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to let us know that this function is the inverse of f. World View Note: The notation used for functions was ο¬rst introduced by the great Swiss mathematician, Leonhard Euler in the 18th century. 1(x) = x For example, if f (x) = x + 5, we could deduce that the inverse function would be 5. If we had an input of 3, we could calculate f (3) = (3) + 5 = 8. Our f β output is 8. If we plug this output into the inverse function we get f β 5 = 3, which is the original input. 1(8) = (8) β β Often the functions are much more involved than those described above. It may be diο¬cult to determine just by looking at the functions if they are inverses. In order to test if two functions, f (x) and g(x) are inverses we will calculate the composition of the two functions at x. If f changes the variable x in some way, then g undoes whatever f did, then we will be back at x again for our ο¬nal solution. In otherwords, if we simplify (f g)(x) the solution will be x. If it is anything but x the functions are not inverses. β¦ 401 Example 524. Are f (x) = 3x + 4 3β and g(x) = x3 4 β 3 f inverses? Caculate composition x3 f (g(x)) Replace g(x) with 4 β 3 3 s 3 x3 β 3 3β x3 β x3 4 β 3 Substitute x3 4 β 3 4 + 4 Divide out the 3β²s 4 + 4 3β x3 x Combine like terms Take cubed root Simpliο¬ed to x! Yes, they are inverses! Our Solution for variable in f Example 525. Are h(x) = 2x + 5 and g(x) = x 2 β 2 5 inverses? Calculate composition x 2 β for variable in h h(g(x)) Replace g(x) with x 2 β 5 + 5 Distrubte 2 Substitute 10 + 5 Combine like terms 5 Did not simplify to x x No, they are not inverses Our Solution β β Example 526. Are f (x) = 3x 2 β 4x + 1 |
and g(x) = x + 2 4x 3 β inverses? Calculate composition f (g(x)) Replace g(x) with x + 2 4x 3 β f Substitute x + 2 4x 3 β x + 2 4x 3 β for variable in f x + 2 4x 3 β x + 2 4x 3 β 3 4 Distribute 3 and 4 into numerators 2 β + 1 402 2 3x + 6 3 β 4x + 8 3 4x β 4x + 1 β (3x + 6)(3 4x 3 (4x + 8)(3 4x 3 β β β β 4x) 4x) 2(3 β + 1(3 3x + 6 2(3 4x + 8 + 1(3 β 4x) 4x) 4x) 4x) β β β β 3x + 6 4x + 8 + 3 6 + 8x 4x β β 11x 11 Multiply each term by LCD: 3 4x β Reduce fractions Distribute Combine like terms Divide out 11 x Simpliο¬ed to x! Yes, they are inverses Our Solution While the composition is useful to show two functions are inverses, a more common problem is to ο¬nd the inverse of a function. If we think of x as our input and y as our output from a function, then the inverse will take y as an input and give x as the output. This means if we switch x and y in our function we will ο¬nd the inverse! This process is called the switch and solve strategy. Switch and solve strategy to ο¬nd an inverse: 1. Replace f (x) with y 2. Switch x and yβs 3. Solve for y 4. Replace y with f β 1(x) Example 527. Find the inverse of f (x) = (x + 4)3 y = (x + 4)3 x = (y + 4)3 + 2 2 Replace f (x) with y β Switch x and y 2 β Solve for y 2 β + 2 Add 2 to both sides 403 x + 2 = (y + 4)3 Cube root both sides 3β Subtract 4 from both sides = β 1(x) = x + 2 3β β β β 4 = y Replace y with f β 4 Our Solution 1(x) f β Example 528. Find the |
inverse of g(x) = 2x 3 β 4x + 2 Replace g(x) with y y = 2x 3 β 4x + 2 x = 2y 3 β 4y + 2 Switch x and y Multiply by (4y + 2) x(4y + 2) = 2y 4xy + 2x = 2y 4xy + 3 2x + 3 = 2y β 4xy + 3 4xy β β 3 Distribute 3 Move all y β²s to one side, rest to other side Subtract 4xy and add 3 to both sides Factor out y β β 2x + 3 = y(2 4x 2 2 4x) Divide by 2 4x 4x β β β β 2x + 3 4x 2 β = y Replace y with gβ 1(x) gβ 1(x) = 2x + 3 4x 2 β Our Solution In this lesson we looked at two diο¬erent things, ο¬rst showing functions are inverses by calculating the composition, and second ο¬nding an inverse when we only have one function. Be careful not to get them backwards. When we already have two functions and are asked to show they are inverses, we do not want to use the switch and solve strategy, what we want to do is calculate the inverse. There may be several ways to represent the same function so the switch and solve strategy may not look the way we expect and can lead us to conclude two functions are not inverses when they are in fact inverses. 404 10.3 Practice - Inverse Functions State if the given functions are inverses. β 5β 3 3 1) g(x) = f (x) = x5 β x β β x 1 3) f (x) = β β x 2 β 2x + 1 g(x) = β x 1 β 10x + 5 5 β 10 β f (x) = x 5) g(x) = β 7) f (x) = 2 x + 3 β g(x) = 3x + 2 x + 2 5 1 9) g(x) = x f (x) = 2x5 + 1 q β 2 Find the inverse of each functions. x 2) g(x) = 4 β x f (x) = 4 x 2 2x β x 4) h(x) = β f (x) = β |
2 x + 2 6) f (x) = x 5 β 10 h(x) = 10x + 5 5 8) f (x) = x + 1 2 g(x) = 2x5 q 1 β 10) g(x) = 8 + 9x f (x) = 5x 9 2 β 2 11) f (x) = (x 2)5 + 3 12) g(x) = x + 1 3β + 2 β x β 5 (x β 1 13) g(x) = 4 x + 2 15) f (x) = β 2x β x + 2 2 17) f (x) = 10 1)3 β 3)3 19) g(x) = β 21) f (x) = (x 23) g(x) = x x β 25) f (x) = x 1 β x + 1 27) g(x) = 8 29) g(x) = 5x β 4 5x + 1 β 31) g(x) = β 33) h(x) = 4 β 1 + x3 4x3β 2 35) f (x) = x + 1 x + 2 3 14) f (x) = β x 3 β 16) g(x) = 9 + x 3 18) f (x) = 5x 20) f (x) = 12 β 2 β 4 15 3x 22) g(x) = β 5 x + 2 2 2x q 3 24) f (x) = β β x + 3 26) h(x) = x x + 2 28) g(x) = β 30) f (x) = 5x x + 2 3 5 β 4 32) f (x) = 3 34) g(x) = (x β 2x5 1)3 + 2 36) f (x) = β β 1 x + 1 37) f (x) = 7 β x β x 39) g(x) = 3x 2 β 38) f (x) = β 40) g(x) = β 3x 4 2x + 1 3 405 10.4 Functions - Exponential Functions Objective: Solve exponential equations by ο¬nding a common base. As our study of algebra gets more advanced we begin to study more involved functions. One pair of inverse functions we will look at are exponential functions and logarithmic functions. Here we |
will look at exponential functions and then we will consider logarithmic functions in another lesson. Exponential functions are functions where the variable is in the exponent such as f (x) = ax. (It is important not to confuse exponential functions with polynomial functions where the variable is in the base such as f (x) = x2). World View Note One common application of exponential functions is population growth. According to the 2009 CIA World Factbook, the country with the highest population growth rate is a tie between the United Arab Emirates (north of Saudi Arabia) and Burundi (central Africa) at 3.69%. There are 32 countries with negative growth rates, the lowest being the Northern Mariana Islands (north of Australia) at 7.08%. β Solving exponetial equations cannot be done using the skill set we have seen in the past. For example, if 3x = 9, we cannot take the x root of 9 because we do not know what the index is and this doesnβt get us any closer to ο¬nding x. However, we may notice that 9 is 32. We can then conclude that if 3x = 32 then x = 2. This is the process we will use to solve exponential functions. If we can re-write a problem so the bases match, then the exponents must also match. β Example 529. 52x+1 = 125 Rewrite 125 as 53 52x+1 = 53 2x + 1 = 3 1 1 Same base, set exponents equal Solve Subtract 1 from both sides β β 2x = 2 Divide both sides by 2 2 x = 1 Our Solution 2 Sometimes we may have to do work on both sides of the equation to get a common base. As we do so, we will use various exponent properties to help. First we will use the exponent property that states (ax)y = axy. 406 Example 530. 83x = 32 Rewrite 8 as 23 and 32 as 25 (23)3x = 25 Multiply exponents 3 and 3x 29x = 25 9x = 5 9 9 5 9 x = Same base, set exponents equal Solve Divide both sides by 9 Our Solution As we multiply exponents we may need to distribute if there are several terms involved. Example 531. 273x+5 = 814x+1 Rewrite 27 as 33 and 81 as 34 (92 would not be same base) (33)3x+5 = ( |
