{{x}^{4} + 4} = \frac{3}{{x}^{2} + 2 x + 2} + \frac{2 x}{{x}^{2} - 2 x + 2}$
Aug 9, 2015
${x}^{4} + 4 = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$ hence
$\frac{2 {x}^{3} + 7 {x}^{2} - 2 x + 6}{{x}^{4} + 4} = \frac{3}{{x}^{2} + 2 x + 2} + \frac{2 x}{{x}^{2} - 2 x + 2}$
#### Explanation:
Here is a quick way to find the quadratic factors of ${x}^{4} + 4$ using complex arithmetic ...
$\sqrt{- 4} = 2 i$
$\sqrt{2 i} = i + 1$
If ${x}_{1}$ is a root of ${x}^{4} + 4 = 0$, then so are $- {x}_{1}$, $\overline{{x}_{1}}$ and $\overline{- {x}_{1}}$, where $\overline{{x}_{1}}$ means the complex conjugate. This four-fold symmetry occurs due to the only power of $x$ being ${x}^{4}$.
Here are the roots in the complex plane:
graph{((x-1)^2+(y-1)^2 - 0.01)((x+1)^2+(y-1)^2 - 0.01)((x-1)^2+(y+1)^2 - 0.01)((x+1)^2+(y+1)^2-0.01) = 0 [-5, 5, -2.5, 2.5]}
So ${x}^{4} + 4 = \left(x - 1 - i\right) \left(x - 1 + i\right) \left(x + 1 - i\right) \left(x + 1 + i\right)$
Now pick pairs of factors to multiply to get Real coefficients.
To do this, pick the ones which are complex conjugates:
$\left(x + 1 - i\right) \left(x + 1 + i\right) = \left(\left(x + 1\right) - i\right) \left(\left(x + 1\right) + i\right)$
$= {\left(x + 1\right)}^{2} - {i}^{2} = {x}^{2} + 2 x + 1 + 1 = {x}^{2} + 2 x + 2$
and
$\left(x - 1 - i\right) \left(x - 1 + i\right) = \left(\left(x - 1\right) - i\right) \left(\left(x - 1\right) + i\right)$
$= {\left(x - 1\right)}^{2} - {i}^{2} = {x}^{2} - 2 x + 1 + 1 = {x}^{2} - 2 x + 2$
For the rest of the partial fraction decomposition, see the other answers.
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# Ellipse (S.N.Dey) | Part-3 | Ex-5
###### In the previous article , we solved few solutions of Short Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11. In this chapter, we will solve few more.
4(i) Find the lengths of axes of the ellipse whose eccentricity is and the distance between focus and directrix is
Solution.
The distance between focus and directrix
By question,
Hence, the length of the major axis and the length of the minor axis
and are two foci of an ellipse whose eccentricity is Find the length of the major axis.
Solution.
Eccentricity
The distance between the two given foci is
Hence, the length of the major axis
5. The length of the latus rectum of an ellipse is unit and that of the major axis, which lies along the axis , is unit. Find its equation in the standard form . Determine the co-ordinates of the foci and the equations of its directrices.
Solution.
The length of the latus rectum of the ellipse is given by
Again, the length of the major axis (2a) is given by
Hence,the equation of the ellipse is
The eccentricity of the ellipse is given by
The co-ordinates of the foci of the ellipse is given by
The equation of the directrix is given by
6. Taking major and minor axes along and axes, find the equation of the ellipse whose
co-ordinates of foci are and the length of minor axis is
Solution.
Since the foci of the given ellipse lie on the -axis , the major axis of the ellipse lies on the axis.
Again, the centre of the ellipse is given by i.e.
So, the ellipse is of the form
So, the co-ordinates of the foci
The length of the minor axis is given by
Hence, using , the equation of the ellipse is
eccentricity and the length of latus rectum
Solution.
Length of latus rectum
From and we get,
So, the equation of the ellipse is
length of minor axis is and the distance between the foci is
Solution.
By the condition, length of minor axis is given by
The distance between the foci is
Hence, the equation of the ellipse is given by
co-ordinates of one vertex are and the co-ordinates of one end of minor axis are
Solution.
By question, the equation of the ellipse can be taken as
Co-ordinates of one vertex is and co-ordinates of one end of minor axis is
Hence, by we get,
co-ordinates of foci are and the eccentricity is
Solution.
By question, the co-ordinates of foci
Hence, the equation of the ellipse is
Find the equation of the ellipse whose
eccentricity is focus is and directrix is
Solution.
Let be any point on the ellipse . The co-ordinates of the focus : and the equation of the directrix is given by
Now, the length of the perpendicular from P on the straight line (1) is given by
Now, for the ellipse, we have
Hence, the equation of the ellipse is given by
eccentricity is focus is and directrix is
Solution.
Let be any point on the ellipse. The co-ordinates of the focus :
The equation of the directrix is
The length of the perpendicular from P on the straight line (1) is
Now, for the ellipse
Hence, the equation of the ellipse is
eccentricity is focus is directrix is
Solution.
Let be any point on the ellipse. The co-ordinates of focus : The equation of the directrix :
The perpendicular distance of the point P from the straight line (1) is
Now, for the ellipse, we know that
focus is directrix is and eccentricity is
Solution.
Let be any point on the ellipse. The co-ordinates of focus : The equation of the directrix :
The perpendicular distance of the point P from the straight line (1) is
Now, for the ellipse, we know that
Hence, the equation represents the required ellipse.
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# What information do you need to complete a rotation?
## What information do you need to complete a rotation?
To perform a geometry rotation, we first need to know the point of rotation, the angle of rotation, and a direction (either clockwise or counterclockwise). A rotation is also the same as a composition of reflections over intersecting lines.
## What is the rule for a 270 degree rotation?
Coordinate Rules for Rotations about the origin: For a rotation of 1800 (x,y) (-x, -y). For a rotation of 2700 (x,y) (-y, x). When a point (x, y) is rotated counterclockwise about the origin, the following rules are true: For a rotation of 900 (x,y) (-y, x).
What are the properties that identify a rotation?
The following are the three basic properties of rotations :
• A rotation maps a line to a line, a ray to a ray, a segment to a segment, and an angle to an angle.
• A rotation preserves lengths of segments.
• A rotation preserves measures of angles.
### What are the examples of rotation?
An example of rotation is the earth’s orbit around the sun. An example of rotation is a group of people holding hands in a circle and walking in the same direction.
### What is the formula for a 90 degree clockwise rotation?
If you want to do a clockwise rotation follow these formulas: 90 = (b, -a); 180 = (-a, -b); 270 = (-b, a); 360 = (a, b).
What are the rules for clockwise rotations?
Terms in this set (9)
• (-y, x) 90 degree rotation counterclockwise around the origin.
• (y, -x) 90 degree rotation clockwise about the origin.
• (-x, -y) 180 degree rotation clockwise and counterclockwise about the origin.
• (-y, x) 270 degree rotation clockwise about the origin.
• (y, -x)
• (x, -y)
• (-x, y)
• (y, x)
#### What are the rules of rotation?
When you rotate the image using the 90 degrees rule, the end points of the image will be (-1, 1) and (-3, 3). The rules for the other common degree rotations are: For 180 degrees, the rule is (x, y) ——–> (-x, -y) For 270 degrees, the rule is (x, y) ——–> (y, -x)
#### What clockwise rotation would be the same as an counterclockwise rotation of 90?
The answer would be, A clockwise rotation of 90 is the same as a counterclockwise rotation of 270.
What are the coordinates of point F after a 90 clockwise rotation about the origin?
if F is rotated 90° about the origin in the clockwise (negative) direction, the image is the point F'(3, 0), as shown with the red line segments. if F is rotated 90° about the origin in the counterclockwise (positive) direction, the image is the point F'(-3, 0), as shown with the green line segments.
## Why is clockwise to the right?
The reason that clocks turn clockwise has to do with sundials, which were the first clocks. In the northern hemisphere, the earth rotates counter-clockwise, which means that from our point of view the sun appears to move across the sky in a clockwise directon.
## Is clockwise rotation positive or negative?
Counterclockwise is the positive rotation direction and clockwise is the negative direction. For example, a torque that rotates an object counterclockwise is a positive torque (see figure 6 below).
Is clockwise tighten or loose?
The easiest way to remember which direction tightens and which one loosens is the old axiom “righty-tighty and lefty-loosey.” This means that turning most threaded things right, or clockwise, tightens them (righty-tighty) and turning them to the left, or counterclockwise, loosens them (lefty-loosey).
### Which way is tighten and loosen?
Typical nuts, screws, bolts, bottle caps, and jar lids are tightened (moved away from the observer) clockwise and loosened (moved towards the observer) counterclockwise in accordance with the right-hand rule.
### What does clockwise mean?
: in the direction in which the hands of a clock rotate as viewed from in front or as if standing on a clock face.
Which way do you turn an allen key to loosen?
Taking care to use the correct size Allen wrench, use the short arm in the socket, rotating counterclockwise unless the screw has left-hand threads, to break the screw loose if it’s very tight.
#### How can I open my Allen key without key?
1. Find a bolt whose head is the required size. Insert the bolt head half-way into the socket.
2. Find a bolt whose head is the required size. Turn two nuts on and counter-tighten them (jam them together).
3. You can get a decent set of hex wrenches for \$10 – \$15; possibly much less if you wait for a sale.
#### How do you remove a rounded nut from a tight space?
Best Methods for Removing Rounded Nuts & Bolts in Tight Spaces:
1. Use a Damaged Bolt Remover Socket Set– the best is the Irwin Damaged Extractor Set on Amazon.
2. Hammer on a smaller socket and remove the rounded bolt.
3. Use Vice Grips and grab the fastener as tight as possible and remove normally.
How is a socket attached to a ratchet?
The socket is placed on the square extension or driver of the ratchet. Once it’s secured to the ratchet the user can loosen or tighten a bolt or nut while the socket remains in place by just turning the handle of the ratchet.
## How do you remove a nut with a rounded head?
Begin with an oscillating tool to cut a groove across the top of the rounded bolt head. Place a flat, screwdriver tip on the impact wrench. Push down on the impact wrench, and pull the trigger to back out the bolt.
## Which wrench is used to loosen hex fasteners?
Flare Nut Wrench
What is another name for a hex wrench?
A hex key, also known as an Allen key or Allen wrench, is a small handheld tool that’s used for for driving bolts and screws with a hexagonal socket. They are available in many different sizes, though they all have the same hexagonal-shaped tip.
### What are the loosening and tightening tools?
Maintenance Tools for Tightening and Loosening
Tool Name Description
Adjustable wrenches with scale Adjustable wrenches with scale to tighten wide range of nuts.
Hook wrenches with square nose for ring nuts Hook wrenches with square nose used mainly to loose and tighten bearing locking nut.
### Can I use pliers instead of a wrench?
Although pliers can be used to tighten and loosen nuts, this is a task better suited to the wrench. Pliers tend to damage the surfaces of fasteners and are usually more difficult to use than a properly-sized wrench.
What to use when you dont have a wrench?
What to Use if You Don’t Have a Wrench
1. Duck tape. Duct tape is useful in almost any situation but you may be surprised to learn that you can use it to loosen bolts.
2. Two coins. Who would think that money can be used as a makeshift tool?
3. Zip-tie.
4. Another nut and bolt.
#### What’s the difference between a socket wrench and a ratchet?
A socket wrench is a type of wrench (spanner in British English) that inserts into a socket to turn a fastener, typically in the form of a nut or bolt. The most prevalent form is the ratcheting socket wrench, often informally called a ratchet.
#### What do you do if you don’t have pliers?
If you need pliers to grip onto something large, you can try a pair of stainless-steel kitchen tongs, although you may struggle with torque. If you think you can manage it with your hands but need a better grip, try using a silicone kitchen mitt or one of those jar-opening pads.
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# Base Ten Blocks
## BASE TEN BLOCKS
### PROPORTIONAL MANIPULATIVES HELP STUDENTS DEVELOP CONCEPTUAL UNDERSTANDING
Base Ten Blocks provide hands-on ways to learn place value, number concepts, operations, measurement, and much more! They help students physically represent what they’re learning so they can develop a deeper understanding of the meaning of each concept. By building number combinations with Base Ten Blocks, students ease into the concept of regrouping or trading, and can see the logical development of each operation.
### SKILLS
• Multiplication and Division
• Number sense
• Place Value
• Counting
• Decimals
## Components
### Units
The unit represents the numeral 1.
Measures 1 cm x 1 cm x 1 cm.
### Rods
The rod represents the numeral 10.
Measures 1 cm x 1 cm x 10 cm
### Flats
The flat represents the numeral 100.
Measures 1 cm x 10 cm x 10 cm.
### Cubes
The cube represents the numeral 1,000.
Measures 10 cm x 10 cm x 10 cm.
### Sets
Starter, Place Value, and Class Sets are available.
### Accessories
A wide variety of accessories are available.
## Learn About Base Ten Blocks
Base Ten Blocks provide a spatial model of our base ten number system. The smallest blocks—cubes that measure 1 cm on a side—are called units. The long, narrow blocks that measure 1 cm by 1 cm by 10 cm are called rods. The flat, square blocks that measure 1 cm by 10 cm by 10 cm are called flats. The largest blocks available that measure 10 cm on a side, are called cubes. When working with base ten place value experiences, we commonly use the unit to represent ones, the rod to represent tens, the flat to represent hundreds, and the cube to represent thousands. Providing names based on the shape rather than on the value allows for the pieces to be renamed when necessary. For example, when studying decimals, a class can use the flat to represent a unit and establish the value of the other pieces from there.
The size relationships among the blocks make them ideal for the investigation of number concepts. Initially students should explore independently with Base Ten Blocks before engaging in structured activities. As they move the blocks around to create designs and build structures, they may be able to discover on their own that it takes 10 of a smaller block to make one of the next larger blocks. Students' designs and structures also lead them to employ spatial visualization and to work intuitively with the geometric concepts of shape, perimeter, area, and volume.
Base Ten Blocks are especially useful in providing students with ways to physically represent the concepts of place value and addition, subtraction, multiplication, and division of whole numbers. By building number combinations with Base Ten Blocks, students ease into the concept of regrouping, or trading, and can see the logical development of each operation. The blocks provide a visual foundation and understanding of the algorithms students use when doing paper-and-pencil computation. Older students can transfer their understanding of whole numbers and whole-number operations to an understanding of decimals and decimal operations.
Place-value mats, available in pads of 50, provide a means for students to organize their work as they explore the relationships among the blocks and determine how groups of blocks can be used to represent numbers. Students may begin by placing unit blocks, one at a time, in the ones column on a mat. For each unit they place, they record the number corresponding to the total number of units placed (1, 2, 3...). They continue this process until they have accumulated 10 units, at which point they match their 10 units to 1 rod and trade those units for the rod, which they place in the tens column. Students continue in the same way, adding units one at a time to the ones column and recording the totals (11, 12, 13...) until it is time to trade for a second rod, which they place in the tens column (20). When they finally come to 99, there are 9 units and 9 rods on the mat. Adding one more unit forces two trades: first 10 units for another rod and then 10 rods for a flat (100). Then it is time to continue adding and recording units and making trades as needed as students work their way through the hundreds and up to the thousands. Combining the placing and trading of rods with the act of recording the corresponding numbers provides students with a connection between concrete and symbolic representations of numbers.
Base Ten Blocks can be used to develop an understanding of the meaning of addition, subtraction, multiplication, and division. Modeling addition on a place-value mat provides students with a visual basis for the concept of regrouping.
Subtraction with regrouping involves trading some of the blocks for smaller blocks of equal value so that the "taking away" can be accomplished. For example, to subtract 15 from 32, a student would trade one of the rods that represent 32 (3 rods and 2 units) for 10 units to form an equivalent representation of 32 (2 rods and 12 units). Then the student would take away 15 (1 rod and 5 units) and be left with a difference of 17 (1 rod and 7 units).
Multiplication can be modeled as repeated addition or with rectangular arrays. Using rectangular arrays can help in understanding the derivation of the partial products, the sum of which is the total product.
Division can be done as repeated subtraction or through building and analyzing rectangular arrays.
By letting the cube, flat, rod, and unit represent 1, 0.1, 0.01, and 0.001, respectively, older students can explore and develop decimal concepts, compare decimals, and perform basic operations with decimal numbers.
The squares along each face make the blocks excellent tools for visualizing and internalizing the concepts of perimeter and surface area of structures. Counting unit blocks in a structure can form the basis for understanding and finding volume.
Base Ten Blocks provide a perfect opportunity for authentic assessment. Watching students work with the blocks gives you a sense of how they approach a mathematical problem. Their thinking can be "seen" through their positioning of the blocks. When a class breaks up into small working groups, you can circulate, listen, and raise questions, all the while focusing on how individuals are thinking.
The challenges that students encounter when working with Base Ten Blocks often elicit unexpected abilities from those whose performance in more symbolic, number-oriented tasks may be weak. On the other hand, some students with good memories for numerical relationships have difficulty with spatial challenges and can more readily learn from freely exploring with Base Ten Blocks. Thus, by observing student's free exploration, you can get a sense of individual styles and intellectual strengths.
Having students describe their creations and share their strategies and thinking with the whole class gives you another opportunity for observational assessment. Furthermore, you may want to gather student's recorded work or invite them to choose pieces to add to their math portfolios.
"These Base Ten Blocks are great and allowed my students to see the difference in place value between hundreds, tens, and ones."
- Diva S. Aurora, CO
## Lessons
Base Ten Blocks provide students with a physical model of our place value system of numeration. By using Base Ten Blocks, students develop an understanding of the processes of addition and subtraction. Their understanding can then be connected to abstract symbolism.
The names of unit, rod, and flat are used to describe the blocks interchangeably with one, ten, and hundred. This is to keep the emphasis on the concrete blocks. While the word ten is likely to trigger the symbolic image of 10, the 1 and the 0, the word "rod" describes the materials without the interference of the symbols.
As with other materials, Base Ten Blocks allow students the chance to build with the blocks before requiring their attention to a specific activity. Students may or may not discover the relationship that 10 of one block makes 10 of the next larger block. Make sure you establish this relationship before starting other instruction.
Here are three activities for getting to know Base Ten Blocks and using them to learn and understand the math that they represent.
How Many Can you Hold? Sum It Up! In this activity, students estimate and then count the number of Base Ten units they can hold in their two hands. Students use Base Ten Blocks to model a number as the sum of two addends. Then they find ways to model the same number with different pairs of addends. Activity Curriculum Strands Topics Number Measurement Counting Estimation Race for a Flat In this game for two pairs of students, players take turns rolling number cubes and finding the sums of the numbers rolled. They use Base Ten Blocks to represent the sums in an effort to be the first to accumulate blocks with a total value of 100. Number Measurement Counting Place, Value Addition Number Counting Place, Value Addition
Working with concrete materials, even after learning abstract procedures, provides students the opportunity to experience a concept in a new way. Though students may be able to solve problems with paper and pencil, perhaps even automatically, being asked to use the Base Ten Blocks requires them to reason in another way. This is beneficial for deepening their understanding and developing their flexibility.
Using Base Ten Blocks during instruction is based on the approach that students’ understanding of a concept is best developed concretely. Once a student demonstrates concrete understanding, the teacher connects that concrete experience to standard symbolism. Students then use abstract symbolism in conjunction with the manipulatives.
It is important when approaching instruction in this way that enough time is provided to develop the concept concretely without the interference of the symbolism. Time used at this level to develop a solid base of understanding is well spent. Allow students to use the materials for as long as they need to. When you encourage them to try the written algorithm without the material, also encourage them to visualize the blocks. It is helpful for them to connect what they write to what they can actually do with the blocks.
Here are three activities using Base Ten Blocks in different contexts.
What Amounts? Modeling Multiplication Race for a Whole In this activity, students look for ways to use a combination of 4 Base Ten Blocks to model as many different numbers as possible. Students use Base Ten Blocks to build rectangular arrays that model the multiplication of two, 2-digit numbers. In this game for two to four players, students roll number cubes to indicate numbers of tenths and hundredths. They collect Base Ten rods and units to represent the tenths and hundredths, respectively, in an effort to be the first to accumulate blocks with a total value of 1. Activity Curriculum Strands Topics Number Place Value, Number Sense Number Measurement Number Measurement, Spatial Visualization, Properties of Rectangles, Area Number Probability & Statistics Number Probability & Statistics, Decimals, Place Value, Addition Game Strategies
Base Ten blocks help students make connections to decimal operations and operations with algebraic expressions in the middle grades. The proportional relationships among the pieces can be used to develop an understanding of the number system and ratio/proportion. In addition, the pieces can be used to model solid figures and develop understanding of volume and surface area.
Here are two activities for using Base Ten Blocks with middle grade students.
Building Boxes Students use Base Ten Blocks to build as many different rectangular prisms as they can from eight rods. Activity Curriculum Strands Topics Number, Measurement, Geometry Multiplication, Volume, Spatial Visualization Double the Dimensions Students use Base Ten Blocks to design and build structures. They determine the volume and surface area of their structures and then predict how these will change when they "double" their structures. Place Value Number Sense Volume, Surface Area, Spatial Visualization
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# Chapter 9 Two-Sample Problems
The previous two chapters treated the questions of estimating and making inferences about a parameter of a single population. In this chapter we consider a comparison of parameters that belong to two different populations. For example, we might wish to compare the average income of all adults in one region of the country with the average income of those in another region, or we might wish to compare the proportion of all men who are vegetarians with the proportion of all women who are vegetarians.
We will study construction of confidence intervals and tests of hypotheses in four situations, depending on the parameter of interest, the sizes of the samples drawn from each of the populations, and the method of sampling. We also examine sample size considerations.
## 9.1 Comparison of Two Population Means: Large, Independent Samples
### Learning Objectives
1. To understand the logical framework for estimating the difference between the means of two distinct populations and performing tests of hypotheses concerning those means.
2. To learn how to construct a confidence interval for the difference in the means of two distinct populations using large, independent samples.
3. To learn how to perform a test of hypotheses concerning the difference between the means of two distinct populations using large, independent samples.
Suppose we wish to compare the means of two distinct populations. Figure 9.1 "Independent Sampling from Two Populations" illustrates the conceptual framework of our investigation in this and the next section. Each population has a mean and a standard deviation. We arbitrarily label one population as Population 1 and the other as Population 2, and subscript the parameters with the numbers 1 and 2 to tell them apart. We draw a random sample from Population 1 and label the sample statistics it yields with the subscript 1. Without reference to the first sample we draw a sample from Population 2 and label its sample statistics with the subscript 2.
Figure 9.1 Independent Sampling from Two Populations
### Definition
Samples from two distinct populations are independent if each one is drawn without reference to the other, and has no connection with the other.
Our goal is to use the information in the samples to estimate the difference $μ1−μ2$ in the means of the two populations and to make statistically valid inferences about it.
## Confidence Intervals
Since the mean $x-1$ of the sample drawn from Population 1 is a good estimator of $μ1$ and the mean $x-2$ of the sample drawn from Population 2 is a good estimator of $μ2$, a reasonable point estimate of the difference $μ1−μ2$ is $x-1−x-2.$ In order to widen this point estimate into a confidence interval, we first suppose that both samples are large, that is, that both $n1≥30$ and $n2≥30.$ If so, then the following formula for a confidence interval for $μ1−μ2$ is valid. The symbols $s12$ and $s22$ denote the squares of s1 and s2. (In the relatively rare case that both population standard deviations $σ1$ and $σ2$ are known they would be used instead of the sample standard deviations.)
### $100(1−α)%$ Confidence Interval for the Difference Between Two Population Means: Large, Independent Samples
$(x-1−x-2)±zα∕2s12n1+s22n2$
The samples must be independent, and each sample must be large: $n1≥30$ and $n2≥30.$
### Example 1
To compare customer satisfaction levels of two competing cable television companies, 174 customers of Company 1 and 355 customers of Company 2 were randomly selected and were asked to rate their cable companies on a five-point scale, with 1 being least satisfied and 5 most satisfied. The survey results are summarized in the following table:
Company 1 Company 2
$n1=174$ $n2=355$
$x-1=3.51$ $x-2=3.24$
$s1=0.51$ $s2=0.52$
Construct a point estimate and a 99% confidence interval for $μ1−μ2$, the difference in average satisfaction levels of customers of the two companies as measured on this five-point scale.
Solution:
The point estimate of $μ1−μ2$ is
$x-1−x-2=3.51−3.24=0.27.$
In words, we estimate that the average customer satisfaction level for Company 1 is 0.27 points higher on this five-point scale than it is for Company 2.
To apply the formula for the confidence interval, proceed exactly as was done in Chapter 7 "Estimation". The 99% confidence level means that $α=1−0.99=0.01$ so that $zα∕2=z0.005.$ From Figure 12.3 "Critical Values of " we read directly that $z0.005=2.576.$ Thus
$(x-1−x-2)±zα∕2s12n1+s22n2=0.27±2.5760.512174+0.522355=0.27±0.12$
We are 99% confident that the difference in the population means lies in the interval $[0.15,0.39]$, in the sense that in repeated sampling 99% of all intervals constructed from the sample data in this manner will contain $μ1−μ2.$ In the context of the problem we say we are 99% confident that the average level of customer satisfaction for Company 1 is between 0.15 and 0.39 points higher, on this five-point scale, than that for Company 2.
## Hypothesis Testing
Hypotheses concerning the relative sizes of the means of two populations are tested using the same critical value and p-value procedures that were used in the case of a single population. All that is needed is to know how to express the null and alternative hypotheses and to know the formula for the standardized test statistic and the distribution that it follows.
The null and alternative hypotheses will always be expressed in terms of the difference of the two population means. Thus the null hypothesis will always be written
$H0:μ1−μ2=D0$
where D0 is a number that is deduced from the statement of the situation. As was the case with a single population the alternative hypothesis can take one of the three forms, with the same terminology:
Form of $Ha$ Terminology
$Ha:μ1−μ2 Left-tailed
$Ha:μ1−μ2>D0$ Right-tailed
$Ha:μ1−μ2≠D0$ Two-tailed
As long as the samples are independent and both are large the following formula for the standardized test statistic is valid, and it has the standard normal distribution. (In the relatively rare case that both population standard deviations $σ1$ and $σ2$ are known they would be used instead of the sample standard deviations.)
### Standardized Test Statistic for Hypothesis Tests Concerning the Difference Between Two Population Means: Large, Independent Samples
$Z=(x-1−x-2)−D0s12n1+s22n2$
The test statistic has the standard normal distribution.
The samples must be independent, and each sample must be large: $n1≥30$ and $n2≥30.$
### Example 2
Refer to Note 9.4 "Example 1" concerning the mean satisfaction levels of customers of two competing cable television companies. Test at the 1% level of significance whether the data provide sufficient evidence to conclude that Company 1 has a higher mean satisfaction rating than does Company 2. Use the critical value approach.
Solution:
• Step 1. If the mean satisfaction levels $μ1$ and $μ2$ are the same then $μ1=μ2$, but we always express the null hypothesis in terms of the difference between $μ1$ and $μ2$, hence H0 is $μ1−μ2=0.$ To say that the mean customer satisfaction for Company 1 is higher than that for Company 2 means that $μ1>μ2$, which in terms of their difference is $μ1−μ2>0.$ The test is therefore
$H0:μ1−μ2=0 vs. Ha:μ1−μ2>0@ α=0.01$
• Step 2. Since the samples are independent and both are large the test statistic is
$Z=(x-1−x-2)−D0s12n1+s22n2$
• Step 3. Inserting the data into the formula for the test statistic gives
$Z=(x-1−x-2)−D0s12n1+s22n2=(3.51−3.24)−00.512174+0.522355=5.684$
• Step 4. Since the symbol in Ha is “>” this is a right-tailed test, so there is a single critical value, $zα=z0.01$, which from the last line in Figure 12.3 "Critical Values of " we read off as 2.326. The rejection region is $[2.326,∞).$
Figure 9.2 Rejection Region and Test Statistic for Note 9.6 "Example 2"
• Step 5. As shown in Figure 9.2 "Rejection Region and Test Statistic for " the test statistic falls in the rejection region. The decision is to reject H0. In the context of the problem our conclusion is:
The data provide sufficient evidence, at the 1% level of significance, to conclude that the mean customer satisfaction for Company 1 is higher than that for Company 2.
### Example 3
Perform the test of Note 9.6 "Example 2" using the p-value approach.
Solution:
The first three steps are identical to those in Note 9.6 "Example 2".
• Step 4. The observed significance or p-value of the test is the area of the right tail of the standard normal distribution that is cut off by the test statistic Z = 5.684. The number 5.684 is too large to appear in Figure 12.2 "Cumulative Normal Probability", which means that the area of the left tail that it cuts off is 1.0000 to four decimal places. The area that we seek, the area of the right tail, is therefore $1−1.0000=0.0000$ to four decimal places. See Figure 9.3. That is, $p -value=0.0000$ to four decimal places. (The actual value is approximately $0.000 000 007.$)
Figure 9.3 P-Value for Note 9.7 "Example 3"
• Step 5. Since 0.0000 < 0.01, $p -value<α$ so the decision is to reject the null hypothesis:
The data provide sufficient evidence, at the 1% level of significance, to conclude that the mean customer satisfaction for Company 1 is higher than that for Company 2.
### Key Takeaways
• A point estimate for the difference in two population means is simply the difference in the corresponding sample means.
• In the context of estimating or testing hypotheses concerning two population means, “large” samples means that both samples are large.
• A confidence interval for the difference in two population means is computed using a formula in the same fashion as was done for a single population mean.
• The same five-step procedure used to test hypotheses concerning a single population mean is used to test hypotheses concerning the difference between two population means. The only difference is in the formula for the standardized test statistic.
### Basic
1. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 90% confidence,
$n1=45$, $x-1=27$, $s1=2$
$n2=60$, $x-2=22$, $s2=3$
2. 99% confidence,
$n1=30$, $x-1=−112$, $s1=9$
$n2=40$, $x-2=−98$, $s2=4$
2. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 95% confidence,
$n1=110$, $x-1=77$, $s1=15$
$n2=85$, $x-2=79$, $s2=21$
2. 90% confidence,
$n1=65$, $x-1=−83$, $s1=12$
$n2=65$, $x-2=−74$, $s2=8$
3. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 99.5% confidence,
$n1=130$, $x-1=27.2$, $s1=2.5$
$n2=155$, $x-2=38.8$, $s2=4.6$
2. 95% confidence,
$n1=68$, $x-1=215.5$, $s1=12.3$
$n2=84$, $x-2=287.8$, $s2=14.1$
4. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 99.9% confidence,
$n1=275$, $x-1=70.2$, $s1=1.5$
$n2=325$, $x-2=63.4$, $s2=1.1$
2. 90% confidence,
$n1=120$, $x-1=35.5$, $s1=0.75$
$n2=146$, $x-2=29.6$, $s2=0.80$
5. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
1. Test $H0:μ1−μ2=3$ vs. $Ha:μ1−μ2≠3$ @ $α=0.05$,
$n1=35$, $x-1=25$, $s1=1$
$n2=45$, $x-2=19$, $s2=2$
2. Test $H0:μ1−μ2=−25$ vs. $Ha:μ1−μ2<−25$ @ $α=0.10$,
$n1=85$, $x-1=188$, $s1=15$
$n2=62$, $x-2=215$, $s2=19$
6. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
1. Test $H0:μ1−μ2=45$ vs. $Ha:μ1−μ2>45$ @ $α=0.001$,
$n1=200$, $x-1=1312$, $s1=35$
$n2=225$, $x-2=1256$, $s2=28$
2. Test $H0:μ1−μ2=−12$ vs. $Ha:μ1−μ2≠−12$ @ $α=0.10$,
$n1=35$, $x-1=121$, $s1=6$
$n2=40$, $x-2=135$, $s2=7$
7. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
1. Test $H0:μ1−μ2=0$ vs. $Ha:μ1−μ2≠0$ @ $α=0.01$,
$n1=125$, $x-1=−46$, $s1=10$
$n2=90$, $x-2=−50$, $s2=13$
2. Test $H0:μ1−μ2=20$ vs. $Ha:μ1−μ2>20$ @ $α=0.05$,
$n1=40$, $x-1=142$, $s1=11$
$n2=40$, $x-2=118$, $s2=10$
8. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach. Compute the p-value of the test as well.