34)4x+1 Multiply exponents 3(3x + 5) and 4(4x + 1) 39x+15 = 316x+4 Same base, set exponents equal 9x + 15 = 16x + 4 Move variables to one side 9x Subtract 9x from both sides Subtract 4 from both sides β β 9x 15 = 7x + 4 4 4 β β 7 11 = 7x Divide both sides by 7 7 11 7 = x Our Solution Another useful exponent property is that negative exponents will give us a reciprocal, 1 n an = aβ Example 532. 2x = 37x β 1 Rewrite 1 9 as 3β 2 (negative exponet to ο¬ip) β 1 Multiply exponents 1 2 and 2x Same base, set exponets equal Subtract 7x from both sides β β 1 1 9 (3β 2)2x = 37x 4x = 37x 3β 4x = 7x 7x 7x β 11x = 11 β β β β β 11 β 1 Divide by β 11 β 1 x = 11 Our Solution If we have several factors with the same base on one side of the equation we can add the exponents using the property that states axay = ax+ y. 407 Example 533. 54x Β· 52x 56x β 1 = 53x+11 Add exponents on left, combing like terms 1 = 53x+11 Same base, set exponents equal β 1 = 3x + 11 Move variables to one sides Subtract 3x from both sides 3x 6x 3x β β β 3x 1 = 11 Add 1 to both sides β + 1 + 1 3x = 12 Divide both sides by 3 3 3 x = 4 Our Solution It may take a bit of practice to get use to knowing which base to use, but as we practice we will get much quicker at knowing which base to use. As we do so, we will use our exponent properties to help us simplify. Again, below are the properties we used to simplify. (ax)y = axy and 1 an = aβn and axay = ax+ y We could see all three properties used in the same problem as we get a common base. This is shown in the next example. Example 534. 162x β 5 Β· (24)2x β 5 3x+1 = 32 1 4 2)3x+1 = 25 |
(2β 6x 20 Β· 28x β Β· 2 = 25 2β Β· x+3 x+3 1 2 1 Β· (2β Write with a common base of 2 Multiply exponents, distributing as needed β 22x x 2β β Β· x+2 22 = 2β 3 Add exponents, combining like terms Same base, set exponents equal β 22 = 2x + x β β + x x + 2 Move variables to one side Add x to both sides 22 = 2 Add 22 to both sides 3x β + 22 + 22 3x = 24 Divide both sides by 3 3 3 x = 8 Our Solution All the problems we have solved here we were able to write with a common base. However, not all problems can be written with a common base, for example, 2 = 10x, we cannot write this problem with a common base. To solve problems like this we will need to use the inverse of an exponential function. The inverse is called a logarithmic function, which we will discuss in another secion. 408 10.4 Practice - Exponential Functions Solve each equation. 1) 31 β 2n = 31 β 3n 3) 42a = 1 5) ( 1 25)β k = 125β 2k β 2 7) 62m+1 = 1 36 9) 6β 3x = 36 11) 64b = 25 13) ( 1 4)x = 16 15) 43a = 43 17) 363x = 2162x+1 19) 92n+3 = 243 21) 33x 2 = 33x+1 β 23) 3β 2x = 33 25) 5m+2 = 5β m 27) ( 1 36)b 1 = 216 β 29) 62 β 2x = 62 31) 4 Β· 3n 2β β 1 = 1 4 33) 43k β 3 35) 9β 2x Β· 42 2k = 16β k β Β· ( 1 243)3x = 243β x 37) 64n β 2 39) 5β 3n β Β· 3 16n+2 = ( 1 4 )3n 1 β 52n = 1 Β· 2) 42x = 1 16 4) 16β 3p = 64β 3p 6) 625β n 2 = 1 125 β 8) 62r 3 = 6r β β 3 10) 52n = 5β n 12) 216β 3v = 363v 14) 27β 2n 1 = 9 β 16 |
) 4β 3v = 64 18) 64x+2 = 16 20) 162k = 1 64 22) 243p = 27β 3p 24) 42n = 42 β 3n 26) 6252x = 25 28) 2162n = 36 30) ( 1 4)3v 2 = 641 β β v 32) 216 6β2a = 63a 34) 322p β 2 8p = ( 1 2 )2p Β· 36) 32m 38) 32 β Β· x 33m = 1 33m = 1 Β· 40) 43r 4β 3r = 1 64 Β· 409 10.5 Functions - Logarithmic Functions Objective: Convert between logarithms and exponents and use that relationship to solve basic logarithmic equations. n The inverse of an exponential function is a new function known as a logarithm. Lograithms are studied in detail in advanced algebra, here we will take an introductory look at how logarithms works. When working with radicals we found that anmβ could be written as there were two ways to write radicals. The expression a m. Each form has its advantages, thus we need to be comfortable using both the radical form and the rational exponent form. Similarly an exponent can be written in two forms, each with its own advantages. The ο¬rst form we are very familiar with, bx = a, where b is the base, a can be thought of as our answer, and x is the exponent. The second way to write this is with a logarithm, logba = x. The word βlogβ tells us that we are in this new form. The variables all still mean the same thing. b is still the base, a can still be thought of as our answer. Using this idea the problem 52 = 25 could also be written as log525 = 2. Both mean the same thing, both are still the same exponent problem, but just as roots can be written in radical form or rational exponent form, both our forms have their own advantages. The most important thing to be comfortable doing with logarithms and exponents is to be able to switch back and forth between the two forms. This is what is shown in the next few examples. Example 535. Write each exponential equation in logarithmic form m3 = 5 Identify base, m, answer, 5, and exponent 3 logm5 = 3 Our Solution 72 = b |
Identify base, 7, answer, b, and exponent, 2 log7b = 2 Our Solution 4 = 2 3 log 2 3 16 81 16 81 Identify base, 2 3, answer, 16 81, and exponent 4 = 4 Our Solution Example 536. Write each logarithmic equation in exponential form log416 = 2 Identify base, 4, answer, 16, and exponent, 2 42 = 16 Our Solution 410 log3x = 7 Identify base, 3, answer, x, and exponent, 7 37 = x Our Solution log93 = 1 1 2 Identify base, 9, answer, 3, and exponent, 1 2 9 2 = 3 Our Solution Once we are comfortable switching between logarithmic and exponential form we are able to evaluate and solve logarithmic expressions and equations. We will ο¬rst evaluate logarithmic expressions. An easy way to evaluate a logarithm is to set the logarithm equal to x and change it into an exponential equation. Example 537. Example 538. Evaluate log264 Set logarithm equal to x log264 = x Change to exponent form 2x = 64 Write as common base, 64 = 26 2x = 26 Same base, set exponents equal x = 6 Our Solution Evaluate log1255 Set logarithm equal to x log1255 = x Change to exponent form 125x = 5 Write as common base, 125 = 53 (53)x = 5 Multiply exponents 53x = 5 3x = 1 3 3 1 3 x = Same base, set exponents equal (5 = 51) Solve Divide both sides by 3 Our Solution Example 539. Evaluate log3 Set logarithm equal to x 1 27 = x Change to exponent form 1 27 Write as common base, = 3β 3 Same base, set exponents equal 3 Our Solution log3 1 27 1 3x = 27 3x = 3β 3 x = β World View Note: Dutch mathematician Adriaan Vlacq published a text in 1628 which listed logarithms calculated out from 1 to 100,000! 411 Solve equations with logarithms is done in a very similar way, we simply will change the equation into exponential form and try to solve the resulting equation. Example 540. Example 541. Example 542. log5x = 2 Change to exponential form 52 = x Evaluate exponent 25 = x Our Solution log2 |