1. Test $H0:μ1−μ2=13$ vs. $Ha:μ1−μ2<13$ @ $α=0.01$,
$n1=35$, $x-1=100$, $s1=2$
$n2=35$, $x-2=88$, $s2=2$
2. Test $H0:μ1−μ2=−10$ vs. $Ha:μ1−μ2≠−10$ @ $α=0.10$,
$n1=146$, $x-1=62$, $s1=4$
$n2=120$, $x-2=73$, $s2=7$
9. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
1. Test $H0:μ1−μ2=57$ vs. $Ha:μ1−μ2<57$ @ $α=0.10$,
$n1=117$, $x-1=1309$, $s1=42$
$n2=133$, $x-2=1258$, $s2=37$
2. Test $H0:μ1−μ2=−1.5$ vs. $Ha:μ1−μ2≠−1.5$ @ $α=0.20$,
$n1=65$, $x-1=16.9$, $s1=1.3$
$n2=57$, $x-2=18.6$, $s2=1.1$
10. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
1. Test $H0:μ1−μ2=−10.5$ vs. $Ha:μ1−μ2>−10.5$ @ $α=0.01$,
$n1=64$, $x-1=85.6$, $s1=2.4$
$n2=50$, $x-2=95.3$, $s2=3.1$
2. Test $H0:μ1−μ2=110$ vs. $Ha:μ1−μ2≠110$ @ $α=0.02$,
$n1=176$, $x-1=1918$, $s1=68$
$n2=241$, $x-2=1782$, $s2=146$
11. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
1. Test $H0:μ1−μ2=50$ vs. $Ha:μ1−μ2>50$ @ $α=0.005$,
$n1=72$, $x-1=272$, $s1=26$
$n2=103$, $x-2=213$, $s2=14$
2. Test $H0:μ1−μ2=7.5$ vs. $Ha:μ1−μ2≠7.5$ @ $α=0.10$,
$n1=52$, $x-1=94.3$, $s1=2.6$
$n2=38$, $x-2=88.6$, $s2=8.0$
12. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach.
1. Test $H0:μ1−μ2=23$ vs. $Ha:μ1−μ2<23$ @ $α=0.20$,
$n1=314$, $x-1=198$, $s1=12.2$
$n2=220$, $x-2=176$, $s2=11.5$
2. Test $H0:μ1−μ2=4.4$ vs. $Ha:μ1−μ2≠4.4$ @ $α=0.05$,
$n1=32$, $x-1=40.3$, $s1=0.5$
$n2=30$, $x-2=35.5$, $s2=0.7$
### Applications
1. In order to investigate the relationship between mean job tenure in years among workers who have a bachelor’s degree or higher and those who do not, random samples of each type of worker were taken, with the following results.
n $x-$ s
Bachelor’s degree or higher 155 5.2 1.3
No degree 210 5.0 1.5
1. Construct the 99% confidence interval for the difference in the population means based on these data.
2. Test, at the 1% level of significance, the claim that mean job tenure among those with higher education is greater than among those without, against the default that there is no difference in the means.
3. Compute the observed significance of the test.
2. Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years.
1. Construct the 95% confidence interval for the difference in the means based on these data.
2. Test the hypothesis that there is a difference in the means against the null hypothesis that there is no difference. Use the 1% level of significance.
3. Compute the observed significance of the test in part (b).
3. In previous years the average number of patients per hour at a hospital emergency room on weekends exceeded the average on weekdays by 6.3 visits per hour. A hospital administrator believes that the current weekend mean exceeds the weekday mean by fewer than 6.3 hours.
1. Construct the 99% confidence interval for the difference in the population means based on the following data, derived from a study in which 30 weekend and 30 weekday one-hour periods were randomly selected and the number of new patients in each recorded.
n $x-$ s
Weekends 30 13.8 3.1
Weekdays 30 8.6 2.7
2. Test at the 5% level of significance whether the current weekend mean exceeds the weekday mean by fewer than 6.3 patients per hour.
3. Compute the observed significance of the test.
4. A sociologist surveys 50 randomly selected citizens in each of two countries to compare the mean number of hours of volunteer work done by adults in each. Among the 50 inhabitants of Lilliput, the mean hours of volunteer work per year was 52, with standard deviation 11.8. Among the 50 inhabitants of Blefuscu, the mean number of hours of volunteer work per year was 37, with standard deviation 7.2.
1. Construct the 99% confidence interval for the difference in mean number of hours volunteered by all residents of Lilliput and the mean number of hours volunteered by all residents of Blefuscu.
2. Test, at the 1% level of significance, the claim that the mean number of hours volunteered by all residents of Lilliput is more than ten hours greater than the mean number of hours volunteered by all residents of Blefuscu.
3. Compute the observed significance of the test in part (b).
5. A university administrator asserted that upperclassmen spend more time studying than underclassmen.
1. Test this claim against the default that the average number of hours of study per week by the two groups is the same, using the following information based on random samples from each group of students. Test at the 1% level of significance.
n $x-$ s
Upperclassmen 35 15.6 2.9
Underclassmen 35 12.3 4.1
2. Compute the observed significance of the test.
6. An kinesiologist claims that the resting heart rate of men aged 18 to 25 who exercise regularly is more than five beats per minute less than that of men who do not exercise regularly. Men in each category were selected at random and their resting heart rates were measured, with the results shown.
n $x-$ s
Regular exercise 40 63 1.0
No regular exercise 30 71 1.2
1. Perform the relevant test of hypotheses at the 1% level of significance.
2. Compute the observed significance of the test.
7. Children in two elementary school classrooms were given two versions of the same test, but with the order of questions arranged from easier to more difficult in Version A and in reverse order in Version B. Randomly selected students from each class were given Version A and the rest Version B. The results are shown in the table.
n $x-$ s
Version A 31 83 4.6
Version B 32 78 4.3
1. Construct the 90% confidence interval for the difference in the means of the populations of all children taking Version A of such a test and of all children taking Version B of such a test.
2. Test at the 1% level of significance the hypothesis that the A version of the test is easier than the B version (even though the questions are the same).
3. Compute the observed significance of the test.
8. The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the city center that departs at 8:15 a.m. or the one that departs at 8:30 a.m. The following sample statistics are assembled by the Transit Authority.
n $x-$ s
8:15 a.m. train 30 323 41
8:30 a.m. train 45 356 45
1. Construct the 90% confidence interval for the difference in the mean number of daily travellers on the 8:15 train and the mean number of daily travellers on the 8:30 train.
2. Test at the 5% level of significance whether the data provide sufficient evidence to conclude that more passengers ride the 8:30 train.
3. Compute the observed significance of the test.
9. In comparing the academic performance of college students who are affiliated with fraternities and those male students who are unaffiliated, a random sample of students was drawn from each of the two populations on a university campus. Summary statistics on the student GPAs are given below.
n $x-$ s
Fraternity 645 2.90 0.47
Unaffiliated 450 2.88 0.42
Test, at the 5% level of significance, whether the data provide sufficient evidence to conclude that there is a difference in average GPA between the population of fraternity students and the population of unaffiliated male students on this university campus.
10. In comparing the academic performance of college students who are affiliated with sororities and those female students who are unaffiliated, a random sample of students was drawn from each of the two populations on a university campus. Summary statistics on the student GPAs are given below.
n $x-$ s
Sorority 330 3.18 0.37
Unaffiliated 550 3.12 0.41
Test, at the 5% level of significance, whether the data provide sufficient evidence to conclude that there is a difference in average GPA between the population of sorority students and the population of unaffiliated female students on this university campus.
11. The owner of a professional football team believes that the league has become more offense oriented since five years ago. To check his belief, 32 randomly selected games from one year’s schedule were compared to 32 randomly selected games from the schedule five years later. Since more offense produces more points per game, the owner analyzed the following information on points per game (ppg).
n $x-$ s
ppg previously 32 20.62 4.17
ppg recently 32 22.05 4.01
Test, at the 10% level of significance, whether the data on points per game provide sufficient evidence to conclude that the game has become more offense oriented.
12. The owner of a professional football team believes that the league has become more offense oriented since five years ago. To check his belief, 32 randomly selected games from one year’s schedule were compared to 32 randomly selected games from the schedule five years later. Since more offense produces more offensive yards per game, the owner analyzed the following information on offensive yards per game (oypg).
n $x-$ s
oypg previously 32 316 40
oypg recently 32 336 35
Test, at the 10% level of significance, whether the data on offensive yards per game provide sufficient evidence to conclude that the game has become more offense oriented.
### Large Data Set Exercises
1. Large Data Sets 1A and 1B list the SAT scores for 1,000 randomly selected students. Denote the population of all male students as Population 1 and the population of all female students as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data1A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data1B.xls
1. Restricting attention to just the males, find n1, $x-1$, and s1. Restricting attention to just the females, find n2, $x-2$, and s2.
2. Let $μ1$ denote the mean SAT score for all males and $μ2$ the mean SAT score for all females. Use the results of part (a) to construct a 90% confidence interval for the difference $μ1−μ2.$
3. Test, at the 5% level of significance, the hypothesis that the mean SAT scores among males exceeds that of females.
2. Large Data Sets 1A and 1B list the GPAs for 1,000 randomly selected students. Denote the population of all male students as Population 1 and the population of all female students as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data1A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data1B.xls
1. Restricting attention to just the males, find n1, $x-1$, and s1. Restricting attention to just the females, find n2, $x-2$, and s2.
2. Let $μ1$ denote the mean GPA for all males and $μ2$ the mean GPA for all females. Use the results of part (a) to construct a 95% confidence interval for the difference $μ1−μ2.$
3. Test, at the 10% level of significance, the hypothesis that the mean GPAs among males and females differ.
3. Large Data Sets 7A and 7B list the survival times for 65 male and 75 female laboratory mice with thymic leukemia. Denote the population of all such male mice as Population 1 and the population of all such female mice as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data7A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data7B.xls
1. Restricting attention to just the males, find n1, $x-1$, and s1. Restricting attention to just the females, find n2, $x-2$, and s2.
2. Let $μ1$ denote the mean survival for all males and $μ2$ the mean survival time for all females. Use the results of part (a) to construct a 99% confidence interval for the difference $μ1−μ2.$
3. Test, at the 1% level of significance, the hypothesis that the mean survival time for males exceeds that for females by more than 182 days (half a year).
4. Compute the observed significance of the test in part (c).
### Answers
1. $(4.20,5.80)$,
2. $(−18.54,−9.46)$
1. $(−12.81,−10.39)$,
2. $(−76.50,−68.10)$
1. Z = 8.753, $±z0.025=±1.960$, reject H0, p-value = 0.0000;
2. $Z=−0.687$, $−z0.10=−1.282$, do not reject H0, p-value = 0.2451
1. Z = 2.444, $±z0.005=±2.576$, do not reject H0, p-value = 0.0146.
2. Z = 1.702, $z0.05=1.645$, reject H0, p-value = 0.0446
1. $Z=−1.19$, p-value = 0.1170, do not reject H0;
2. $Z=−0.92$, p-value = 0.3576, do not reject H0
1. Z = 2.68, p-value = 0.0037, reject H0;
2. $Z=−1.34$, p-value = 0.1802, do not reject H0
1. $0.2±0.4$,
2. Z = 1.360, $z0.01=2.326$, do not reject H0 (not greater)
3. p-value = 0.0869
1. $5.2±1.9$,
2. $Z=−1.466$, $−z0.050=−1.645$, do not reject H0 (exceeds by 6.3 or more)
3. p-value = 0.0708
1. Z = 3.888, $z0.01=2.326$, reject H0 (upperclassmen study more)
2. p-value = 0.0001
1. $5±1.8$,
2. Z = 4.454, $z0.01=2.326$, reject H0 (Test A is easier)
3. p-value = 0.0000
1. Z = 0.738, $±z0.025=±1.960$, do not reject H0 (no difference)
2. $Z=−1.398$, $−z0.10=−1.282$, reject H0 (more offense oriented)
1. $n1=419$, $x-1=1540.33$, $s1=205.40$, $n2=581$, $x-2=1520.38$, and $s2=217.34.$
2. $(−2.24,42.15)$
3. $H0:μ1−μ2=0$ vs. $Ha:μ1−μ2>0.$ Test Statistic: Z = 1.48. Rejection Region: $[1.645,∞).$ Decision: Fail to reject H0.
1. $n1=65$, $x-1=665.97$, $s1=41.60$, $n2=75$, $x-2=455.89$, and $s2=63.22.$
2. $(187.06,233.09)$
3. $H0:μ1−μ2=182$ vs. $Ha:μ1−μ2>182.$ Test Statistic: Z = 3.14. Rejection Region: $[2.33,∞).$ Decision: Reject H0.
4. $p−value=0.0008$
## 9.2 Comparison of Two Population Means: Small, Independent Samples
### Learning Objectives
1. To learn how to construct a confidence interval for the difference in the means of two distinct populations using small, independent samples.
2. To learn how to perform a test of hypotheses concerning the difference between the means of two distinct populations using small, independent samples.
When one or the other of the sample sizes is small, as is often the case in practice, the Central Limit Theorem does not apply. We must then impose conditions on the population to give statistical validity to the test procedure. We will assume that both populations from which the samples are taken have a normal probability distribution and that their standard deviations are equal.
## Confidence Intervals
When the two populations are normally distributed and have equal standard deviations, the following formula for a confidence interval for $μ1−μ2$ is valid.
### $100(1−α)%$ Confidence Interval for the Difference Between Two Population Means: Small, Independent Samples
$(x-1−x-2)±tα∕2sp2(1n1+1n2)wheresp2=(n1−1)s12+(n2−1)s22n1+n2−2$
The number of degrees of freedom is $df=n1+n2−2.$
The samples must be independent, the populations must be normal, and the population standard deviations must be equal. “Small” samples means that either $n1<30$ or $n2<30.$
The quantity $sp2$ is called the pooled sample variance. It is a weighted average of the two estimates $s12$ and $s22$ of the common variance $σ12=σ22$ of the two populations.
### Example 4
A software company markets a new computer game with two experimental packaging designs. Design 1 is sent to 11 stores; their average sales the first month is 52 units with sample standard deviation 12 units. Design 2 is sent to 6 stores; their average sales the first month is 46 units with sample standard deviation 10 units. Construct a point estimate and a 95% confidence interval for the difference in average monthly sales between the two package designs.
Solution:
The point estimate of $μ1−μ2$ is
$x-1−x-2=52−46=6$
In words, we estimate that the average monthly sales for Design 1 is 6 units more per month than the average monthly sales for Design 2.
To apply the formula for the confidence interval, we must find $tα∕2.$ The 95% confidence level means that α = 1 − 0.95 = 0.05 so that $tα∕2=t0.025.$ From Figure 12.3 "Critical Values of ", in the row with the heading df = 11 + 6 − 2 = 15 we read that $t0.025=2.131.$ From the formula for the pooled sample variance we compute
$sp2=(n1−1)s12+(n2−1)s22n1+n2−2=(10)(12)2+(5)(10)215=129.3-$
Thus
$(x-1−x-2)±tα∕2sp2(1n1+1n2)=6±(2.131)129.3-(111+16)≈6±12.3$
We are 95% confident that the difference in the population means lies in the interval $[−6.3,18.3]$, in the sense that in repeated sampling 95% of all intervals constructed from the sample data in this manner will contain $μ1−μ2.$ Because the interval contains both positive and negative values the statement in the context of the problem is that we are 95% confident that the average monthly sales for Design 1 is between 18.3 units higher and 6.3 units lower than the average monthly sales for Design 2.
## Hypothesis Testing
Testing hypotheses concerning the difference of two population means using small samples is done precisely as it is done for large samples, using the following standardized test statistic. The same conditions on the populations that were required for constructing a confidence interval for the difference of the means must also be met when hypotheses are tested.
### Standardized Test Statistic for Hypothesis Tests Concerning the Difference Between Two Population Means: Small, Independent Samples
$T=(x-1−x-2)−D0sp2(1n1+1n2)wheresp2=(n1−1)s12+(n2−1)s22n1+n2−2$
The test statistic has Student’s t-distribution with $df=n1+n2−2$ degrees of freedom.
The samples must be independent, the populations must be normal, and the population standard deviations must be equal. “Small” samples means that either $n1<30$ or $n2<30.$
### Example 5
Refer to Note 9.11 "Example 4" concerning the mean sales per month for the same computer game but sold with two package designs. Test at the 1% level of significance whether the data provide sufficient evidence to conclude that the mean sales per month of the two designs are different. Use the critical value approach.
Solution:
• Step 1. The relevant test is
$H0:μ1−μ2=0 vs. Ha:μ1−μ2≠0@ α=0.01$
• Step 2. Since the samples are independent and at least one is less than 30 the test statistic is
$T=(x-1−x-2)−D0sp2(1n1+1n2)$
which has Student’s t-distribution with $df=11+6−2=15$ degrees of freedom.
• Step 3. Inserting the data and the value $D0=0$ into the formula for the test statistic gives
$T=(x-1−x-2)−D0sp2(1n1+1n2)=(52−46)−0129.3-(111+16)=1.040$
• Step 4. Since the symbol in Ha is “≠” this is a two-tailed test, so there are two critical values, $±tα∕2=±t0.005.$ From the row in Figure 12.3 "Critical Values of " with the heading $df=15$ we read off $t0.005=2.947.$ The rejection region is $(−∞,−2.947]∪[2.947,∞).$
Figure 9.4 Rejection Region and Test Statistic for Note 9.13 "Example 5"
• Step 5. As shown in Figure 9.4 "Rejection Region and Test Statistic for " the test statistic does not fall in the rejection region. The decision is not to reject H0. In the context of the problem our conclusion is:
The data do not provide sufficient evidence, at the 1% level of significance, to conclude that the mean sales per month of the two designs are different.
### Example 6
Perform the test of Note 9.13 "Example 5" using the p-value approach.
Solution:
The first three steps are identical to those in Note 9.13 "Example 5".
• Step 4. Because the test is two-tailed the observed significance or p-value of the test is the double of the area of the right tail of Student’s t-distribution, with 15 degrees of freedom, that is cut off by the test statistic T = 1.040. We can only approximate this number. Looking in the row of Figure 12.3 "Critical Values of " headed $df=15$, the number 1.040 is between the numbers 0.866 and 1.341, corresponding to t0.200 and t0.100.
The area cut off by t = 0.866 is 0.200 and the area cut off by t = 1.341 is 0.100. Since 1.040 is between 0.866 and 1.341 the area it cuts off is between 0.200 and 0.100. Thus the p-value (since the area must be doubled) is between 0.400 and 0.200.
• Step 5. Since $p>0.200>0.01$, $p>α$, so the decision is not to reject the null hypothesis:
The data do not provide sufficient evidence, at the 1% level of significance, to conclude that the mean sales per month of the two designs are different.
### Key Takeaways
• In the context of estimating or testing hypotheses concerning two population means, “small” samples means that at least one sample is small. In particular, even if one sample is of size 30 or more, if the other is of size less than 30 the formulas of this section must be used.
• A confidence interval for the difference in two population means is computed using a formula in the same fashion as was done for a single population mean.
### Basic
In all exercises for this section assume that the populations are normal and have equal standard deviations.
1. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 95% confidence,
$n1=10$, $x-1=120$, $s1=2$
$n2=15$, $x-2=101$, $s2=4$
2. 99% confidence,
$n1=6$, $x-1=25$, $s1=1$
$n2=12$, $x-2=17$, $s2=3$
2. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 90% confidence,
$n1=28$, $x-1=212$, $s1=6$
$n2=23$, $x-2=198$, $s2=5$
2. 99% confidence,
$n1=14$, $x-1=68$, $s1=8$
$n2=20$, $x-2=43$, $s2=3$
3. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 99.9% confidence,
$n1=35$, $x-1=6.5$, $s1=0.2$
$n2=20$, $x-2=6.2$, $s2=0.1$
2. 99% confidence,
$n1=18$, $x-1=77.3$, $s1=1.2$
$n2=32$, $x-2=75.0$, $s2=1.6$
4. Construct the confidence interval for $μ1−μ2$ for the level of confidence and the data from independent samples given.
1. 99.5% confidence,
$n1=40$, $x-1=85.6$, $s1=2.8$
$n2=20$, $x-2=73.1$, $s2=2.1$
2. 99.9% confidence,
$n1=25$, $x-1=215$, $s1=7$
$n2=35$, $x-2=185$, $s2=12$
5. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach.
1. Test $H0:μ1−μ2=11$ vs. $Ha:μ1−μ2>11$ @ $α=0.025$,
$n1=6$, $x-1=32$, $s1=2$
$n2=11$, $x-2=19$, $s2=1$
2. Test $H0:μ1−μ2=26$ vs. $Ha:μ1−μ2≠26$ @ $α=0.05$,
$n1=17$, $x-1=166$, $s1=4$
$n2=24$, $x-2=138$, $s2=3$
6. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach.
1. Test $H0:μ1−μ2=40$ vs. $Ha:μ1−μ2<40$ @ $α=0.10$,
$n1=14$, $x-1=289$, $s1=11$
$n2=12$, $x-2=254$, $s2=9$
2. Test $H0:μ1−μ2=21$ vs. $Ha:μ1−μ2≠21$ @ $α=0.05$,
$n1=23$, $x-1=130$, $s1=6$
$n2=27$, $x-2=113$, $s2=8$
7. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach.
1. Test $H0:μ1−μ2=−15$ vs. $Ha:μ1−μ2<−15$ @ $α=0.10$,
$n1=30$, $x-1=42$, $s1=7$
$n2=12$, $x-2=60$, $s2=5$
2. Test $H0:μ1−μ2=103$ vs. $Ha:μ1−μ2≠103$ @ $α=0.10$,
$n1=17$, $x-1=711$, $s1=28$
$n2=32$, $x-2=598$, $s2=21$
8. Perform the test of hypotheses indicated, using the data from independent samples given. Use the critical value approach.
1. Test $H0:μ1−μ2=75$ vs. $Ha:μ1−μ2>75$ @ $α=0.025$,
$n1=45$, $x-1=674$, $s1=18$
$n2=29$, $x-2=591$, $s2=13$
2. Test $H0:μ1−μ2=−20$ vs. $Ha:μ1−μ2≠−20$ @ $α=0.005$,
$n1=30$, $x-1=137$, $s1=8$
$n2=19$, $x-2=166$, $s2=11$
9. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach. (The p-value can be only approximated.)
1. Test $H0:μ1−μ2=12$ vs. $Ha:μ1−μ2>12$ @ $α=0.01$,
$n1=20$, $x-1=133$, $s1=7$
$n2=10$, $x-2=115$, $s2=5$
2. Test $H0:μ1−μ2=46$ vs. $Ha:μ1−μ2≠46$ @ $α=0.10$,
$n1=24$, $x-1=586$, $s1=11$
$n2=27$, $x-2=535$, $s2=13$
10. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach. (The p-value can be only approximated.)
1. Test $H0:μ1−μ2=38$ vs. $Ha:μ1−μ2<38$ @ $α=0.01$,
$n1=12$, $x-1=464$, $s1=5$
$n2=10$, $x-2=432$, $s2=6$
2. Test $H0:μ1−μ2=4$ vs. $Ha:μ1−μ2≠4$ @ $α=0.005$,
$n1=14$, $x-1=68$, $s1=2$
$n2=17$, $x-2=67$, $s2=3$
11. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach. (The p-value can be only approximated.)
1. Test $H0:μ1−μ2=50$ vs. $Ha:μ1−μ2>50$ @ $α=0.01$,
$n1=30$, $x-1=681$, $s1=8$
$n2=27$, $x-2=625$, $s2=8$
2. Test $H0:μ1−μ2=35$ vs. $Ha:μ1−μ2≠35$ @ $α=0.10$,
$n1=36$, $x-1=325$, $s1=11$
$n2=29$, $x-2=286$, $s2=7$
12. Perform the test of hypotheses indicated, using the data from independent samples given. Use the p-value approach. (The p-value can be only approximated.)
1. Test $H0:μ1−μ2=−4$ vs. $Ha:μ1−μ2<−4$ @ $α=0.05$,
$n1=40$, $x-1=80$, $s1=5$
$n2=25$, $x-2=87$, $s2=5$
2. Test $H0:μ1−μ2=21$ vs. $Ha:μ1−μ2≠21$ @ $α=0.01$,
$n1=15$, $x-1=192$, $s1=12$
$n2=34$, $x-2=180$, $s2=8$
### Applications
1. A county environmental agency suspects that the fish in a particular polluted lake have elevated mercury level. To confirm that suspicion, five striped bass in that lake were caught and their tissues were tested for mercury. For the purpose of comparison, four striped bass in an unpolluted lake were also caught and tested. The fish tissue mercury levels in mg/kg are given below.
$Sample 1 Sample 2(from polluted lake)(from unpolluted lake)0.5800.3820.7110.2760.5710.5700.6660.3660.598$
1. Construct the 95% confidence interval for the difference in the population means based on these data.
2. Test, at the 5% level of significance, whether the data provide sufficient evidence to conclude that fish in the polluted lake have elevated levels of mercury in their tissue.
2. A genetic engineering company claims that it has developed a genetically modified tomato plant that yields on average more tomatoes than other varieties. A farmer wants to test the claim on a small scale before committing to a full-scale planting. Ten genetically modified tomato plants are grown from seeds along with ten other tomato plants. At the season’s end, the resulting yields in pound are recorded as below.
$Sample 1Sample 2(genetically modified)(regular)2021232127222518252025202718232524232220$
1. Construct the 99% confidence interval for the difference in the population means based on these data.
2. Test, at the 1% level of significance, whether the data provide sufficient evidence to conclude that the mean yield of the genetically modified variety is greater than that for the standard variety.
3. The coaching staff of a professional football team believes that the rushing offense has become increasingly potent in recent years. To investigate this belief, 20 randomly selected games from one year’s schedule were compared to 11 randomly selected games from the schedule five years later. The sample information on rushing yards per game (rypg) is summarized below.
n $x-$ s
rypg previously 20 112 24
rypg recently 11 114 21
1. Construct the 95% confidence interval for the difference in the population means based on these data.
2. Test, at the 5% level of significance, whether the data on rushing yards per game provide sufficient evidence to conclude that the rushing offense has become more potent in recent years.
4. The coaching staff of professional football team believes that the rushing offense has become increasingly potent in recent years. To investigate this belief, 20 randomly selected games from one year’s schedule were compared to 11 randomly selected games from the schedule five years later. The sample information on passing yards per game (pypg) is summarized below.
n $x-$ s
pypg previously 20 203 38
pypg recently 11 232 33
1. Construct the 95% confidence interval for the difference in the population means based on these data.
2. Test, at the 5% level of significance, whether the data on passing yards per game provide sufficient evidence to conclude that the passing offense has become more potent in recent years.
5. A university administrator wishes to know if there is a difference in average starting salary for graduates with master’s degrees in engineering and those with master’s degrees in business. Fifteen recent graduates with master’s degree in engineering and 11 with master’s degrees in business are surveyed and the results are summarized below.
n $x-$ s
Engineering 15 68,535 1627
Business 11 63,230 2033
1. Construct the 90% confidence interval for the difference in the population means based on these data.
2. Test, at the 10% level of significance, whether the data provide sufficient evidence to conclude that the average starting salaries are different.
6. A gardener sets up a flower stand in a busy business district and sells bouquets of assorted fresh flowers on weekdays. To find a more profitable pricing, she sells bouquets for 15 dollars each for ten days, then for 10 dollars each for five days. Her average daily profit for the two different prices are given below.
n $x-$ s
$15 10 171 26$10 5 198 29
1. Construct the 90% confidence interval for the difference in the population means based on these data.
2. Test, at the 10% level of significance, whether the data provide sufficient evidence to conclude the gardener’s average daily profit will be higher if the bouquets are sold at \$10 each.
### Answers
1. $(16.16,21.84)$,
2. $(4.28,11.72)$
1. $(0.13,0.47)$,
2. $(1.14,3.46)$
1. T = 2.787, $t0.025=2.131$, reject H0,
2. T = 1.831, $±t0.025=±2.023$, do not reject H0
1. $T=−1.349$, $−t0.10=−1.303$, reject H0,
2. T = 1.411, $±t0.05=±1.678$, do not reject H0
1. T = 2.411, $df=28$, $p -value>0.01$, do not reject H0,
2. T = 1.473, $df=49$, $p -value<0.10$, reject H0
1. T = 2.827, $df=55$, $p -value<0.01$, reject H0.
2. T = 1.699, $df=63$, $p -value <0.10$, reject H0
1. $0.2267±0.2182$,
2. T = 3.635, $df=7$, $t0.05=1.895$, reject H0 (elevated levels)
1. $−2±17.7$,
2. $T=−0.232$, $df=29$, $−t0.05=−1.699$, do not reject H0 (not more potent)
1. $5305±1227$,
2. T = 7.395, $df=24$, $±t0.05=±1.711$, reject H0 (different)
## 9.3 Comparison of Two Population Means: Paired Samples
### Learning Objectives
1. To learn the distinction between independent samples and paired samples.
2. To learn how to construct a confidence interval for the difference in the means of two distinct populations using paired samples.
3. To learn how to perform a test of hypotheses concerning the difference in the means of two distinct populations using paired samples.
Suppose chemical engineers wish to compare the fuel economy obtained by two different formulations of gasoline. Since fuel economy varies widely from car to car, if the mean fuel economy of two independent samples of vehicles run on the two types of fuel were compared, even if one formulation were better than the other the large variability from vehicle to vehicle might make any difference arising from difference in fuel difficult to detect. Just imagine one random sample having many more large vehicles than the other. Instead of independent random samples, it would make more sense to select pairs of cars of the same make and model and driven under similar circumstances, and compare the fuel economy of the two cars in each pair. Thus the data would look something like Table 9.1 "Fuel Economy of Pairs of Vehicles", where the first car in each pair is operated on one formulation of the fuel (call it Type 1 gasoline) and the second car is operated on the second (call it Type 2 gasoline).