(3x + 5) = 4 Change to exponential form 24 = 3x + 5 Evaluate exponent 16 = 3x + 5 Solve 5 5 Subtract 5 from both sides β β 3 11 = 3x Divide both sides by 3 3 11 3 = x Our Solution logx8 = 3 Change to exponential form x3 = 8 Cube root of both sides x = 2 Our Solution There is one base on a logarithm that gets used more often than any other base, base 10. Similar to square roots not writting the common index of 2 in the radical, we donβt write the common base of 10 in the logarithm. So if we are working on a problem with no base written we will always assume that base is base 10. Example 543. 2 Rewrite as exponent, 10 is base 2 = x Evaluate, remember negative exponent is fraction β log x = 10β 1 100 = x Our Solution This lesson has introduced the idea of logarithms, changing between logs and exponents, evaluating logarithms, and solving basic logarithmic equations. In an advanced algebra course logarithms will be studied in much greater detail. 412 10.5 Practice - Logarithmic Functions Rewrite each equation in exponential form. 1) log9 81 = 2 3) log7 1 49 = 2 β 5) log13 169 = 2 2) logb a = 16 β 4) log16 256 = 2 6) log11 1 = 0 Rewrite each equations in logarithmic form. 7) 80 = 1 9) 152 = 225 1 11) 64 6 = 2 Evaluate each expression. 13) log125 5 15) log343 1 7 17) log4 16 19) log6 36 21) log2 64 Solve each equation. 23) log5 x = 1 25) log2 x = β 27) log11 k = 2 2 29) log9 (n + 9) = 4 31) log5 ( 3m) = 3 β 33) log11 (x + 5) = β 35) log4 (6b + 4) = 0 1 37) log5 ( β 39) log2 (10 10x + 4) = 4 5a) = 3 β 8) 17β 2 = 1 289 1 10) 144 2 = 12 12) 192 = 361 14) log5 125 16) log7 1 18) log4 1 64 20) log36 6 22) |
log3 243 24) log8 k = 3 26) log n = 3 28) log4 p = 4 β 4) = β 8r = 1 30) log11 (x 32) log2 β 34) log7 β 36) log11 (10v + 1) = 3n = 4 38) log9 (7 β 40) log8 (3k 6x) = β 1) = 1 β 1 1 β 2 413 10.6 Functions - Compound Interest Objective: Calculate ο¬nal account balances using the formulas for compound and continuous interest. An application of exponential functions is compound interest. When money is invested in an account (or given out on loan) a certain amount is added to the balance. This money added to the balance is called interest. Once that interest is added to the balance, it will earn more interest during the next compounding period. This idea of earning interest on interest is called compound interest. For example, if you invest S100 at 10% interest compounded annually, after one year you will earn S10 in interest, giving you a new balance of S110. The next year you will earn another 10% or S11, giving you a new balance of S121. The third year you will earn another 10% or S12.10, giving you a new balance of S133.10. This pattern will continue each year until you close the account. There are several ways interest can be paid. The ο¬rst way, as described above, is compounded annually. In this model the interest is paid once per year. But interest can be compounded more often. Some common compounds include compounded semi-annually (twice per year), quarterly (four times per year, such as quarterly taxes), monthly (12 times per year, such as a savings account), weekly (52 times per year), or even daily (365 times per year, such as some student loans). When interest is compounded in any of these ways we can calculate the balance after any amount of time using the following formula: Compound Interest Formula: A = P 1 + nt r n A = Final Amount P = Principle (starting balance) r = Interest rate (as a decimal) n = number of compounds per year t = time (in years) Example 544. If you take a car loan for S25000 with an interest rate of 6.5% compounded quarterly, no payments required for the ο¬rst ο¬ve years, what |
will your balance be at the end of those ο¬ve years? P = 25000, r = 0.065, n = 4, t = 5 5 4 A = 25000 1 + Β· 0.065 4 A = 25000(1.01625)4 Identify each variable Plug each value into formula, evaluate parenthesis 5 Multiply exponents Β· 414 A = 25000(1.01625)20 Evaluate exponent A = 25000(1.38041977 ) Multiply A = 34510.49 S34, 510.49 Our Solution We can also ο¬nd a missing part of the equation by using our techniques for solving equations. Example 545. What principle will amount to S3000 if invested at 6.5% compounded weekly for 4 years? A = 3000, r = 0.065, n = 52, t = 4 4 52 3000 = P 1 + 0.065 52 Β· Identify each variable Evaluate parentheses 3000 = P (1.00125)52 4 Multiply exponent 3000 = P (1.00125)208 Evaluate exponent Β· 3000 = P (1.296719528 ) Divide each side by 1.296719528 1.296719528 1.296719528 2313.53 = P Solution for P S2313.53 Our Solution It is interesting to compare equal investments that are made at several diο¬erent types of compounds. The next few examples do just that. Example 546. If S4000 is invested in an account paying 3% interest compounded monthly, what is the balance after 7 years? P = 4000, r = 0.03, n = 12, t = 7 7 12 A = 4000 1 + Β· 0.03 12 Identify each variable Plug each value into formula, evaluate parentheses A = 4000(1.0025)12 7 Multiply exponents A = 4000(1.0025)84 Evaluate exponent Β· A = 4000(1.2333548) Multiply A = 4933.42 S4933.42 Our Solution 415 To investigate what happens to the balance if the compounds happen more often, we will consider the same problem, this time with interest compounded daily. Example 547. If S4000 is invested in an account paying 3% interest compounded daily, what is the balance after 7 years? P = 4000, r = 0.03, n = 365, t = 7 7 365 |
A = 4000 1 + Β· 0.03 365 Identify each variable Plug each value into formula, evaluate parenthesis A = 4000(1.00008219 )365 7 Multiply exponent A = 4000(1.00008219 )2555 Evaluate exponent Β· A = 4000(1.23366741.) Multiply A = 4934.67 S4934.67 Our Solution While this diο¬erence is not very large, it is a bit higher. The table below shows the result for the same problem with diο¬erent compounds. Compound Annually Balance S4919.50 Semi-Annually S4927.02 S4930.85 S4933.42 S4934.41 S4934.67 Quarterly Monthly Weekly Daily As the table illustrates, the more often interest is compounded, the higher the ο¬nal balance will be. The reason is, because we are calculating compound interest or interest on interest. So once interest is paid into the account it will start earning interest for the next compound and thus giving a higher ο¬nal balance. The next question one might consider is what is the maximum number of compounds possible? We actually have a way to calculate interest compounded an inο¬nite number of times a year. This is when the interest is compounded continuously. When we see the word βcontinuouslyβ we will know that we cannot use the ο¬rst formula. Instead we will use the following formula: Interest Compounded Continuously: A = Pert A = Final Amount P = Principle (starting balance) e = a constant approximately 2.71828183. r = Interest rate (written as a decimal) t = time (years) 416 The variable e is a constant similar in idea to pi (Ο) in that it goes on forever without repeat or pattern, but just as pi (Ο) naturally occurs in several geometry applications, so does e appear in many exponential applications, continuous interest being one of them. If you have a scientiο¬c calculator you probably have an e button (often using the 2nd or shift key, then hit ln) that will be useful in calculating interest compounded continuously. World View Note: e ο¬rst appeared in 1618 in Scottish mathematicianβs Napierβs work on logarithms. However it was Euler in Switzerland who used the letter e ο¬rst to represent this value. Some |
say he used e because his name begins with E. Others, say it is because exponent starts with e. Others say it is because Eulerβs work already had the letter a in use, so e would be the next value. Whatever the reason, ever since he used it in 1731, e became the natural base. Example 548. If S4000 is invested in an account paying 3% interest compounded continuously, what is the balance after 7 years? P = 4000, r = 0.03, t = 7 Identify each of the variables A = 4000e0.03 Β· 7 Multiply exponent A = 4000e0.21 Evaluate e0.21 A = 4000(1.23367806 ) Multiply A = 4934.71 S4934.71 Our Solution Albert Einstein once said that the most powerful force in the universe is compound interest. Consider the following example, illustrating how powerful compound interest can be. Example 549. If you invest S6.16 in an account paying 12% interest compounded continuously for 100 years, and that is all you have to leave your children as an inheritance, what will the ο¬nal balance be that they will receive? P = 6.16, r = 0.12, t = 100 Identify each of the variables A = 6.16e0.12 Β· 100 Multiply exponent A = 6.16e12 Evaluate A = 6.16(162, 544.79) Multiply A = 1, 002, 569.52 S1, 002, 569.52 Our Solution In 100 years that one time investment of S6.16 investment grew to over one million dollars! Thatβs the power of compound interest! 417 10.6 Practice - Compound Interest Solve 1) Find each of the following: a. S500 invested at 4% compounded annually for 10 years. b. S600 invested at 6% compounded annually for 6 years. c. S750 invested at 3% compounded annually for 8 years. d. S1500 invested at 4% compounded semiannually for 7 years. e. S900 invested at 6% compounded semiannually for 5 years. f. S950 invested at 4% compounded semiannually for 12 years. g. S2000 invested at 5% compounded quarterly for 6 years. h. S2250 invested at 4% compounded quarterly for 9 years. i. S3500 invested at 6% compounded quarterly |
for 12 years. 418 j. All of the above compounded continuously. 2) What principal will amount to S2000 if invested at 4% interest compounded semiannually for 5 years? 3) What principal will amount to S3500 if invested at 4% interest compounded quarterly for 5 years? 4) What principal will amount to S3000 if invested at 3% interest compounded semiannually for 10 years? 5) What principal will amount to S2500 if invested at 5% interest compounded semiannually for 7.5 years? 6) What principal will amount to S1750 if invested at 3% interest compounded quarterly for 5 years? 7) A thousand dollars is left in a bank savings account drawing 7% interest, compounded quarterly for 10 years. What is the balance at the end of that time? 8) A thousand dollars is left in a credit union drawing 7% compounded monthly. What is the balance at the end of 10 years? 9) S1750 is invested in an account earning 13.5% interest compounded monthly for a 2 year period. What is the balance at the end of 9 years? 10) You lend out S5500 at 10% compounded monthly. If the debt is repaid in 18 months, what is the total owed at the time of repayment? 11) A S10, 000 Treasury Bill earned 16% compounded monthly. If the bill matured in 2 years, what was it worth at maturity? 12) You borrow S25000 at 12.25% interest compounded monthly. If you are unable to make any payments the ο¬rst year, how much do you owe, excluding penalties? 13) A savings institution advertises 7% annual interest, compounded daily, How much more interest would you earn over the bank savings account or credit union in problems 7 and 8? 14) An 8.5% account earns continuous interest. If S2500 is deposited for 5 years, what is the total accumulated? 15) You lend S100 at 10% continuous interest. If you are repaid 2 months later, what is owed? 419 10.7 Functions - Trigonometric Functions Objective: Solve for a missing side of a right triangle using trigonometric ratios. There are six special functions that describe the relationship between the sides of a right triangle and the angles of the triangle. We will discuss three of the functions here. The three functions are called the sine, cosine, and tangent (the three others are cosecant, secant, and cotang |
ent, but we will not need to use them here). To the right is a picture of a right triangle. Based on which angle we are interested in on a given problem we will name the three sides in relationship to that angle. In the picture, angle A is the angle we will use to name the other sides. The longest side, the side opposite the right angle is always called the hypotenouse. The side across from the angle A is called the opposite side. Opposite Hypotenuse Adjacent A The third side, the side between our angle and the right angle is called the adjacent side. It is important to remember that the opposite and adjacent sides are named in relationship to the angle A or the angle we are using in a problem. If the angle had been the top angle, the opposite and adjacent sides would have been switched. The three trigonometric funtions are functions taken of angles. When an angle goes into the function, the output is a ratio of two of the triangle sides. The ratios are as describe below: sinΞΈ = opposite hypotenuse cosΞΈ = adjacent hypotenuse tanΞΈ = opposite adjacent The βweirdβ variable ΞΈ is a greek letter, pronounced βthetaβ and is close in idea to our letter βtβ. Often working with triangles, the angles are repesented with Greek letters, in honor of the Ancient Greeks who developed much of Geometry. Some students remember the three ratios by remembering the word βSOH CAH TOAβ where each letter is the ο¬rst word of: βSine: Opposite over Hypotenuse; Cosine: Adjacent over Hypotenuse; and Tangent: Opposite over Adjacent.β Knowing how to use each of these relationships is fundamental to solving problems using trigonometry. World View Note: The word βsineβ comes from a mistranslation of the Arab word jayb Example 550. 420 Using the diagram at right, ο¬nd each of the following: sinΞΈ, cosΞΈ, tanΞΈ, sinΞ±, cosΞ±, and tanΞ±. First we will ο¬nd the three ratios of ΞΈ. from ΞΈ, the The hypotenuse is 10, opposite side is 6 and the adjacent side is 8. So we ο¬ll in the following: sinΞΈ = opposite hypotenuse = |