Table 9.1 Fuel Economy of Pairs of Vehicles
Make and Model Car 1 Car 2
Buick LaCrosse 17.0 17.0
Dodge Viper 13.2 12.9
Honda CR-Z 35.3 35.4
Hummer H 3 13.6 13.2
Lexus RX 32.7 32.5
Mazda CX-9 18.4 18.1
Saab 9-3 22.5 22.5
Toyota Corolla 26.8 26.7
Volvo XC 90 15.1 15.0
The first column of numbers form a sample from Population 1, the population of all cars operated on Type 1 gasoline; the second column of numbers form a sample from Population 2, the population of all cars operated on Type 2 gasoline. It would be incorrect to analyze the data using the formulas from the previous section, however, since the samples were not drawn independently. What is correct is to compute the difference in the numbers in each pair (subtracting in the same order each time) to obtain the third column of numbers as shown in Table 9.2 "Fuel Economy of Pairs of Vehicles" and treat the differences as the data. At this point, the new sample of differences $d1=0.0,…,d9=0.1$ in the third column of Table 9.2 "Fuel Economy of Pairs of Vehicles" may be considered as a random sample of size n = 9 selected from a population with mean $μd=μ1−μ2.$ This approach essentially transforms the paired two-sample problem into a one-sample problem as discussed in the previous two chapters.
Table 9.2 Fuel Economy of Pairs of Vehicles
Make and Model Car 1 Car 2 Difference
Buick LaCrosse 17.0 17.0 0.0
Dodge Viper 13.2 12.9 0.3
Honda CR-Z 35.3 35.4 −0.1
Hummer H 3 13.6 13.2 0.4
Lexus RX 32.7 32.5 0.2
Mazda CX-9 18.4 18.1 0.3
Saab 9-3 22.5 22.5 0.0
Toyota Corolla 26.8 26.7 0.1
Volvo XC 90 15.1 15.0 0.1
Note carefully that although it does not matter what order the subtraction is done, it must be done in the same order for all pairs. This is why there are both positive and negative quantities in the third column of numbers in Table 9.2 "Fuel Economy of Pairs of Vehicles".
## Confidence Intervals
When the population of differences is normally distributed the following formula for a confidence interval for $μd=μ1−μ2$ is valid.
### $100(1−α)%$ Confidence Interval for the Difference Between Two Population Means: Paired Difference Samples
$d-±tα∕2sdn$
where there are n pairs, $d-$ is the mean and sd is the standard deviation of their differences.
The number of degrees of freedom is $df=n−1.$
The population of differences must be normally distributed.
### Example 7
Using the data in Table 9.1 "Fuel Economy of Pairs of Vehicles" construct a point estimate and a 95% confidence interval for the difference in average fuel economy between cars operated on Type 1 gasoline and cars operated on Type 2 gasoline.
Solution:
We have referred to the data in Table 9.1 "Fuel Economy of Pairs of Vehicles" because that is the way that the data are typically presented, but we emphasize that with paired sampling one immediately computes the differences, as given in Table 9.2 "Fuel Economy of Pairs of Vehicles", and uses the differences as the data.
The mean and standard deviation of the differences are
$d-=Σdn=1.39=0.14-andsd=Σd2−1n(Σd)2n−1=0.41−19(1.3)28=0.16-$
The point estimate of $μ1−μ2=μd$ is
$d-=0.14$
In words, we estimate that the average fuel economy of cars using Type 1 gasoline is 0.14 mpg greater than the average fuel economy of cars using Type 2 gasoline.
To apply the formula for the confidence interval, we must find $tα∕2.$ The 95% confidence level means that $α=1−0.95=0.05$ so that $tα∕2=t0.025.$ From Figure 12.3 "Critical Values of ", in the row with the heading $df=9−1=8$ we read that $t0.025=2.306.$ Thus
$d-±tα∕2sdn=0.14±2.3060.16-9≈0.14±0.13$
We are 95% confident that the difference in the population means lies in the interval $[0.01,0.27]$, in the sense that in repeated sampling 95% of all intervals constructed from the sample data in this manner will contain $μd=μ1−μ2.$ Stated differently, we are 95% confident that mean fuel economy is between 0.01 and 0.27 mpg greater with Type 1 gasoline than with Type 2 gasoline.
## Hypothesis Testing
Testing hypotheses concerning the difference of two population means using paired difference samples is done precisely as it is done for independent samples, although now the null and alternative hypotheses are expressed in terms of $μd$ instead of $μ1−μ2.$ Thus the null hypothesis will always be written
$H0:μd=D0$
The three forms of the alternative hypothesis, with the terminology for each case, are:
Form of $Ha$ Terminology
$Ha:μd Left-tailed
$Ha:μd>D0$ Right-tailed
$Ha:μd≠D0$ Two-tailed
The same conditions on the population of differences that was required for constructing a confidence interval for the difference of the means must also be met when hypotheses are tested. Here is the standardized test statistic that is used in the test.
### Standardized Test Statistic for Hypothesis Tests Concerning the Difference Between Two Population Means: Paired Difference Samples
$T=d-−D0sd∕n$
where there are n pairs, $d-$ is the mean and sd is the standard deviation of their differences.
The test statistic has Student’s t-distribution with $df=n−1$ degrees of freedom.
The population of differences must be normally distributed.
### Example 8
Using the data of Table 9.2 "Fuel Economy of Pairs of Vehicles" test the hypothesis that mean fuel economy for Type 1 gasoline is greater than that for Type 2 gasoline against the null hypothesis that the two formulations of gasoline yield the same mean fuel economy. Test at the 5% level of significance using the critical value approach.
Solution:
The only part of the table that we use is the third column, the differences.
• Step 1. Since the differences were computed in the order $Type 1 mpg− Type 2 mpg$, better fuel economy with Type 1 fuel corresponds to $μd=μ1−μ2>0.$ Thus the test is
$H0:μd=0 vs. Ha:μd>0@ α=0.05$
(If the differences had been computed in the opposite order then the alternative hypotheses would have been $Ha:μd<0.$)
• Step 2. Since the sampling is in pairs the test statistic is
$T=d-−D0sd∕n$
• Step 3. We have already computed $d-$ and sd in the previous example. Inserting their values and $D0=0$ into the formula for the test statistic gives
$T=d-−D0sd∕n=0.14-0.16-∕3=2.600$
• Step 4. Since the symbol in Ha is “>” this is a right-tailed test, so there is a single critical value, $tα=t0.05$ with 8 degrees of freedom, which from the row labeled $df=8$ in Figure 12.3 "Critical Values of " we read off as 1.860. The rejection region is $[1.860,∞).$
• Step 5. As shown in Figure 9.5 "Rejection Region and Test Statistic for " the test statistic falls in the rejection region. The decision is to reject H0. In the context of the problem our conclusion is:
Figure 9.5 Rejection Region and Test Statistic for Note 9.20 "Example 8"
The data provide sufficient evidence, at the 5% level of significance, to conclude that the mean fuel economy provided by Type 1 gasoline is greater than that for Type 2 gasoline.
### Example 9
Perform the test of Note 9.20 "Example 8" using the p-value approach.
Solution:
The first three steps are identical to those in Note 9.20 "Example 8".
• Step 4. Because the test is one-tailed the observed significance or p-value of the test is just the area of the right tail of Student’s t-distribution, with 8 degrees of freedom, that is cut off by the test statistic T = 2.600. We can only approximate this number. Looking in the row of Figure 12.3 "Critical Values of " headed $df=8$, the number 2.600 is between the numbers 2.306 and 2.896, corresponding to t0.025 and t0.010.
The area cut off by t = 2.306 is 0.025 and the area cut off by t = 2.896 is 0.010. Since 2.600 is between 2.306 and 2.896 the area it cuts off is between 0.025 and 0.010. Thus the p-value is between 0.025 and 0.010. In particular it is less than 0.025. See Figure 9.6.
Figure 9.6 P-Value for Note 9.21 "Example 9"
• Step 5. Since 0.025 < 0.05, $p<α$ so the decision is to reject the null hypothesis:
The data provide sufficient evidence, at the 5% level of significance, to conclude that the mean fuel economy provided by Type 1 gasoline is greater than that for Type 2 gasoline.
The paired two-sample experiment is a very powerful study design. It bypasses many unwanted sources of “statistical noise” that might otherwise influence the outcome of the experiment, and focuses on the possible difference that might arise from the one factor of interest.
If the sample is large (meaning that n ≥ 30) then in the formula for the confidence interval we may replace $tα∕2$ by $zα∕2.$ For hypothesis testing when the number of pairs is at least 30, we may use the same statistic as for small samples for hypothesis testing, except now it follows a standard normal distribution, so we use the last line of Figure 12.3 "Critical Values of " to compute critical values, and p-values can be computed exactly with Figure 12.2 "Cumulative Normal Probability", not merely estimated using Figure 12.3 "Critical Values of ".
### Key Takeaways
• When the data are collected in pairs, the differences computed for each pair are the data that are used in the formulas.
• A confidence interval for the difference in two population means using paired sampling is computed using a formula in the same fashion as was done for a single population mean.
• The same five-step procedure used to test hypotheses concerning a single population mean is used to test hypotheses concerning the difference between two population means using pair sampling. The only difference is in the formula for the standardized test statistic.
### Basic
In all exercises for this section assume that the population of differences is normal.
1. Use the following paired sample data for this exercise.
$Population 135323535363536Population 228262726292729$
1. Compute $d-$ and sd.
2. Give a point estimate for $μ1−μ2=μd.$
3. Construct the 95% confidence interval for $μ1−μ2=μd$ from these data.
4. Test, at the 10% level of significance, the hypothesis that $μ1−μ2>7$ as an alternative to the null hypothesis that $μ1−μ2=7.$
2. Use the following paired sample data for this exercise.
$Population 110312796110Population 2811067388$ $Population 190118130106Population 2709510983$
1. Compute $d-$ and sd.
2. Give a point estimate for $μ1−μ2=μd.$
3. Construct the 90% confidence interval for $μ1−μ2=μd$ from these data.
4. Test, at the 1% level of significance, the hypothesis that $μ1−μ2<24$ as an alternative to the null hypothesis that $μ1−μ2=24.$
3. Use the following paired sample data for this exercise.
$Population 140275534Population 253426850$
1. Compute $d-$ and sd.
2. Give a point estimate for $μ1−μ2=μd.$
3. Construct the 99% confidence interval for $μ1−μ2=μd$ from these data.
4. Test, at the 10% level of significance, the hypothesis that $μ1−μ2≠−12$ as an alternative to the null hypothesis that $μ1−μ2=−12.$
4. Use the following paired sample data for this exercise.
$Population 1196165181201190Population 2212182199210205$
1. Compute $d-$ and sd.
2. Give a point estimate for $μ1−μ2=μd.$
3. Construct the 98% confidence interval for $μ1−μ2=μd$ from these data.
4. Test, at the 2% level of significance, the hypothesis that $μ1−μ2≠−20$ as an alternative to the null hypothesis that $μ1−μ2=−20.$
### Applications
1. Each of five laboratory mice was released into a maze twice. The five pairs of times to escape were:
Mouse 1 2 3 4 5
First release 129 89 136 163 118
Second release 113 97 139 85 75
1. Compute $d-$ and sd.
2. Give a point estimate for $μ1−μ2=μd.$
3. Construct the 90% confidence interval for $μ1−μ2=μd$ from these data.
4. Test, at the 10% level of significance, the hypothesis that it takes mice less time to run the maze on the second trial, on average.
2. Eight golfers were asked to submit their latest scores on their favorite golf courses. These golfers were each given a set of newly designed clubs. After playing with the new clubs for a few months, the golfers were again asked to submit their latest scores on the same golf courses. The results are summarized below.
Golfer 1 2 3 4 5 6 7 8
Own clubs 77 80 69 73 73 72 75 77
New clubs 72 81 68 73 75 70 73 75
1. Compute $d-$ and sd.
2. Give a point estimate for $μ1−μ2=μd.$
3. Construct the 99% confidence interval for $μ1−μ2=μd$ from these data.
4. Test, at the 1% level of significance, the hypothesis that on average golf scores are lower with the new clubs.
3. A neighborhood home owners association suspects that the recent appraisal values of the houses in the neighborhood conducted by the county government for taxation purposes is too high. It hired a private company to appraise the values of ten houses in the neighborhood. The results, in thousands of dollars, are
House County Government Private Company
1 217 219
2 350 338
3 296 291
4 237 237
5 237 235
6 272 269
7 257 239
8 277 275
9 312 320
10 335 335
1. Give a point estimate for the difference between the mean private appraisal of all such homes and the government appraisal of all such homes.
2. Construct the 99% confidence interval based on these data for the difference.
3. Test, at the 1% level of significance, the hypothesis that appraised values by the county government of all such houses is greater than the appraised values by the private appraisal company.
4. In order to cut costs a wine producer is considering using duo or 1 + 1 corks in place of full natural wood corks, but is concerned that it could affect buyers’s perception of the quality of the wine. The wine producer shipped eight pairs of bottles of its best young wines to eight wine experts. Each pair includes one bottle with a natural wood cork and one with a duo cork. The experts are asked to rate the wines on a one to ten scale, higher numbers corresponding to higher quality. The results are:
Wine Expert Duo Cork Wood Cork
1 8.5 8.5
2 8.0 8.5
3 6.5 8.0
4 7.5 8.5
5 8.0 7.5
6 8.0 8.0
7 9.0 9.0
8 7.0 7.5
1. Give a point estimate for the difference between the mean ratings of the wine when bottled are sealed with different kinds of corks.
2. Construct the 90% confidence interval based on these data for the difference.
3. Test, at the 10% level of significance, the hypothesis that on the average duo corks decrease the rating of the wine.
5. Engineers at a tire manufacturing corporation wish to test a new tire material for increased durability. To test the tires under realistic road conditions, new front tires are mounted on each of 11 company cars, one tire made with a production material and the other with the experimental material. After a fixed period the 11 pairs were measured for wear. The amount of wear for each tire (in mm) is shown in the table:
Car Production Experimental
1 5.1 5.0
2 6.5 6.5
3 3.6 3.1
4 3.5 3.7
5 5.7 4.5
6 5.0 4.1
7 6.4 5.3
8 4.7 2.6
9 3.2 3.0
10 3.5 3.5
11 6.4 5.1
1. Give a point estimate for the difference in mean wear.
2. Construct the 99% confidence interval for the difference based on these data.
3. Test, at the 1% level of significance, the hypothesis that the mean wear with the experimental material is less than that for the production material.
6. A marriage counselor administered a test designed to measure overall contentment to 30 randomly selected married couples. The scores for each couple are given below. A higher number corresponds to greater contentment or happiness.
Couple Husband Wife
1 47 44
2 44 46
3 49 44
4 53 44
5 42 43
6 45 45
7 48 47
8 45 44
9 52 44
10 47 42
11 40 34
12 45 42
13 40 43
14 46 41
15 47 45
16 46 45
17 46 41
18 46 41
19 44 45
20 45 43
21 48 38
22 42 46
23 50 44
24 46 51
25 43 45
26 50 40
27 46 46
28 42 41
29 51 41
30 46 47
1. Test, at the 1% level of significance, the hypothesis that on average men and women are not equally happy in marriage.
2. Test, at the 1% level of significance, the hypothesis that on average men are happier than women in marriage.
### Large Data Set Exercises
1. Large Data Set 5 lists the scores for 25 randomly selected students on practice SAT reading tests before and after taking a two-week SAT preparation course. Denote the population of all students who have taken the course as Population 1 and the population of all students who have not taken the course as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data5.xls
1. Compute the 25 differences in the order $after− before$, their mean $d-$, and their sample standard deviation sd.
2. Give a point estimate for $μd=μ1−μ2$, the difference in the mean score of all students who have taken the course and the mean score of all who have not.
3. Construct a 98% confidence interval for $μd.$
4. Test, at the 1% level of significance, the hypothesis that the mean SAT score increases by at least ten points by taking the two-week preparation course.