6 10 = 3 5 5 adjacent = 6 tanΞΈ = opposite cosΞΈ = adjacent hypotenuse = 8 10 = 4 8 = 3 Now we will ο¬nd the three ratios of Ξ±. from Ξ±, the The hypotenuse is 10, opposite side is 8 and the adjacent side is 6. So we ο¬ll in the following: 4 8 ΞΈ 6 Ξ± 10 6 Opposite of ΞΈ Adjacent of Ξ± Ξ± Adjacent of ΞΈ Opposite of Ξ± 8 ΞΈ 10 Hypotenuse sinΞ± = opposite cosΞ± = adjacent 5 10 = 4 hypotenuse = 8 hypotenuse = 6 10 = 3 6 = 4 adjacent = 8 5 3 tanΞ± = opposite We can either use a trigonometry table or a calculator to ο¬nd decimal values for sine, cosine, or tangent of any angle. We only put angle values into the trigonometric functions, never values for sides. Using either a table or a calculator, we can solve the next example. Example 551. sin 42β¦ Use calculator or table 0.669 Our Solution tan 12β¦ Use calculator or table 0.213 Our Solution cos 18β¦ Use calculator or table 0.951 Our Solution By combining the ratios together with the decimal approximations the calculator or table gives us, we can solve for missing sides of a triangle. The trick will be to determine which angle we are working with, naming the sides we are working with, and deciding which trig function can be used with the sides we have. 421 Example 552. Find the measure of the missing side. x 25β¦ 4 We will be using the angle marked 25β¦, from this angle, the side marked 4 is the opposite side and the side marked x is the adjacent side. The trig ratio that uses the opposite and adjacent sides is tangent. So we will take the tangent of our angle. tan25β¦ = 0.466 1 = 4 x 4 x Tangent is opposite over adjacent Evaluate tan25β¦, put over 1 so we have proportion 0.466x = 4 0.466 0.466 Divide both sides by 0.466 Find cross product x = 8.58 Our Solution Example 553. Find the measure of the missing side. We will be using the angle marked 70β¦. From this angle, the x is the adjacent the hypotenuse. and the side is 9 The trig ratio that uses adjacent and hypotenuse |
is the cosine. So we will take the cosine of our angle. Cosine is adjacent over hypotenuse Evaluate cos70β¦, put over 1 so we have a proportion x 70β¦ 9 cos70β¦ = 0.342 1 = x 9 x 9 3.08 = 1x 3.08 = x Our Solution. Find the cross product. 422 10.7 Practice - Trigonometric Functions Find the value of each. Round your answers to the nearest ten-thousandth. 1) cos 71β¦ 3) sin 75β¦ 2) cos 23β¦ 4) sin 50β¦ Find the value of the trig function indicated. 5) sin ΞΈ 6) tan ΞΈ 7) sin ΞΈ 24 ΞΈ 25 8) sin ΞΈ 7 5 ΞΈ 4 3 8 15 ΞΈ 17 9) sin ΞΈ 8 2β 8 ΞΈ 8 3 23β 7 ΞΈ 16 10) cos ΞΈ 25 ΞΈ 20 15 Find the measure of each side indicated. Round to the nearest tenth. 11) 12) x A C 13 51β¦ B B 5 C 56β¦ x A 423 13) 15) 17) 19) B 13 C x 24β¦ A A x 71β¦ C9 B C x 68β¦ B A 6 B C6 71.4β¦ x A x 52.3β¦ B 5 C A A x B 63β¦ 7.6 C C 6 52.2β¦ B x A B 11 69β¦ C x A 14) 16) 18) 20) 424 21) 23) 25) 27) A 5 38β¦x B C B x 67β¦ A 4 C B x C 67.2β¦ 4 A B x 32β¦ A 4 C C B x 28β¦ 12 A B 12 48β¦ C x A A 7 C 16β¦ x B A x 13.1 B 64β¦ C 22) 24) 26) 28) 425 29) A 2.4 22β¦ 31) B 61β¦ x 3 C x B A C B 33) 35) 11 A 30β¦ x C x B C 75β¦ 11 A x B 29β¦ A 3.9 C B 12 53β¦ A x C C x 47β¦ B 3 A B 43β¦ x A 3 C 30) 32) 34) 36) 426 37) 39) A C 11 37.1β¦ x |
A B C 13.1 40β¦ x B 38) A 40) C 1.4 65β¦ B x C 18.1 A 35.5β¦ x B 427 10.8 Functions - Inverse Trigonometric Functions Objective: Solve for missing angles of a right triangle using inverse trigonometry. We used a special function, one of the trig functions, to take an angle of a triangle and ο¬nd the side length. Here we will do the opposite, take the side lengths and ο¬nd the angle. Because this is the opposite operation, we will use the inverse function of each of the trig ratios we saw before. The notation we will use for the inverse trig functions will be similar to the inverse notation we used with functions. sinβ1 opposite hypotenuse = ΞΈ cosβ1 adjacent hypotenuse = ΞΈ tanβ1 opposite adjacent = ΞΈ Just as with inverse functions, the 1 is not an exponent, it is a notation to tell us that these are inverse functions. While the regular trig functions take angles as inputs, these inverse functions will always take a ratio of sides as inputs. We can calculate inverse trig values using a table or a calculator (usually pressing shift or 2nd ο¬rst). β Example 554. sinA = 0.5 We donβ²t know the angle so we use an inverse trig function 1(0.5) = A Evaluate using table or calculator sinβ 30β¦ = A Our Solution cosB = 0.667 We donβ²t know the angle so we use an inverse trig function 1(0.667) = B Evaluate using table or calculator cosβ 48β¦ = B Our Solution tanC = 1.54 We donβ²t know the angle so we use an inverse trig function 1(1.54) = C Evaluate using table or calculator tanβ 57β¦ = C Our Solution If we have two sides of a triangle, we can easily calculate their ratio as a decimal and then use one of the inverse trig functions to ο¬nd a missing angle. Example 555. Find the indicated angle. 428 ΞΈ 17 12 From angle ΞΈ the given sides are the opposite (12) and the hypotenuse (17). The trig function that uses opposite and hypotenuse is the sine Because we are looking for an angle we use the inverse sine sinβ 1 12 17 Sine is opposite over hyptenuse, use inverse to οΏ½ |
οΏ½nd angle sinβ 1(0.706) Evaluate fraction, take sine inverse using table or calculator 45β¦ Our Solution Example 556. Find the indicated angle 5 3 Ξ± From the angle Ξ±, the given sides are the opposite (5) and the adjacent (3) The trig function that uses opposite and adjacent is the tangent As we are looking for an angle we will use the inverse tangent. tanβ 1 5 3 Tangent is opposite over adjacent. Use inverse to ο¬nd angle tanβ 1(1.667) Evaluate fraction, take tangent inverse on table or calculator 59β¦ Our Solution Using a combination of trig functions and inverse trig functions, if we are given two parts of a right triangle (two sides or a side and an angle), we can ο¬nd all the other sides and angles of the triangle. This is called solving a triangle. When we are solving a triangle, we can use trig ratios to solve for all the missing parts of it, but there are some properties from geometry that may be helpful along the way. The angles of a triangle always add up to 180β¦, because we have a right triangle, 90β¦ are used up in the right angle, that means there are another 90β¦ left in the two acute angles. In other words, the smaller two angles will always add to 90, if we know one angle, we can quickly ο¬nd the other by subtracting from 90. Another trick is on the sides of the angles. If we know two sides of the right triangle, we can use the Pythagorean Theorem to ο¬nd the third side. The Pythagorean Theorem states that if c is the hypotenuse of the triangle, and a and 429 b are the other two sides (legs), then we can use the following formula, a2 + b2 = c2 to ο¬nd a missing side. Often when solving triangles we use trigonometry to ο¬nd one part, then use the angle sum and/or the Pythagorean Theorem to ο¬nd the other two parts. Example 557. Solve the triangle 5 We have one angle and one side. We can use these to ο¬nd either other side. We will ο¬nd the other leg, the adjacent side to the 35β¦ angle. 35β¦ The 5 is the opposite side, so we will use the tangent to ο¬ |
nd the leg. Tangent is opposite over adjacent 5 tan35β¦ = x 5 0.700 x 1 0.700x = 5 0.700 0.700 Divide both sides by 0.700 Find cross product = Evaluate tangent, put it over one so we have a proportion x = 7.1 The missing leg. a2 + b2 = c2 We can now use pythagorean thorem to ο¬nd hypotenuse, c 52 + 7.12 = c2 Evaluate exponents 25 + 50.41 = c2 Add 75.41 = c2 Square root both sides 8.7 = c The hypotenuse 90β¦ β 35β¦ To ο¬nd the missing angle we subtract from 90β¦ 55β¦ The missing angle 55β¦ 5 8.7 35β¦ 7.1 Our Solution In the previous example, once we found the leg to be 7.1 we could have used the sine function on the 35β¦ angle to get the hypotenuse and then any inverse trig 430 function to ο¬nd the missing angle and we would have found the same answers. The angle sum and pythagorean theorem are just nice shortcuts to solve the problem quicker. Example 558. Solve the triangle In this triangle we have two sides. We will ο¬rst ο¬nd the angle on the right side, adjacent to 3 and opposite from the 9. 3 Tangent uses opposite and adjacent To ο¬nd an angle we use the inverse tangent. 9 tanβ 1 tanβ Evaluate fraction 9 3 1(3) Evaluate tangent 71.6β¦ The angle on the right side 90β¦ β Subtract angle from 90β¦ to get other angle 71.6β¦ 18.4β¦ The angle on the left side a2 + b2 = c2 Pythagorean theorem to ο¬nd hypotenuse 92 + 32 = c2 Evaluate exponents 81 + 9 = c2 Add 90 = c2 3 10β or 9.5 = c Square root both sides The hypotenuse 9.5 18.4β¦ 9 71.6β¦ 3 Our Solution World View Note: Ancient Babylonian astronomers kept detailed records of the starts, planets, and eclipses using trigonometric ratios as early as 1900 BC! 431 10.8 Practice - Inverse Trigonometric Functions Find each angle measure to the nearest |
degree. 1) sin Z = 0.4848 3) sin Y = 0.6561 2) sin Y = 0.6293 4) cos Y = 0.6157 Find the measure of the indicated angle to the nearest degree. 5) 7) 9) 35? 32 30? 31 3? 6 6) 8) 39? 46 24? 8 10) 16 23? 432 Find the measure of each angle indicated. Round to the nearest tenth. 11) B 8 A ΞΈ 8 C 13) 15) A ΞΈ 11 C B 11 B 7 ΞΈ A 4 C 17) A ΞΈ 10 C 16 B 12) 14) 16) 18) ΞΈ B 7 C 13 C 7 B ΞΈ 9.7 A A B 12 ΞΈ A 13 C ΞΈ 15.3 A B 5.6 C 433 19) 21) 23) 25) C 9.3 ΞΈ A 13.2 B B 5 C ΞΈ 4 A C 10 B B ΞΈ A 12 C 9 ΞΈ 15.7 A 20) A ΞΈ 7 B A B A ΞΈ 15 C 9 ΞΈ 6.8 2 C B 22) 24) 26) B 4 15 A ΞΈ 6 C C 434 ΞΈ B 14 C 6 A Solve each triangle. Round answers to the nearest tenth. 27) A ΞΈ 14 C 15 B 29) C 7 A 31) 33) C 28.4 B 62β¦ A C 2.9 7 28) 30) B ΞΈ 15 B ΞΈ 7 32) 34) B 15 C 4 A A 6.3 C C 51β¦ 9.3 B A B A 435 35) 7 B 3 C 37) B 16 39) A C 52β¦ A A B 45β¦ 8 C 36) A 7 38) C 21β¦ B A 48β¦ B 10.4 C 40) B 14 C 6.8 A 436 437 0.1 2 1) β 2) 5 3) 2 4) 2 6 5 5) β 6) β 7) 8 8) 0 9) 2 β 10) β 11) 4 12) β 13) 3 14) 15) 16) 17) 18) β β β β β 19) β 20 21) 7 β 0.2 1) 7 2 2) 5 4 3) 7 5 Answers - Chapter 0 Answers - Integers 20 43) β 44) 27 45) 24 β 3 46) β 47) |
7 48) 3 49) 2 50) 5 51) 2 52) 9 53) 7 54) 10 β 55) 4 56) 10 8 57) β 58) 6 59) 60) 6 9 β β 22) 0 23) 11 24) 9 25) 26) β β 27) β 28) 4 29) 0 30) 31) 32) β β β 33) β 34) 14 35) 8 36) 6 3 4 3 8 4 35 80 56 6 36 10 37) 38) β β 39) β 40) 63 41) β 42) 4 Answers - Fractions 4) 8 3 5) 3 2 6) 5 4 438 7) 5 4 8) 4 3 9) 3 2 10) 8 3 11) 5 2 12) 8 7 13) 7 2 14) 4 3 15) 4 3 16) 3 2 17) 6 5 18) 7 6 19) 3 2 20) 8 7 21) 8 22) 5 3 4 9 2 3 13 4 23) 24) β β 25) β 26) 3 4 27) 33 20 28) 33 56 29) 4 30) 18 7 31) 1 2 32) 19 20 β 59) 37 20 60) β 61) 33 20 62) 3 7 63) 47 56 64) β 65) 2 3 66) β 67) 1 68) 7 8 69) 19 20 70) 71) β β 72) β 73) 34 7 74) 75) 76) β β β 77) β 78) 39 14 5 3 7 6 4 3 2 5 145 56 29 15 23 3 3 8 2 3 5 24 5 6 79) β 80) 1 10 81) 2 82) 62 21 17 15 7 10 8 7 21 26 3 2 5 27 1 10 45 7 1 10 7 33) 3 34) β 35) β 36) 5 14 37) β 38) 20 21 39) 2 9 40) 4 3 41) β 42) 25 21 43) β 44) β 45) 40 9 46) β 47) β 48) 13 15 49) 4 27 50) 32 65 51) 1 15 52) 1 53) β 54) β 55) 2 7 56) 2 57) 3 58) 31 8 β 439 0.3 1) 24 1 2) β 3) 5 4) 180 5) 4 6) 8 7) 1 8) 8 9) 6 0.4 1) 7 2) 29 3) 1 4) 3 5) 23 6) 14 |
7) 25 8) 46 9) 7 10) 8 11) 5 12) 10 13) 1 14) 6 15) 1 16) 2 17) 36 18) 54 Answers - Order of Operation 6 10 9 22 10) 11) β β 12) β 13) 20 14) β 15) 2 16) 28 17) 18) 40 15 β β Answers - Properties of Algebra 19) 7 20) 38 21) r + 1 4x 2 β 22) β 23) 2n 24) 11b + 7 25) 15v 26) 7x 27) 9x β 7a 1 β 28) β 29) k + 5 30) 31) 32) 33) β β β β 3p 5x β 9 m 9 β 10n 34) 5 r β 35) 10n + 3 β 19) 3 20) 0 21) 22) 23) 18 3 4 β β β 24) 3 25) 2 37) 8x + 32 β 38) 24v + 27 39) 8n2 + 72n 40) 5 9a β 7k2 + 42k 41) β 42) 10x + 20x2 6 β 2n 43) β 44) β 45) 40m 36x 2 β 8m2 β 46) 47) β β 18p2 + 2p 36x + 9x2 48) 32n 8 β 9b2 + 90b 49) 50) β β 28r 4 β 40n 51) β 52) 16x2 80n2 β 20x β 36) 5b 53) 14b + 90 440 54) 60v 7 β 3x + 8x2 89x + 81 68k2 19 β 8k β 90a 34 49p β 10x + 17 55) 56) 57) 58) 59) β β β β β 60) β 61) 10 4n β 30 + 9m 62) β 63) 12x + 60 64) 30r 16r2 74) 2x2 6x 8n2 75) 4p 48 β β 4b2 45 β β 72n 42b β 79v β 8x + 22 20n2 + 80n 42 β 12 + 57a + 54a2 65) β 66) β 67) 79 68) 69) 70) 71) β β β β 20k 75 β 128x 121 72) β 73) 4n2 β 3n β 5 β 3 β 7 β 76) 3x2 + 7x 77) |
78) β β v2 + 2v + 2 7b2 + 3b 8 β 79) 4k2 + 12 β 80) a2 + 3a 81) 3x2 15 β n2 + 6 β 5 82) β β 1.1 1) 7 2) 11 5 3) β 4) 4 5) 10 6) 6 7) 19 β 6 8) β 9) 18 10) 6 11) 12) 20 7 β β 108 13) β 14) 5 1.2 1) 4 β Answers - Chapter 1 Answers to One-Step Equations 8 15) β 16) 4 17) 17 18) 4 19) 20 20) β 21) 3 22) 16 208 23) 13 β 9 24) β 25) 15 26) 8 27) 28) 10 204 β β 29) 5 30) 2 11 14 31) β 32) β 33) 14 34) 1 35) 36) 37) 38) 39) 40) 11 15 240 135 16 380 β β β β β β Answers to Two-Step Equations 2) 7 441 3) 14 β 2 4) β 5) 10 12 6) β 7) 0 8) 12 9) 10 β 16 10) β 11) 14 7 12) β 13) 4 5 14) β 15) 16 1.3 3 1) β 2) 6 3) 7 4) 0 5) 1 6) 3 7) 5 4 8) β 9) 0 10) 3 11) 1 12) All real numbers 13) 8 14) 1 15) 7 β 15 16) β 17) 7 18) 12 19) 9 20) 0 21) 11 22) 6 β 10 23) β 24) 13 25) 1 26) 4 9 27) β 28) 15 6 29) β 30) 6 31) 16 β 4 32) β 33) 8 34) 13 β 2 35) β 36) 10 12 37) β 38) 0 39) 12 40) 9 β Answers to General Linear Equations 16) 0 17) 2 18) β 19) β 20) 3 21) 3 22) 23) β β 24) β 25) 8 26) 0 27) β 28) 5 29) β 30 31) β 32) 0 33) β 34) 0 35) 0 36) 37) β β 38) β 39) 5 40) 6 41) 0 42) 2 β 43) No Solution 44) 0 442 45) 12 46) All real |
numbers 47) No Solution 48) 1 9 49) β 50) 0 1.4 1) 3 4 2) β 3) 6 5 4) 1 6 5) β 6) 25 8 7) 8) β β 9) β 10) 3 2 1.5 4 3 19 6 7 9 1 3 2 1. b = c a 2. h = gi 3. x = gb f 4. y = pq 3 5. x = a 3b 6. y = cb dm 7. m = E c2 8. D = ds S 9. Ο = 3V 4r3 10. m = 2E v2 Answers to Solving with Fractions 3 2 4 3 5 3 11) 0 12) 4 3 13) β 14) 1 2 15) β 16) 1 17) 0 18) β 19) 1 20) 1 21) 1 2 Answers - Formulas 11. c = b a β 12. x = g + f 13. y = cm + cn 4 3) 14. r = k(a β 5 15. D = 12V Οn 16. k = F R β 17. n = P p β 18. L = S c L 2B β 19. D = TL + d 20. Ea = IR + Eg 21. Lo = L 1 + at 443 22) 23) 24 25) 16 26) 27) β β 28) β 29) 4 3 30) 3 2 22. x = c b β a 23. m = p q β 2 24. L = q + 6p 6 25. k = qr + m 26. T = R b β a 27. v = 16t2 + h t 28. h = s Οr2 β Οr 29. Q2 = Q1 + PQ1 P 30. r1 = L β 2d Ο β Οr2 31. T1 = Rd kAT2 β kA 32. v2 = Pg + V1 V1 2 33. a = c b β x 34. r = d t 35. w = V βh 36. h = 3v Οr2 1 1 β a β b 37. a = c 38. b = c 39. t = 5 + bw a s 40. w = at 41. x = c 42. x = 3 43. y = 3 β x β b bx 5y |