2. Large Data Set 12 lists the scores on one round for 75 randomly selected members at a golf course, first using their own original clubs, then two months later after using new clubs with an experimental design. Denote the population of all golfers using their own original clubs as Population 1 and the population of all golfers using the new style clubs as Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data12.xls
1. Compute the 75 differences in the order $original clubs− new clubs$, their mean $d-$, and their sample standard deviation sd.
2. Give a point estimate for $μd=μ1−μ2$, the difference in the mean score of all golfers using their original clubs and the mean score of all golfers using the new kind of clubs.
3. Construct a 90% confidence interval for $μd.$
4. Test, at the 1% level of significance, the hypothesis that the mean golf score decreases by at least one stroke by using the new kind of clubs.
3. Consider the previous problem again. Since the data set is so large, it is reasonable to use the standard normal distribution instead of Student’s t-distribution with 74 degrees of freedom.
1. Construct a 90% confidence interval for $μd$ using the standard normal distribution, meaning that the formula is $d-±zα∕2sdn.$ (The computations done in part (a) of the previous problem still apply and need not be redone.) How does the result obtained here compare to the result obtained in part (c) of the previous problem?
2. Test, at the 1% level of significance, the hypothesis that the mean golf score decreases by at least one stroke by using the new kind of clubs, using the standard normal distribution. (All the work done in part (d) of the previous problem applies, except the critical value is now $zα$ instead of $tα$ (or the p-value can be computed exactly instead of only approximated, if you used the p-value approach).) How does the result obtained here compare to the result obtained in part (c) of the previous problem?
3. Construct the 99% confidence intervals for $μd$ using both the $t-$ and $z-$distributions. How much difference is there in the results now?
### Answers
1. $d-=7.4286$, $sd=0.9759$,
2. $d-=7.4286$,
3. $(6.53,8.33)$,
4. T = 1.162, $df=6$, $t0.10=1.44$, do not reject H0
1. $d-=−14.25$, $sd=1.5$,
2. $d-=−14.25$,
3. $(−18.63,−9.87)$,
4. $T=−3.000$, $df=3$, $±t0.05=±2.353$, reject H0
1. $d-=25.2$, $sd=35.6609$,
2. 25.2,
3. $25.2±34.0$
4. T = 1.580, $df=4$, $t0.10=1.533$, reject H0 (takes less time)
1. 3.2,
2. $3.2±7.5$
3. T = 1.392, $df=9$, $t0.10=2.821$, do not reject H0 (government appraisals not higher)
1. 0.65,
2. $0.65±0.69$,
3. T = 3.014, $df=10$, $t0.01=2.764$, reject H0 (experimental material wears less)
1. $d-=16.68$ and $sd=10.77$
2. $d-=16.68$
3. $(11.31,22.05)$
4. $H0:μ1−μ2=10$ vs. $Ha:μ1−μ2>10.$ Test Statistic: T = 3.1014. $d.f.=24.$ Rejection Region: $[2.492,∞).$ Decision: Reject H0.
1. $(1.6266,2.6401).$ Endpoints change in the third deciaml place.
2. $H0:μ1−μ2=1$ vs. $Ha:μ1−μ2>1.$ Test Statistic: Z = 3.6791. Rejection Region: $[2.33,∞).$ Decision: Reject H0. The decision is the same as in the previous problem.
3. Using the $t−$distribution, $(1.3188,2.9478).$ Using the $z−$distribution, $(1.3401,2.9266).$ There is a difference.
## 9.4 Comparison of Two Population Proportions
### Learning Objectives
1. To learn how to construct a confidence interval for the difference in the proportions of two distinct populations that have a particular characteristic of interest.
2. To learn how to perform a test of hypotheses concerning the difference in the proportions of two distinct populations that have a particular characteristic of interest.
Suppose we wish to compare the proportions of two populations that have a specific characteristic, such as the proportion of men who are left-handed compared to the proportion of women who are left-handed. Figure 9.7 "Independent Sampling from Two Populations In Order to Compare Proportions" illustrates the conceptual framework of our investigation. Each population is divided into two groups, the group of elements that have the characteristic of interest (for example, being left-handed) and the group of elements that do not. We arbitrarily label one population as Population 1 and the other as Population 2, and subscript the proportion of each population that possesses the characteristic with the number 1 or 2 to tell them apart. We draw a random sample from Population 1 and label the sample statistic it yields with the subscript 1. Without reference to the first sample we draw a sample from Population 2 and label its sample statistic with the subscript 2.
Figure 9.7 Independent Sampling from Two Populations In Order to Compare Proportions
Our goal is to use the information in the samples to estimate the difference $p1−p2$ in the two population proportions and to make statistically valid inferences about it.
## Confidence Intervals
Since the sample proportion $p^1$ computed using the sample drawn from Population 1 is a good estimator of population proportion p1 of Population 1 and the sample proportion $p^2$ computed using the sample drawn from Population 2 is a good estimator of population proportion p2 of Population 2, a reasonable point estimate of the difference $p1−p2$ is $p^1−p^2.$ In order to widen this point estimate into a confidence interval we suppose that both samples are large, as described in Section 7.3 "Large Sample Estimation of a Population Proportion" in Chapter 7 "Estimation" and repeated below. If so, then the following formula for a confidence interval for $p1−p2$ is valid.
### $100(1−α)%$ Confidence Interval for the Difference Between Two Population Proportions
$(p^1−p^2)±zα∕2p^1(1−p^1)n1+p^2(1−p^2)n2$
The samples must be independent, and each sample must be large: each of the intervals
$[p^1−3 p^1(1−p^1)n1, p^1+3 p^1(1−p^1)n1 ]$
and
$[p^2−3 p^2(1−p^2)n2, p^2+3 p^2(1−p^2)n2 ]$
must lie wholly within the interval $[0,1 ].$
### Example 10
The department of code enforcement of a county government issues permits to general contractors to work on residential projects. For each permit issued, the department inspects the result of the project and gives a “pass” or “fail” rating. A failed project must be re-inspected until it receives a pass rating. The department had been frustrated by the high cost of re-inspection and decided to publish the inspection records of all contractors on the web. It was hoped that public access to the records would lower the re-inspection rate. A year after the web access was made public, two samples of records were randomly selected. One sample was selected from the pool of records before the web publication and one after. The proportion of projects that passed on the first inspection was noted for each sample. The results are summarized below. Construct a point estimate and a 90% confidence interval for the difference in the passing rate on first inspection between the two time periods.
$No public web accessn1=500p^1=0.67Public web accessn2=100p^2=0.80$
Solution:
The point estimate of $p1−p2$ is
$p^1−p^2=0.67−0.80=−0.13$
Because the “No public web access” population was labeled as Population 1 and the “Public web access” population was labeled as Population 2, in words this means that we estimate that the proportion of projects that passed on the first inspection increased by 13 percentage points after records were posted on the web.
The sample sizes are sufficiently large for constructing a confidence interval since for sample 1:
$3 p^1(1−p^1)n1=3 (0.67)(0.33)500=0.06$
so that
$p^1−3 p^1(1−p^1)n1, p^1+3 p^1(1−p^1)n1=[0.67−0.06,0.67+0.06 ]=[0.61,0.73 ]⊂[0,1 ]$
and for sample 2:
$3 p^1(1−p^1)n1=3 (0.8)(0.2)100=0.12$
so that
$p^2−3 p^2(1−p^2)n2, p^2+3 p^2(1−p^2)n2=[0.8−0.12,0.8+0.12 ]=[0.68,0.92 ]⊂[0,1 ]$
To apply the formula for the confidence interval, we first observe that the 90% confidence level means that $α=1−0.90=0.10$ so that $zα∕2=z0.05.$ From Figure 12.3 "Critical Values of " we read directly that $z0.05=1.645.$ Thus the desired confidence interval is
$(p^1−p^2)±zα∕2p^1(1−p^1)n1+p^2(1−p^2)n2=−0.13±1.645(0.67)(0.33)500+(0.8)(0.2)100 =−0.13±0.07$
The 90% confidence interval is $[−0.20,−0.06 ].$ We are 90% confident that the difference in the population proportions lies in the interval $[−0.20,−0.06 ]$, in the sense that in repeated sampling 90% of all intervals constructed from the sample data in this manner will contain $p1−p2.$ Taking into account the labeling of the two populations, this means that we are 90% confident that the proportion of projects that pass on the first inspection is between 6 and 20 percentage points higher after public access to the records than before.
## Hypothesis Testing
In hypothesis tests concerning the relative sizes of the proportions p1 and p2 of two populations that possess a particular characteristic, the null and alternative hypotheses will always be expressed in terms of the difference of the two population proportions. Hence the null hypothesis is always written
$H0:p1−p2=D0$
The three forms of the alternative hypothesis, with the terminology for each case, are:
Form of $Ha$ Terminology
$Ha:p1−p2 Left-tailed
$Ha:p1−p2>D0$ Right-tailed
$Ha:p1−p2≠D0$ Two-tailed
As long as the samples are independent and both are large the following formula for the standardized test statistic is valid, and it has the standard normal distribution.
### Standardized Test Statistic for Hypothesis Tests Concerning the Difference Between Two Population Proportions
$Z=(p^1−p^2)−D0p^1(1−p^1)n1+p^2(1−p^2)n2$
The test statistic has the standard normal distribution.
The samples must be independent, and each sample must be large: each of the intervals
$[p^1−3 p^1(1−p^1)n1, p^1+3 p^1(1−p^1)n1 ]$
and
$[p^2−3 p^2(1−p^2)n2, p^2+3 p^2(1−p^2)n2 ]$
must lie wholly within the interval $[0,1 ].$
### Example 11
Using the data of Note 9.25 "Example 10", test whether there is sufficient evidence to conclude that public web access to the inspection records has increased the proportion of projects that passed on the first inspection by more than 5 percentage points. Use the critical value approach at the 10% level of significance.
Solution:
• Step 1. Taking into account the labeling of the populations an increase in passing rate at the first inspection by more than 5 percentage points after public access on the web may be expressed as $p2>p1+0.05$, which by algebra is the same as $p1−p2<−0.05.$ This is the alternative hypothesis. Since the null hypothesis is always expressed as an equality, with the same number on the right as is in the alternative hypothesis, the test is
$H0:p1−p2=−0.05 vs. Ha:p1−p2<−0.05@ α=0.10$
• Step 2. Since the test is with respect to a difference in population proportions the test statistic is
$Z=(p^1−p^2)−D0p^1(1−p^1)n1+p^2(1−p^2)n2$
• Step 3. Inserting the values given in Note 9.25 "Example 10" and the value $D0=−0.05$ into the formula for the test statistic gives
$Z=(p^1−p^2)−D0p^1(1−p^1)n1+p^2(1−p^2)n2=(−0.13)−(−0.05)(0.67)(0.33)500+(0.8)(0.2)100=−1.770$
• Step 4. Since the symbol in Ha is “<” this is a left-tailed test, so there is a single critical value, $zα=−z0.10.$ From the last row in Figure 12.3 "Critical Values of " $z0.10=1.282$, so $−z0.10=−1.282.$ The rejection region is $(−∞,−1.282 ].$
• Step 5. As shown in Figure 9.8 "Rejection Region and Test Statistic for " the test statistic falls in the rejection region. The decision is to reject H0. In the context of the problem our conclusion is:
The data provide sufficient evidence, at the 10% level of significance, to conclude that the rate of passing on the first inspection has increased by more than 5 percentage points since records were publicly posted on the web.
Figure 9.8 Rejection Region and Test Statistic for Note 9.27 "Example 11"
### Example 12
Perform the test of Note 9.27 "Example 11" using the p-value approach.
Solution:
The first three steps are identical to those in Note 9.27 "Example 11".
• Step 4. Because the test is left-tailed the observed significance or p-value of the test is just the area of the left tail of the standard normal distribution that is cut off by the test statistic $Z=−1.770.$ From Figure 12.2 "Cumulative Normal Probability" the area of the left tail determined by −1.77 is 0.0384. The p-value is 0.0384.
• Step 5. Since the p-value 0.0384 is less than $α=0.10$, the decision is to reject the null hypothesis: The data provide sufficient evidence, at the 10% level of significance, to conclude that the rate of passing on the first inspection has increased by more than 5 percentage points since records were publicly posted on the web.
Finally a common misuse of the formulas given in this section must be mentioned. Suppose a large pre-election survey of potential voters is conducted. Each person surveyed is asked to express a preference between, say, Candidate A and Candidate B. (Perhaps “no preference” or “other” are also choices, but that is not important.) In such a survey, estimators $p^A$ and $p^B$ of pA and pB can be calculated. It is important to realize, however, that these two estimators were not calculated from two independent samples. While $p^A−p^B$ may be a reasonable estimator of $pA−pB$, the formulas for confidence intervals and for the standardized test statistic given in this section are not valid for data obtained in this manner.
### Key Takeaways
• A confidence interval for the difference in two population proportions is computed using a formula in the same fashion as was done for a single population mean.
• The same five-step procedure used to test hypotheses concerning a single population proportion is used to test hypotheses concerning the difference between two population proportions. The only difference is in the formula for the standardized test statistic.
### Basic
1. Construct the confidence interval for $p1−p2$ for the level of confidence and the data given. (The samples are sufficiently large.)
1. 90% confidence,
$n1=1670$, $p^1=0.42$
$n2=900$, $p^2=0.38$
2. 95% confidence,
$n1=600$, $p^1=0.84$
$n2=420$, $p^2=0.67$
2. Construct the confidence interval for $p1−p2$ for the level of confidence and the data given. (The samples are sufficiently large.)
1. 98% confidence,
$n1=750$, $p^1=0.64$
$n2=800$, $p^2=0.51$
2. 99.5% confidence,
$n1=250$, $p^1=0.78$
$n2=250$, $p^2=0.51$
3. Construct the confidence interval for $p1−p2$ for the level of confidence and the data given. (The samples are sufficiently large.)
1. 80% confidence,
$n1=300$, $p^1=0.255$
$n2=400$, $p^2=0.193$
2. 95% confidence,
$n1=3500$, $p^1=0.147$
$n2=3750$, $p^2=0.131$
4. Construct the confidence interval for $p1−p2$ for the level of confidence and the data given. (The samples are sufficiently large.)