β x β 5 45. y = 7 3x β 2 46. a = 7b + 4 5 47. b = 5a 4 β 7 48. x = 8 + 5y 4 49. y = 4x 8 β 5 44. x = 7 2y β 3 50. f = 9c + 160 5 Answers to Absolute Value Equations 1.6 1) 8, 2) 7, 3) 1, 4) 2, 5) 6, 6) 38 9 7) β 8) β 9) 3 29 4 6, β 2, β 3, 9 39 7 β, β 6 29 3 β 1, 3 β 9, 15 10) 16 5 11) 7, 12) 13) 1.7 β β 5 3 β 2, 0 14) 3, 15) β 16) 0, 2 β 6, 0 7 4, 4 3, 7 2 17 2, 6 5 6, 2 8 β β 25 3 β β 13 7 21 β 2, 10 7 5, 1 21) β 22) 6, 23) 1, 24) 7, 17) 18) 19) 20) β β β β 10 3 1 25) 26) β β 1) c 2) x a = k yz = k 3) wx = k 4) r s2 = k 5) f xy = k Answers - Variation 6) jm3 = k 7) h b = k x a b2β = k 8) 9) ab = k 10) a b = 3 444 27) 6, 16 3 β 28) 2 5, 0 13 7, 1 3, 5 2 7 4 3, β 6, 2 5 29) 30) 31) β β β 32) β 33) 7, 1 5 34) 35) 22 5, 19 22, β β 2 13 11 38 β β 36) 0, 12 5 β 11) P rq = 0.5 12) cd = 28 13) t u2 = 0.67 14) e fg = 4 15) wx3 = 1458 23) 241,920,000 cans 32) 1600 km 16) h j = 1.5 a x y2β 17) = 0.33 18) mn = 3.78 19) 6 k 20) 5.3 k 21) 33.3 cm 22) 160 kg/cm3 1.8 1) 11 2) 5 4 3) β 4) 32 13 5) β |
6) 62 7) 16 8) 17 4 9) 35, 36, 37 10) β 11) 43, 14, β 12) 52, 54 42, 13, 41 12 β β β β 13) 61, 63, 65 14) 83, 85, 87 15) 9, 11, 13 16) 56, 56, 68 1.9 1) 6, 16 2) 10, 40 3) 18, 38 24) 3.5 hours 25) 4.29 dollars 26) 450 m 27) 40 kg 28) 5.7 hr 29) 40 lb 30) 100 N 31) 27 min 33) r = 36 34) 8.2 mph 35) 2.5 m 36) V = 100.5 cm3 37) 6.25 km 38) I = 0.25 Answer Set - Number and Geometry 33) 40, 200 34) 60, 180 35) 20, 200 36) 30, 15 37) 76, 532 38) 110, 880 39) 2500, 5000 40) 4, 8 41) 2, 4 42) 3, 5 43) 14, 16 44) 1644 45) 325, 950 17) 64, 64, 52 18) 36, 36, 108 19) 30, 120, 30 20) 30, 90, 60 21) 40, 80, 60 22) 28, 84, 68 23) 24, 120, 36 24) 32, 96, 52 25) 25, 100, 55 26) 45, 30 27) 96, 56 28) 27, 49 29) 57, 83 30) 17, 31 31) 6000, 24000 32) 1000, 4000 Answers - Age Problems 4) 17, 40 5) 27, 31 6) 12, 48 445 7) 31, 36 8) 16, 32 9) 12, 20 10) 40, 16 11) 10, 6 12) 12, 8 13) 26 14) 8 15) 4 16) 3 17) 10, 20 18) 14 1.10 1) 1 1 3 2) 25 1 2 3) 3 4) 10, 20 1 2 5) 30, 45 6) 3 7) 300 13 8) 10 9) 7 10) 30 11) 150 12) 360 13) 8 2.1 19) 9, 18 20) 15, 20 21) 50, 22 22) 12 23) 72, 16 24) 6 25) 37, 46 26) 15 27) 45 28) 14, 54 29) 8, 4 30 |
) 16, 32 31) 10, 28 32) 12,20 33) 141, 67 34) 16, 40 35) 84, 52 36) 14, 42 37) 10 38) 10, 6 39) 38, 42 40) 5 Answers - Distance, Rate, and Time Problems 14) 10 15) 2 16) 3 17) 48 18) 600 19) 6 20) 120 21) 36 22) 2 23) 570 24) 24, 18 25) 300 26) 8, 16 27) 56 28) 95, 120 29) 180 30) 105, 130 31) 2:15 PM 32) 200 33) 1 3 34) 15 35) 27 4 36) 1 2 37) 3, 2 38) 90 Answers - Chapter 2 446 1) B(4, E( H( K( β β β 3) 6) 9) 12) Answers - Points and Lines 3) C(1, 2) D( 1, 4) 2) β 5, 0) F(2, 1, β 4, 3) 4) I( β β 3) G(1, 3) 1) J(0, 2) β 2) 8) 11) 14) 4) 7) 10) 13) 447 15) 18) 21) 2.2 1) 3 2 2) 5 3) Undeο¬ned 1 2 4) β 5) 5 6 2 3 1 6) β 7) β 8) 5 4 1 9) β 10) 0 11) Undeο¬ned 12) 16 7 17 31 3 2 β β 13) 14) 2.3 17) 20) 7 13 5 8 5 4 28) 1 16 29) β 30) 2 7 31) β 32) 2 33) β 34) 3 35) β 36) 6 37) β 38) 1 39) 2 40) 1 16) 19) 22) Answers - Slope 7 17 15) 4 3 16) β 17) 0 18) 5 11 19) 1 2 20) 1 16 21) 22) 11 2 12 31 β β 23) Undeο¬ned 24) 24 11 25) 26) 27) 26 27 19 10 1 3 β β β 448 1) y = 2x + 5 2) y = β 3) y = x 4) y = 5) y = β β 6) y = β 7) y = 1 3 8) y = 2 5 6x + ) y = β 10 |
) y = β 11) y = x 12) y = 13) y = 14 4x 3 β 3 4 x + 2 1 10 x 15) y = β 3 10 β 16) y = 1 10x β 2x 1 β x + 70 11 17) y = β 18) y = 6 11 19 20) y = β 21) x = β 22) y = 1 7 8 x + 6 37 10 x 23) y = β 24) y = 5 x 2 1 β 25) y = 4x x + 1 2 3 4x + 3 26) y = β 27) y = β 28) x = 4 2.4 Answers - Slope-Intercept 1 2 x + 1 37) 29) y = β 30) y = 6 5 x + 4 38) 39) 40) 41) 42) 31) 32) 33) 34) 35) 36) 449 1) x = 2 2) x = 1 2) 2) 4) 3) y β 2 = 1 2 (x 4) y 1 = β β 5) y + 5 = 9(x + 1) β β 2 (x 1 6) y + 2 = 2(x β β 1 = 3 4 (x + 4) 2(x 7) y β 8) y + 3 = β β 9) y + 2 = 3x β 10) y 1 = 4(x + 1) β 11 12) y 2 = β β 13) y + 3 = 1 5 (x + 5) 14) y + 4 = 4 = 15) y β 16) y + 4 = 17) y = 2x 2 3(x + 1) 5 4 (x + 1) β β 3 2 x(x 1) β β 3 β 2x + 2 18) y = β 2.5 2 3 10 3 1) 2 2) β 3) 4 4) β 5) 1 6) 6 5 7) 8) 7 3 4 β β Answers - Point-Slope Form 19) 20) y = β 21 10 3 37) y + 2 = 3 2(x + 4) 38) y β 1 = 3 8 (x + 4) 5 = 1 4 (x 39) y β 40) y + 4 = 3) β (x + 1) 41) y + 3 = 42) y + 5 = β β β 3) 8 7(x β 4(x + 1) 1 3 4 x β |
11 4 3 2 1 10 43) y = 44) y = β β 45) y = β 46) y = 1 2 x 47) y = β 48) y = 1 3 x + 1 49) y = x + 2 β 50) y = x + 2 51) y = 4x + 3 52 22) y = 23) y = 24) y = 7 4 3 2 5 2 β β β 25) y = β 26) y = 7 x 3 27 28) y = β 3 β 3 29) x = β 30) y = 2x 31) y = β 32) y = 6 x 5 3 = 33) y β 34(x + 4) β 35) y 36) y β β 1 = 1 8(x 5 = β 5) β 8(x + 4) 1 Answers - Parallel and Perpendicular Lines 9) 0 10) 2 11) 3 12) 13) β β 14) β 15) 2 16) β 5 4 3 1 3 3 8 17) x = 2 18) y 19) y 2 = 7 5 (x 4 = 9 2 (x 5) 3) β β β β 20x 3 4(x β 1) β 2) 3 = β 3(x + 1) 21) y 22) y β β 450 23) x = 4 1) β 4 = 7 24) y β 25) y + 5 = 5 (x (x β 26) y + 2 = 1) 1) β 2(x 27) y 28x (x 3 = β β β 5) 1) β 4 (x 1 29) y 2 = β β β 30) y + 5 = 7 3 (x + 3) 3(x 31) y + 2 = 4) 2) β β 1 2(x + 2) 5 = 32) y β 33) y = β 34) y = 3 5 β 2x + 5 x + 5 x x x 35) y = 36 37) y = β 38) y = 5 x 2 39) y = β 40 41) y = x 1 β 42) y = 2x + 1 43) y = 2 44) y = 45) y = 46) y = β β β 47) y = β 48 2x + 5 x + 4 β Answers - Chapter 3 Answers - Solve and Graph Inequalities 18) x < 6: (, 6) β β 19) a < 12: (, 12 |
) 20) v > 1: [1, β 21) x > 11: [11, β β ) ) β 2] 22) x 6 18: ( β 23) k > 19: (19, β β ) β 18], β 10: (, 10] 24) n 6 25) p < β β 1: ( 26) x 6 20: ( β β, β 1) β β β, 20] 27) m > 2: [2, 28) n 6 5: (, 5] 29) r > 8: (8], β 30) x 6 3: ( β 31) b > 1: (1, 32) n > 0: [0, β β ) β ) β β 26, ) ) β 13) x > 110: [110, 26: [ 14) n > β 15) r < 1: ( β, 1) β β 6: ( 16) m 6 β 17) n > 6: [ β β β β 6, β ) β, 6] 33) v < 0: (, 0) 34) x > 2: (2, β β ) β 451 3.1 1) ( 2) ( 3) ( 4) ( 5) ( 5, 4], 5] ) β 2 β β 5, 6) ( β 7) m < β 8) m 6 1 9) x > 5 5 2 10) a 6 β 11) b > β 12) x > 1 β ) 35) No solution: 36) n > 1: (1, β 3.2 37) {All real numbers.} : R 38) p 6 3: (, 3] β β Answers - Compound Inequalities 4, ) β β ) 1) n 6 β 9 or n > 2: ( 2) m > 4 or m < β 3) x > 5 or x < β 5: ( β 4) r > 0 or r < β 7 : ( 5) x < 6) n < β β, 7: ( β β β 7 or n > 8 : ( ) β [ 9], β [2,, 5) S β β β, 5) [5, S β 7), S (0) 7), β β β S S (8, ) β 7) β 8 < v < |
3: ( β 8) 7 < x < 4: ( 8, 3) 7, 4) β 9) b < 5: ( β, 5) β β 2 6 n 6 6: [ 10) β 11) 7 6 a 6 6: [ β 12) v > 6: [6, ) β 2, 6] 7, 6] β β 13) β 14) 2: [ 2] β 6, β 9, 0] β β 15) 3 < k 6 4: (3, 4] 16) 17) β β 2 6 n 6 4: [ 2 < x < 2: ( 2, 4] 2, 2) β β 18) No solution : β 1 6 m < 4: [ 19) β 20) r > 8 or r < β 21) No solution : β 1, 4) β 6: : ( 6), β β β (8, ) β S 22) x 6 0 or x > 8 : (, 0] β β (8, ) β 23) No solution : β 24) n > 5 or n < 1 : ( S, 1) β β [5, ) β S 452 25) 5 6 x < 19: [5, 19) 26) n < 14 or n > 17 : ( β 27) 1 6 v 6 8: [1, 8] 14), β β β [17, ) β S 28) a 6 1 or a > 19 : (, 1] β β [19, ) β 29) k > 2 or k < 20 : ( β β 30) {All real numbers.} : R β, 20) S β [2, ) β S 31) β 1 < x 6 1: ( 32) m > 4 or m 6 β β 1, 1] 1 : ( 1], β β β (4, ) β S Answers - Absolute Value Inequalities 3.3 1) 2) 3) 4) 5) 6) 7) 8) 9) 3, 3 8, 8 3, 3 7, 1 4, 8 4, 20 β β β β β β 2, 4 7, 1 7 3, 11 3 β β β 10) 7, 2 β 3, 5 11) β 12) 0, 4 13) 1, 4 14) (, 5) β β 15) ( 16) |