1. 99% confidence,
$n1=2250$, $p^1=0.915$
$n2=2525$, $p^2=0.858$
2. 95% confidence,
$n1=120$, $p^1=0.650$
$n2=200$, $p^2=0.505$
5. Perform the test of hypotheses indicated, using the data given. Use the critical value approach. Compute the p-value of the test as well. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0$ vs. $Ha:p1−p2>0$ @ $α=0.10$,
$n1=1200$, $p^1=0.42$
$n2=1200$, $p^2=0.40$
2. Test $H0:p1−p2=0$ vs. $Ha:p1−p2≠0$ @ $α=0.05$,
$n1=550$, $p^1=0.61$
$n2=600$, $p^2=0.67$
6. Perform the test of hypotheses indicated, using the data given. Use the critical value approach. Compute the p-value of the test as well. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0.05$ vs. $Ha:p1−p2>0.05$ @ $α=0.05$,
$n1=1100$, $p^1=0.57$
$n2=1100$, $p^2=0.48$
2. Test $H0:p1−p2=0$ vs. $Ha:p1−p2≠0$ @ $α=0.05$,
$n1=800$, $p^1=0.39$
$n2=900$, $p^2=0.43$
7. Perform the test of hypotheses indicated, using the data given. Use the critical value approach. Compute the p-value of the test as well. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0.25$ vs. $Ha:p1−p2<0.25$ @ $α=0.005$,
$n1=1400$, $p^1=0.57$
$n2=1200$, $p^2=0.37$
2. Test $H0:p1−p2=0.16$ vs. $Ha:p1−p2≠0.16$ @ $α=0.02$,
$n1=750$, $p^1=0.43$
$n2=600$, $p^2=0.22$
8. Perform the test of hypotheses indicated, using the data given. Use the critical value approach. Compute the p-value of the test as well. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0.08$ vs. $Ha:p1−p2>0.08$ @ $α=0.025$,
$n1=450$, $p^1=0.67$
$n2=200$, $p^2=0.52$
2. Test $H0:p1−p2=0.02$ vs. $Ha:p1−p2≠0.02$ @ $α=0.001$,
$n1=2700$, $p^1=0.837$
$n2=2900$, $p^2=0.854$
9. Perform the test of hypotheses indicated, using the data given. Use the p-value approach. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0$ vs. $Ha:p1−p2<0$ @ $α=0.005$,
$n1=1100$, $p^1=0.22$
$n2=1300$, $p^2=0.27$
2. Test $H0:p1−p2=0$ vs. $Ha:p1−p2≠0$ @ $α=0.01$,
$n1=650$, $p^1=0.35$
$n2=650$, $p^2=0.41$
10. Perform the test of hypotheses indicated, using the data given. Use the p-value approach. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0.15$ vs. $Ha:p1−p2>0.15$ @ $α=0.10$,
$n1=950$, $p^1=0.41$
$n2=500$, $p^2=0.23$
2. Test $H0:p1−p2=0.10$ vs. $Ha:p1−p2≠0.10$ @ $α=0.10$,
$n1=220$, $p^1=0.92$
$n2=160$, $p^2=0.78$
11. Perform the test of hypotheses indicated, using the data given. Use the p-value approach. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0.22$ vs. $Ha:p1−p2>0.22$ @ $α=0.05$,
$n1=90$, $p^1=0.72$
$n2=75$, $p^2=0.40$
2. Test $H0:p1−p2=0.37$ vs. $Ha:p1−p2≠0.37$ @ $α=0.02$,
$n1=425$, $p^1=0.772$
$n2=425$, $p^2=0.331$
12. Perform the test of hypotheses indicated, using the data given. Use the p-value approach. (The samples are sufficiently large.)
1. Test $H0:p1−p2=0.50$ vs. $Ha:p1−p2<0.50$ @ $α=0.10$,
$n1=40$, $p^1=0.65$
$n2=55$, $p^2=0.24$
2. Test $H0:p1−p2=0.30$ vs. $Ha:p1−p2≠0.30$ @ $α=0.10$,
$n1=7500$, $p^1=0.664$
$n2=1000$, $p^2=0.319$
### Applications
In all the remaining exercsises the samples are sufficiently large (so this need not be checked).
1. Voters in a particular city who identify themselves with one or the other of two political parties were randomly selected and asked if they favor a proposal to allow citizens with proper license to carry a concealed handgun in city parks. The results are:
Party A Party B
Sample size, n 150 200
Number in favor, x 90 140
1. Give a point estimate for the difference in the proportion of all members of Party A and all members of Party B who favor the proposal.
2. Construct the 95% confidence interval for the difference, based on these data.
3. Test, at the 5% level of significance, the hypothesis that the proportion of all members of Party A who favor the proposal is less than the proportion of all members of Party B who do.
4. Compute the p-value of the test.
2. To investigate a possible relation between gender and handedness, a random sample of 320 adults was taken, with the following results:
Men Women
Sample size, n 168 152
Number of left-handed, x 24 9
1. Give a point estimate for the difference in the proportion of all men who are left-handed and the proportion of all women who are left-handed.
2. Construct the 95% confidence interval for the difference, based on these data.
3. Test, at the 5% level of significance, the hypothesis that the proportion of men who are left-handed is greater than the proportion of women who are.
4. Compute the p-value of the test.
3. A local school board member randomly sampled private and public high school teachers in his district to compare the proportions of National Board Certified (NBC) teachers in the faculty. The results were:
Private Schools Public Schools
Sample size, n 80 520
Proportion of NBC teachers, $p^$ 0.175 0.150
1. Give a point estimate for the difference in the proportion of all teachers in area public schools and the proportion of all teachers in private schools who are National Board Certified.
2. Construct the 90% confidence interval for the difference, based on these data.
3. Test, at the 10% level of significance, the hypothesis that the proportion of all public school teachers who are National Board Certified is less than the proportion of private school teachers who are.
4. Compute the p-value of the test.
4. In professional basketball games, the fans of the home team always try to distract free throw shooters on the visiting team. To investigate whether this tactic is actually effective, the free throw statistics of a professional basketball player with a high free throw percentage were examined. During the entire last season, this player had 656 free throws, 420 in home games and 236 in away games. The results are summarized below.
Home Away
Sample size, n 420 236
Free throw percent, $p^$ 81.5% 78.8%
1. Give a point estimate for the difference in the proportion of free throws made at home and away.
2. Construct the 90% confidence interval for the difference, based on these data.
3. Test, at the 10% level of significance, the hypothesis that there exists a home advantage in free throws.
4. Compute the p-value of the test.
5. Randomly selected middle-aged people in both China and the United States were asked if they believed that adults have an obligation to financially support their aged parents. The results are summarized below.
China USA
Sample size, n 1300 150
Number of yes, x 1170 110
Test, at the 1% level of significance, whether the data provide sufficient evidence to conclude that there exists a cultural difference in attitude regarding this question.
6. A manufacturer of walk-behind push mowers receives refurbished small engines from two new suppliers, A and B. It is not uncommon that some of the refurbished engines need to be lightly serviced before they can be fitted into mowers. The mower manufacturer recently received 100 engines from each supplier. In the shipment from A, 13 needed further service. In the shipment from B, 10 needed further service. Test, at the 10% level of significance, whether the data provide sufficient evidence to conclude that there exists a difference in the proportions of engines from the two suppliers needing service.
### Large Data Set Exercises
1. Large Data Sets 6A and 6B record results of a random survey of 200 voters in each of two regions, in which they were asked to express whether they prefer Candidate A for a U.S. Senate seat or prefer some other candidate. Let the population of all voters in region 1 be denoted Population 1 and the population of all voters in region 2 be denoted Population 2. Let p1 be the proportion of voters in Population 1 who prefer Candidate A, and p2 the proportion in Population 2 who do.
http://www.gone.2012books.lardbucket.org/sites/all/files/data6A.xls
http://www.gone.2012books.lardbucket.org/sites/all/files/data6B.xls
1. Find the relevant sample proportions $p^1$ and $p^2.$
2. Construct a point estimate for $p1−p2.$
3. Construct a 95% confidence interval for $p1−p2.$
4. Test, at the 5% level of significance, the hypothesis that the same proportion of voters in the two regions favor Candidate A, against the alternative that a larger proportion in Population 2 do.
2. Large Data Set 11 records the results of samples of real estate sales in a certain region in the year 2008 (lines 2 through 536) and in the year 2010 (lines 537 through 1106). Foreclosure sales are identified with a 1 in the second column. Let all real estate sales in the region in 2008 be Population 1 and all real estate sales in the region in 2010 be Population 2.
http://www.gone.2012books.lardbucket.org/sites/all/files/data11.xls
1. Use the sample data to construct point estimates $p^1$ and $p^2$ of the proportions p1 and p2 of all real estate sales in this region in 2008 and 2010 that were foreclosure sales. Construct a point estimate of $p1−p2.$
2. Use the sample data to construct a 90% confidence for $p1−p2.$
3. Test, at the 10% level of significance, the hypothesis that the proportion of real estate sales in the region in 2010 that were foreclosure sales was greater than the proportion of real estate sales in the region in 2008 that were foreclosure sales. (The default is that the proportions were the same.)
### Answers
1. $(0.0068,0.0732)$,
2. $(0.1163,0.2237)$
1. $(0.0210,0.1030)$,
2. $(0.0001,0.0319)$
1. Z = 0.996, $z0.10=1.282$, $p -value=0.1587$, do not reject H0,
2. $Z=−2.120$, $±z0.025=±1.960$, $p -value=0.0340$, reject H0
1. $Z=−2.602$, $−z0.005=−2.576$, $p -value=0.0047$, reject H0,
2. Z = 2.020, $±z0.01=±2.326$, $p -value=0.0434$, do not reject H0
1. $Z=−2.85$, $p -value=0.0022$, reject H0,
2. $Z=−2.23$, $p -value=0.0258$, do not reject H0
1. Z = 1.36, $p -value=0.0869$, do not reject H0,
2. Z = 2.32, $p -value=0.0204$, do not reject H0
1. −0.10,
2. $−0.10±0.101$,
3. $Z=−1.943$, $−z0.05=−1.645$, reject H0 (fewer in Party A favor),
4. p-value = 0.0262
1. 0.025,
2. $0.025±0.0745$,
3. Z = 0.552, $z0.10=1.282$, do not reject H0 (as many public school teachers are certified),
4. p-value = 0.2912
1. Z = 4.498, $±z0.005=±2.576$, reject H0 (different)
1. $p^1=0.355$ and $p^2=0.41$
2. $p^1−p^2=−0.055$
3. $(−0.1501,0.0401)$
4. $H0:p1−p2=0$ vs. $Ha:p1−p2<0.$ Test Statistic: $Z=−1.1335.$ Rejection Region: $(−∞,−1.645 ].$ Decision: Fail to reject H0.
## 9.5 Sample Size Considerations
### Learning Objective
1. To learn how to apply formulas for estimating the size samples that will be needed in order to construct a confidence interval for the difference in two population means or proportions that meets given criteria.
As was pointed out at the beginning of Section 7.4 "Sample Size Considerations" in Chapter 7 "Estimation", sampling is typically done with definite objectives in mind. For example, a physician might wish to estimate the difference in the average amount of sleep gotten by patients suffering a certain condition with the average amount of sleep got by healthy adults, at 90% confidence and to within half an hour. Since sampling costs time, effort, and money, it would be useful to be able to estimate the smallest size samples that are likely to meet these criteria.
## Estimating $μ1−μ2$ with Independent Samples
Assuming that large samples will be required, the confidence interval formula for estimating the difference $μ1−μ2$ between two population means using independent samples is $(x-1−x-2)±E$, where
$E=zα∕2s12n1+s22n2$
To say that we wish to estimate the mean to within a certain number of units means that we want the margin of error E to be no larger than that number. The number $zα∕2$ is determined by the desired level of confidence.
The numbers s1 and s2 are estimates of the standard deviations $σ1$ and $σ2$ of the two populations. In analogy with what we did in Section 7.4 "Sample Size Considerations" in Chapter 7 "Estimation" we will assume that we either know or can reasonably approximate $σ1$ and $σ2.$
We cannot solve for both n1 and n2, so we have to make an assumption about their relative sizes. We will specify that they be equal. With these assumptions we obtain the minimum sample sizes needed by solving the equation displayed just above for $n1=n2.$
### Minimum Equal Sample Sizes for Estimating the Difference in the Means of Two Populations Using Independent Samples
The estimated minimum equal sample sizes $n1=n2$ needed to estimate the difference $μ1−μ2$ in two population means to within E units at $100(1−α)$% confidence is
$n1=n2=(zα∕2)2(σ12+σ22)E2(rounded up)$
In all the examples and exercises the population standard deviations $σ1$ and $σ2$ will be given.
### Example 13
A law firm wishes to estimate the difference in the mean delivery time of documents sent between two of its offices by two different courier companies, to within half an hour and with 99.5% confidence. From their records it will randomly sample the same number n of documents as delivered by each courier company. Determine how large n must be if the estimated standard deviations of the delivery times are 0.75 hour for one company and 1.15 hours for the other.
Solution:
Confidence level 99.5% means that $α=1−0.995=0.005$ so $α∕2=0.0025.$ From the last line of Figure 12.3 "Critical Values of " we obtain $z0.0025=2.807.$
To say that the estimate is to be “to within half an hour” means that E = 0.5. Thus
$n=(zα∕2)2(σ12+σ22)E2=(2.807)2(0.752+1.152)0.52=59.40953746$
which we round up to 60, since it is impossible to take a fractional observation. The law firm must sample 60 document deliveries by each company.
## Estimating $μ1−μ2$ with Paired Samples
As we mentioned at the end of Section 9.3 "Comparison of Two Population Means: Paired Samples", if the sample is large (meaning that n ≥ 30) then in the formula for the confidence interval we may replace $tα∕2$ by $zα∕2$, so that the confidence interval formula becomes $d-±E$ for
$E=zα∕2sdn$
The number sd is an estimate of the standard deviations $σd$ of the population of differences. We must assume that we either know or can reasonably approximate $σd.$ Thus, assuming that large samples will be required to meet the criteria given, we can solve the displayed equation for n to obtain an estimate of the number of pairs needed in the sample.
### Minimum Sample Size for Estimating the Difference in the Means of Two Populations Using Paired Difference Samples
The estimated minimum number of pairs n needed to estimate the difference $μd=μ1−μ2$ in two population means to within E units at $100(1−α)$% confidence using paired difference samples is
$n=(zα∕2)2σd2E2(rounded up)$
In all the examples and exercises the population standard deviation of the differences $σd$ will be given.
### Example 14
A automotive tire manufacturer wishes to compare the mean lifetime of two tread designs under actual driving conditions. They will mount one of each type of tire on n vehicles (both on the front or both on the back) and measure the difference in remaining tread after 20,000 miles of driving. If the standard deviation of the differences is assumed to be 0.025 inch, find the minimum samples size needed to estimate the difference in mean depth (at 20,000 miles use) to within 0.01 inch at 99.9% confidence.
Solution:
Confidence level 99.9% means that $α=1−0.999=0.001$ so $α∕2=0.0005.$ From the last line of Figure 12.3 "Critical Values of " we obtain $z0.0005=3.291.$
To say that the estimate is to be “to within 0.01 inch” means that E = 0.01. Thus
$n=(zα∕2)2σd2E2=(3.291)2(0.025)2(0.01)2=67.69175625$
which we round up to 68. The manufacturer must test 68 pairs of tires.
## Estimating $p1−p2$
The confidence interval formula for estimating the difference $p1−p2$ between two population proportions is $p^1−p^2±E$, where
$E=zα∕2p^1(1−p^1)n1+p^2(1−p^2)n2$
To say that we wish to estimate the mean to within a certain number of units means that we want the margin of error E to be no larger than that number. The number $zα∕2$ is determined by the desired level of confidence.
We cannot solve for both n1 and n2, so we have to make an assumption about their relative sizes. We will specify that they be equal. With these assumptions we obtain the minimum sample sizes needed by solving the displayed equation for $n1=n2.$
### Minimum Equal Sample Sizes for Estimating the Difference in Two Population Proportions
The estimated minimum equal sample sizes $n1=n2$ needed to estimate the difference $p1−p2$ in two population proportions to within E percentage points at $100(1−α)$% confidence is
$n1=n2=(zα∕2)2(p^1(1−p^1)+p^2(1−p^2))E2(rounded up)$
Here we face the same dilemma that we encountered in the case of a single population proportion: the formula for estimating how large a sample to take contains the numbers $p^1$ and $p^2$, which we know only after we have taken the sample. There are two ways out of this dilemma. Typically the researcher will have some idea as to the values of the population proportions p1 and p2, hence of what the sample proportions $p^1$ and $p^2$ are likely to be. If so, those estimates can be used in the formula.
The second approach to resolving the dilemma is simply to replace each of $p^1$ and $p^2$ in the formula by 0.5. As in the one-population case, this is the most conservative estimate, since it gives the largest possible estimate of n. If we have an estimate of only one of p1 and p2 we can use that estimate for it, and use the conservative estimate 0.5 for the other.
### Example 15
Find the minimum equal sample sizes necessary to construct a 98% confidence interval for the difference $p1−p2$ with a margin of error E = 0.05,
1. assuming that no prior knowledge about p1 or p2 is available; and
2. assuming that prior studies suggest that $p1≈0.2$ and $p2≈0.3.$
Solution:
Confidence level 98% means that $α=1−0.98=0.02$ so $α∕2=0.01.$ From the last line of Figure 12.3 "Critical Values of " we obtain $z0.01=2.326.$
1. Since there is no prior knowledge of p1 or p2 we make the most conservative estimate that $p^1=0.5$ and $p^2=0.5.$ Then
$n1=n2=(zα∕2)2(p^1(1−p^1)+p^2(1−p^2))E2=(2.326)2((0.5)(0.5)+(0.5)(0.5))0.052=1082.0552$
which we round up to 1,083. We must take a sample of size 1,083 from each population.
2. Since $p1≈0.2$ we estimate $p^1$ by 0.2, and since $p2≈0.3$ we estimate $p^2$ by 0.3. Thus we obtain
$n1=n2=(zα∕2)2(p^1(1−p^1)+p^2(1−p^2))E2=(2.326)2((0.2)(0.8)+(0.3)(0.7))0.052=800.720848$
which we round up to 801. We must take a sample of size 801 from each population.
### Key Takeaways
• If the population standard deviations $σ1$ and $σ2$ are known or can be estimated, then the minimum equal sizes of independent samples needed to obtain a confidence interval for the difference $μ1−μ2$ in two population means with a given maximum error of the estimate E and a given level of confidence can be estimated.
• If the standard deviation $σd$ of the population of differences in pairs drawn from two populations is known or can be estimated, then the minimum number of sample pairs needed under paired difference sampling to obtain a confidence interval for the difference $μd=μ1−μ2$ in two population means with a given maximum error of the estimate E and a given level of confidence can be estimated.
• The minimum equal sample sizes needed to obtain a confidence interval for the difference in two population proportions with a given maximum error of the estimate and a given level of confidence can always be estimated. If there is prior knowledge of the population proportions p1 and p2 then the estimate can be sharpened.
### Basic
1. Estimate the common sample size n of equally sized independent samples needed to estimate $μ1−μ2$ as specified when the population standard deviations are as shown.
1. 90% confidence, to within 3 units, $σ1=10$ and $σ2=7$
2. 99% confidence, to within 4 units, $σ1=6.8$ and $σ2=9.3$
3. 95% confidence, to within 5 units, $σ1=22.6$ and $σ2=31.8$
2. Estimate the common sample size n of equally sized independent samples needed to estimate $μ1−μ2$ as specified when the population standard deviations are as shown.
1. 80% confidence, to within 2 units, $σ1=14$ and $σ2=23$
2. 90% confidence, to within 0.3 units, $σ1=1.3$ and $σ2=0.8$
3. 99% confidence, to within 11 units, $σ1=42$ and $σ2=37$
3. Estimate the number n of pairs that must be sampled in order to estimate $μd=μ1−μ2$ as specified when the standard deviation sd of the population of differences is as shown.
1. 80% confidence, to within 6 units, $σd=26.5$
2. 95% confidence, to within 4 units, $σd=12$
3. 90% confidence, to within 5.2 units, $σd=11.3$
4. Estimate the number n of pairs that must be sampled in order to estimate $μd=μ1−μ2$ as specified when the standard deviation sd of the population of differences is as shown.
1. 90% confidence, to within 20 units, $σd=75.5$
2. 95% confidence, to within 11 units, $σd=31.4$
3. 99% confidence, to within 1.8 units, $σd=4$
5. Estimate the minimum equal sample sizes $n1=n2$ necessary in order to estimate $p1−p2$ as specified.
1. 80% confidence, to within 0.05 (five percentage points)
1. when no prior knowledge of p1 or p2 is available
2. when prior studies indicate that $p1≈0.20$ and $p2≈0.65$
2. 90% confidence, to within 0.02 (two percentage points)
1. when no prior knowledge of p1 or p2 is available
2. when prior studies indicate that $p1≈0.75$ and $p2≈0.63$
3. 95% confidence, to within 0.10 (ten percentage points)
1. when no prior knowledge of p1 or p2 is available
2. when prior studies indicate that $p1≈0.11$ and $p2≈0.37$
6. Estimate the minimum equal sample sizes $n1=n2$ necessary in order to estimate $p1−p2$ as specified.
1. 80% confidence, to within 0.02 (two percentage points)
1. when no prior knowledge of p1 or p2 is available
2. when prior studies indicate that $p1≈0.78$ and $p2≈0.65$
2. 90% confidence, to within 0.05 (two percentage points)
1. when no prior knowledge of p1 or p2 is available
2. when prior studies indicate that $p1≈0.12$ and $p2≈0.24$
3. 95% confidence, to within 0.10 (ten percentage points)
1. when no prior knowledge of p1 or p2 is available
2. when prior studies indicate that $p1≈0.14$ and $p2≈0.21$
### Applications
1. An educational researcher wishes to estimate the difference in average scores of elementary school children on two versions of a 100-point standardized test, at 99% confidence and to within two points. Estimate the minimum equal sample sizes necessary if it is known that the standard deviation of scores on different versions of such tests is 4.9.
2. A university administrator wishes to estimate the difference in mean grade point averages among all men affiliated with fraternities and all unaffiliated men, with 95% confidence and to within 0.15. It is known from prior studies that the standard deviations of grade point averages in the two groups have common value 0.4. Estimate the minimum equal sample sizes necessary to meet these criteria.
3. An automotive tire manufacturer wishes to estimate the difference in mean wear of tires manufactured with an experimental material and ordinary production tire, with 90% confidence and to within 0.5 mm. To eliminate extraneous factors arising from different driving conditions the tires will be tested in pairs on the same vehicles. It is known from prior studies that the standard deviations of the differences of wear of tires constructed with the two kinds of materials is 1.75 mm. Estimate the minimum number of pairs in the sample necessary to meet these criteria.
4. To assess to the relative happiness of men and women in their marriages, a marriage counselor plans to administer a test measuring happiness in marriage to n randomly selected married couples, record the their test scores, find the differences, and then draw inferences on the possible difference. Let $μ1$ and $μ2$ be the true average levels of happiness in marriage for men and women respectively as measured by this test. Suppose it is desired to find a 90% confidence interval for estimating $μd=μ1−μ2$ to within two test points. Suppose further that, from prior studies, it is known that the standard deviation of the differences in test scores is $σd≈10.$ What is the minimum number of married couples that must be included in this study?
5. A journalist plans to interview an equal number of members of two political parties to compare the proportions in each party who favor a proposal to allow citizens with a proper license to carry a concealed handgun in public parks. Let p1 and p2 be the true proportions of members of the two parties who are in favor of the proposal. Suppose it is desired to find a 95% confidence interval for estimating $p1−p2$ to within 0.05. Estimate the minimum equal number of members of each party that must be sampled to meet these criteria.
6. A member of the state board of education wants to compare the proportions of National Board Certified (NBC) teachers in private high schools and in public high schools in the state. His study plan calls for an equal number of private school teachers and public school teachers to be included in the study. Let p1 and p2 be these proportions. Suppose it is desired to find a 99% confidence interval that estimates $p1−p2$ to within 0.05.
1. Supposing that both proportions are known, from a prior study, to be approximately 0.15, compute the minimum common sample size needed.
2. Compute the minimum common sample size needed on the supposition that nothing is known about the values of p1 and p2.
### Answers
1. $n1=n2=45$,
2. $n1=n2=56.$
3. $n1=n2=234$
1. $n1=n2=33.$
2. $n1=n2=35.$
3. $n1=n2=13$
1. $n1=n2=329$,
2. $n1=n2=255.$
1. $n1=n2=3383$,
2. $n1=n2=2846.$
1. $n1=n2=193$,
2. $n1=n2=128$
1. $n1=n2≈80$
2. $n1=n2≈34$
3. $n1=n2≈769$
|
Full of Math Examples
FULL OF MATH EXAMPLES
# Integration by substitution Examples
Here are the all examples in Integration by substitution method.
Integration by substitution is a general method for solving integration problems.
## Example #1
#### Find the integration of sin mx using substitution method.
We know that derivative of mx is m. Thus, we make the substitution mx=t so that mdx=dt.
Therefore,
$\quad \int \sin m x d x=\frac{1}{m} \int \sin t d t $$=-\frac{1}{m} \cos t+C$$=-\frac{1}{m} \cos m x+C$
## Example #4
We know that derivative of $\tan ^{-1} x=\frac{1}{1+x^{2}} .$ So, we use the substitution $\tan ^{-1} x=t$ so that $\frac{d x}{1+x^{2}}=d t$ Therefore, $\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x$$=\int \sin t d t$$=-\cos t+C$$=-\cos \left(\tan ^{-1} x\right)+\mathrm{C} ## Example #5 #### Find the integration of \int \sin ^{3} x \cos ^{2} x d x using substitution method. \int \sin ^{3} x \cos ^{2} x d x$$=\int \sin ^{2} x \cos ^{2} x(\sin x) d x $$=\int\left(1-\cos ^{2} x\right) \cos ^{2} x(\sin x) d x ## Example #6 #### Find the integration of \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} using substitution method. Derivative of \sqrt{x} is \frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}} By using substitution \sqrt{x}=t So that \frac{1}{2 \sqrt{x}} d x=d t giving d x=2 t d t ## Example #7 #### Find \int \frac{\sin \sqrt{x}}{\sqrt{x}} \mathrm{d} x using substitution method. Substitute \sqrt{x}=t Then, dt =\frac{1}{2} x^{-1 / 2} \mathrm{d} x$$ =\frac{1}{2 x^{1 / 2}} \mathrm{d} x$ $=\frac{1}{2 \sqrt{x}} \mathrm{d} x$
$\int \frac{\sin \sqrt{x}}{\sqrt{x}} \mathrm{d} x=2 \int \sin t \mathrm{d} t$
Then ,
\begin{aligned} 2 \int \sin t \mathrm{d} t &=-2 \cos t+c \\ &=-2 \cos \sqrt{x}+c \end{aligned}
## Example #8
#### Find $\int_{1 / 4}^{1 / 2} \frac{\cos (\pi t)}{\sin ^{2}(\pi t)} d t$ using substitution method.
Let $u=\sin (\pi t)$ so $d u=\pi \cos (\pi t) d t$
Then, $d u / \pi=\cos (\pi t) d t$
We change the limits to $\sin (\pi / 4)=\sqrt{2} / 2$ and $\sin (\pi / 2)=1$
Then ,
$\int_{1 / 4}^{1 / 2} \frac{\cos (\pi t)}{\sin ^{2}(\pi t)} d t$$=\int_{\sqrt{2} / 2}^{1} \frac{1}{\pi} \frac{1}{u^{2}} d u$$=\int_{\sqrt{2} / 2}^{1} \frac{1}{\pi} u^{-2} d u$$=\frac{1}{\pi}\left.\frac{u^{-1}}{-1}\right|_{\sqrt{2} / 2} ^{1}=-\frac{1}{\pi}+\frac{\sqrt{2}}{\pi}$
|
# Percentage Error
The difference between Approximate and Exact Values,
as a percentage of the Exact Value.
## Comparing Approximate to Exact
First find the Error:
Subtract one value from the other. Ignore any minus sign.
Example: I estimated 260 people, but 325 came.
260 − 325 = −65, ignore the "−" sign, so my error is 65
Then find the Percentage Error:
Show the error as a percent of the exact value, so divide by the exact value and make it a percentage:
Example continued: 65/325 = 0.2 = 20%
Percentage Error is all about comparing a guess or estimate to an exact value. See percentage change, difference and error for other options.
## How to Calculate
Step 1: Calculate the error (subtract one value from the other) ignore any minus sign. Step 2: Divide the error by the exact value (we get a decimal number) Step 3: Convert that to a percentage (by multiplying by 100 and adding a "%" sign)
## As A Formula
This is the formula for "Percentage Error":
|Approximate Value − Exact Value| × 100% |Exact Value|
(The "|" symbols mean absolute value, so negatives become positive)
Example: I thought 70 people would turn up to the concert, but in fact 80 did!
|70 − 80| |80| × 100% = 10 80 × 100% = 12.5%
I was in error by 12.5%
Example: The report said the carpark held 240 cars, but we counted only 200 parking spaces.
|240 − 200| |200| × 100% = 40 200 × 100% = 20%
The report had a 20% error.
We can also use a theoretical value (when it is well known) instead of an exact value.
### Example: Sam does an experiment to find how long it takes an apple to drop 2 meters.
The theoretical value (using physics formulas) is 0.64 seconds.
But Sam measures 0.62 seconds, which is an approximate value.
|0.62 − 0.64| |0.64| × 100% = 0.02 0.64 × 100% = 3% (to nearest 1%)
So Sam was only 3% off.
## Without "Absolute Value"
We can also use the formula without "Absolute Value". This can give a positive or negative result, which may be useful to know.
Approximate Value − Exact Value × 100% Exact Value
Example: They forecast 20 mm of rain, but we really got 25 mm.
20 − 25 25 × 100% = −5 25 × 100% = −20%
They were in error by −20% (their estimate was too low)
## In Measurement
Measuring instruments are not exact!
And we can use Percentage Error to estimate the possible error when measuring.
### Example: You measure the plant to be 80 cm high (to the nearest cm)
This means you could be up to 0.5 cm wrong (the plant could be between 79.5 and 80.5 cm high)
|
# In a recent survey, 248 students or 32% of the sample said they worked part time during the summer. How many students were surveyed?
Dec 8, 2016
There were 775 student surveyed.
#### Explanation:
This question can be rewritten as:
248 is 32% of what?
"Percent" or "%" means "out of 100" or "per 100", Therefore 32% can be written as $\frac{32}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
$248 = \frac{32}{100} \times n$
$\frac{100}{32} \times 248 = \frac{100}{32} \times \frac{32}{100} \times n$
$\frac{24800}{32} = \frac{\cancel{100}}{\cancel{32}} \times \frac{\cancel{32}}{\cancel{100}} \times n$
$775 = n$
|
# Common Core: 3rd Grade Math : Multiplying Within 100 to Solve Word Problems
## Example Questions
### Example Question #1 : Multiply And Divide Within 100 To Solve Word Problems: Ccss.Math.Content.3.Oa.A.3
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #2 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #3 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #4 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #5 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #6 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #801 : Operations & Algebraic Thinking
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #1 : Multiply And Divide Within 100 To Solve Word Problems: Ccss.Math.Content.3.Oa.A.3
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #9 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
Explanation:
Looking at the picture below, we have groups, each with oranges. To find the total we could add to itself times, or we could do multiplication. Remember, multiplication is the same thing as repeated addition.
Let's let equal the total number of oranges.
Our equation is
### Example Question #10 : Multiplying Within 100 To Solve Word Problems
There are boxes of oranges. Each box contains oranges. How many oranges are there?
|
# What are the first and second derivatives of f(x)=x^8*ln(x)?
Dec 24, 2015
Let's remember product rule: be $y = f \left(x\right) g \left(x\right)$, then $y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$
#### Explanation:
Solving:
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 8 {x}^{7} \ln x + {x}^{8} \left(\frac{1}{x}\right) = 8 {x}^{7} \ln x + {x}^{7} = {x}^{7} \left(8 \ln x + 1\right)$
Using product rule again:
${\left(\mathrm{df} \left(x\right)\right)}^{2} / \left({d}^{2} x\right) = 7 {x}^{6} \left(8 \ln x + 1\right) + {x}^{7} \left(\frac{8}{x}\right)$
${\left(\mathrm{df} \left(x\right)\right)}^{2} / \left({d}^{2} x\right) = 56 {x}^{6} \ln x + 7 {x}^{6} + 8 {x}^{6} = 56 {x}^{6} \ln x + 15 {x}^{6}$
${\left(\mathrm{df} \left(x\right)\right)}^{2} / \left({d}^{2} x\right) = {x}^{6} \left(56 \ln x + 15\right)$
|
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