( 175, ) S 5 3] S h [9, 1) S (0, ) β S 1) (54, β ) β [3, ) β S [14 ) β 18) ( 19) ( 20) ( 21) ( 22) [ 23) ( 24) (, β, 2 3) β β β β β β, 0) S 1],, β β 4] 5 2] β β β β 25) [1, 3] 26) [ 1 2, 1] 27) ( 4), β β β ( 3, ) β β 28) [3, 7] 29) [1, 3 2 ] 30) [ 2, β 31) ( β β 4 3] β, 3 2) S ( 5 2 32) ( 1 S 2), β β β 33) [2, 4] S 34) [ 3, 2] β β 453 ), β (1, β ) 4.1 1) ( 2) ( 3) ( 4) ( 1, 2) 4, 3) 1, β 3, 1) 3) β β β β 5) No Solution 2, β 3, 1) 6) ( 2) β 7) ( β 8) (4, 4) 3, 9) ( 1) β β 10) No Solution 11) (3, 4) β 4.2 1) (1, 2) ( β 3) β 3, 2) 2, 3) ( β 4) (0, 3) 5) ( 1, β 7, 6) ( β 7) (1, 5) 4, 8) ( β 9) (3, 3) 5) β 2) 8) β β 1) β 10) (4, 4) 11) (2, 6) 12) ( 13) ( β β 3, 3) 6) 2, β 14) (0, 2) 4.3 Answers - Chapter 4 Answers - Graphing 12) (4, 13) (1, 14) ( β 15) (3, 4) β 3) β 1, 3) 4) β 16) No Solution 17) (2, β 18) (4, 1) 2) 23) ( 1, β β 1) 24) (2, 3) 25) ( 26) ( 1, 4, β β β β 2) 3) 27 |
) No Solution 3, 1) 2) 28) ( β 29) (4, β 30) (1, 4) 29) (4, 30) ( β 3) β 1, 5) 31) (0, 2) 7) 32) (0, β 33) (0, 3) 4) 2) 3) 34) (1, 35) (4, β β 36) (8, β 37) (2, 0) 38) (2, 5) 39) ( 4, 8) β 40) (2, 3) 3, 4) 1) 19) ( β 20) (2, β 21) (3, 2) 4, 22) ( 4) β β Answers - Substitution 15) (1, 16) ( 17) ( β β 5) β 1, 0) 1, 8) 18) (3, 7) 19) (2, 3) 8) 20) (8, β 21) (1, 7) 22) (1, 7) 3, 23) ( β 24) (1, β 25) (1, 3) β 3) 26) (2, 1) 27) ( 28) ( 2, 8) 4, 3) β β 2) 454 2, 4) 1) ( β 2) (2, 4) Answers - Addition/Elimination 2) 12) (1, β 13) (0, 4) 3) No solution 14) ( 4) Inο¬nite number of solutions 5) No solution 6) Inο¬nite number of solutions 7) No solution 8) (2, 2) 9) ( β 10) ( 5) β 3, 6) β 3, β 11) ( 2, β β 9) 4.4 1) (1, β 2) (5, β 3) (2, 3, 4) (3, 2) β 2, 1) β 2, 1, 4) 5) ( 6) ( β β 3, 2, 1) β 7) (1, 2, 3) 8) solutions β 9) (0, 0, 0) 10) solutions β 11) (19, 0, 1, 0) β 15) (8, 2) 16) (0, 3) 17) (4, 6) 18) ( 19) ( β β 8) 6, β 2, 3) 20) (1, 2 |
) 4) 21) (0, β 22) (0, 1) 23) ( β 24) (2, 2, 0) 2) β 14) solutions β 15) (2, 1 2, 2) β solutions 16) β 17)( β 1, 2, 3) 18)( 1, 2, 2) β 19) (0, 2, 1) β 20) no solution 21) (10, 2, 3) Answers - Three Variables 1, 2) 3, 2) 12) solutions β 13) (0, 0, 0) 2) 3) 25) ( 26) ( 27) ( 28) ( 29) ( 1, β 3, 0) 1, β 3, 0) 8, 9) β β β β β 30) (1, 2) 31) ( 32) ( 2, 1) 1, 1) β β 33) (0, 0) 34) Inο¬nite number of solutions 23) (2, 3, 1) 24) β solutions 25) no solutions 26) (1, 2, 4) 25, 18, 25) β 27) ( β, 3 28) ( 2 7 7, 29) (1, β β 30) (7, 4, 5, 6) 1) 2, β 2 7) β 3, 31) (1, 2, 4, 1) β 1, 0, 4) β 3, 13) β 22) no solution 32) ( β β 4.5 1) 33Q, 70D Answers - Value Problems 455 2) 26 h, 8 n 19) 13 d, 19 q 3) 236 adult, 342 child 20) 28 q 4) 9d, 12q 5) 9, 18 6) 7q, 4h 7) 9, 18 8) 25, 20 9) 203 adults, 226 child 10) 130 adults, 70 students 11) 128 card, 75 no card 12) 73 hotdogs, 58 hamburgers 13) 135 students, 97 non-students 14) 12d, 15q 15) 13n, 5d 16) 8 20c, 32 25c 17) 6 15c, 9 25c 18) 5 4.6 1) 2666.7 2) 2 3) 30 4) 1, 8 5) 8 6) 10 7) 20 8) 16 9) 17.25 21) 15 n, 20 d 22) 20 S1, 6 S |
5 23) 8 S20, 4 S10 24) 27 25) S12500 @ 12% S14500 @ 13% 26) S20000 @ 5% S30000 @ 7.5% 27) S2500 @ 10% S6500 @ 12% 28) S12400 @ 6% S5600 @ 9% 29) S4100 @ 9.5% S5900 @ 11% 30) S7000 @ 4.5% S9000 @ 6.5% 31) S1600 @ 4%; S2400 @ 8% 32) S3000 @ 4.6% S4500 @ 6.6% Answers - Mixture Problems 10) 1.5 11) 10 12) 8 13) 9.6 14) 36 15) 40, 60 16) 30, 70 17) 40, 20 18) 40, 110 456 33) S3500 @ 6%; S5000 @ 3.5% 34) S7000 @ 9% S5000 @ 7.5% 35) S6500 @ 8%; S8500 @ 11% 36) S12000 @ 7.25% S5500 @ 6.5% 37) S3000 @ 4.25%; S3000 @ 5.75% 38) S10000 @ 5.5% S4000 @ 9% 39) S7500 @ 6.8%; S3500 @ 8.2% 40) S3000 @ 11%; S24000 @ 7% 41) S5000 @ 12% S11000 @ 8% 42) 26n, 13d, 10q 43) 18, 4, 8 44) 20n, 15d, 10q 19) 20, 30 20) 100, 200 21) 40, 20 22) 10, 5 23) 250, 250 24) 21, 49 25) 20, 40 26) 2, 3 27) 56, 144 28) 1.5, 3.5 29) 30 30) 10 31) 75, 25 32) 55, 20 5.1 1) 49 2) 47 3) 24 4) 36 5) 12m2n 6) 12x3 7) 8m6n3 8) x3y6 9) 312 10) 412 11) 48 12) 36 13) 4u6v4 14) x3y3 15) 16a16 16) 16x4y4 5.2 1) 32x8y10 2) 32b13 a2 33) 440, 160 |
34) 20 35) 35, 63 36) 3, 2 37) 1.2 38) 150 39) 10 40) 30, 20 41) 75 42) 20, 60 43) 25 Answers - Chapter 5 Answers to Exponent Properties 17) 42 18) 34 19) 3 20) 33 21) m2 22) xy3 4 23) 4x2y 3 24) y2 4 25) 4x10y14 26) 8u18v6 27) 2x17y16 28) 3uv 29) x2y 6 30) 4a2 3 31) 64 32) 2a 33) y3 512x24 34) y5x2 2 35) 64m12n12 36) n10 2m 37) 2x2y 38) 2y2 39) 2q7r8p 40) 4x2y4z2 41) x4y16z4 42) 256q4r8 43) 4y4z Answers to Negative Exponents 3) 2a15 b11 4) 2x3y2 457 5) 16x4y8 6) 1 7) y16 x5 8) 32 m5n15 9) 2 9y 10) y5 2x7 11) 1 y2x3 12) y8x5 4 13) u 4v6 14) x7y2 2 15) u2 12v5 y 2x4 17) 2 y7 16) 5.3 1) 8.85 2) 7.44 102 Γ 4 10β Γ 10β 2 3) 8.1 Γ 4) 1.09 5) 3.9 6) 1.5 100 Γ 10β 2 Γ 104 Γ 7) 870000 8) 256 9) 0.0009 10) 50000 11) 2 12) 0.00006 13) 1.4 14) 1.76 3 10β Γ 10 10β Γ 18) a16 2b 19) 16a12b12 20) y8x4 4 21) 1 8m4n7 22) 2x16y2 23) 16n6m4 24) 2x y3 25) 1 x15y 26) 4y4 27) u 2v 28) 4y5 29) 8 30) 1 2u3v5 Answers to Scientiο¬c Notation 15) 1.662 16) 5.018 17) 1.56 18) 4.353 19) 1 |
.815 20) 9.836 21) 5.541 22) 6.375 23) 3.025 24) 1.177 25) 2.887 26) 6.351 27) 2.405 6 10β 106 Γ Γ 3 10β Γ 108 104 1 10β Γ Γ Γ 5 10β 4 10β 9 10β 16 10β 6 10β 21 10β Γ Γ Γ Γ Γ Γ 20 10β Γ 10β 2 28) 2.91 Γ 458 31) 2y5x4 32) a3 2b3 33) 1 x2y11z 34) a2 8c10b12 35) 1 h3k j6 36) x30z6 16y4 37) 2b14 a12c7 38) m14q8 4p4 39) x2 y4z4 40) mn7 p5 29) 1.196 30) 1.2 Γ 31) 2.196 2 2 10β Γ 107 10β Γ 103 1014 Γ 32) 2.52 Γ 33) 1.715 34) 8.404 35) 1.149 36) 3.939 37) 4.6 Γ 38) 7.474 39) 3.692 40) 1.372 41) 1.034 101 106 Γ Γ 109 Γ 102 103 7 10β 103 106 Γ Γ Γ Γ 42) 1.2 106 Γ 5.4 1) 3 2) 7 3) 4) β β 5) β 6) 8 7) 5 10 6 7 1 8) β 9) 12 10) 1 β 11) 3p4 12) β 3p β m3 + 12m2 n3 + 10n2 13) β 14) 8x3 + 8x2 15) 5n4 + 5n Answers to Introduction to Polynomials 16) 2v4 + 6 17) 13p3 1 1 β β β β 3x 22) 18) β 19) 3n3 + 8 20) x4 + 9x2 5 21) 2b4 + 2b + 10 3r4 + 12r2 5x4 + 14x3 4n + 7 3a2 23) β 24) 5n4 25) 7a4 2a β β 26) 12v3 + 3v + 3 27) p2 + 4p β 28) 3m4 2m + 6 β 29) 5b3 + 12b2 + 5 15n4 + 4n 30 |
) β 6 β 6 β 31) n3 32) 33) β β 5n2 + 3 β 6x4 + 13x3 12n4 + n2 + 7 34) 9x2 + 10x2 3r3 + 7r2 + 1 35) r4 β 36) 10x3 6x2 + 3x 8 β β 37) 9n4 + 2n3 + 6n2 38) 2b4 39) β b3 + 4b2 + 4b β 3b4 + 13b3 11b + 19 7b2 β β 40) 12n4 n3 6n2 + 10 β x3 β 4x + 2 41) 2x4 β 42) 3x4 + 9x2 + 4x β 5.5 Answers to Multiply Polynomials 1) 6p 42 β 2) 32k2 + 16k 3) 12x + 6 4) 18n3 + 21n2 5) 20m5 + 20m4 6) 12r 21 β 7) 32n2 + 80n + 48 8) 2x2 β 9) 56b2 7x 4 β 19b β 10) 4r2 + 40r+64 β 15 13) 15v2 26v + 8 β 44a 14) 6a2 β 15) 24x2 β 16) 20x2 17) 30x2 β β β 22x 32 7 β 29x + 6 14xy 18) 16u2 + 10uv 4y2 21v2 β β 19) 3x2 + 13xy + 12y2 20) 40u2 34uv 48v2 β β 21) 56x2 + 61xy + 15y2 11) 8x2 + 22x + 15 12) 7n2 + 43n 42 β 22) 5a2 23) 6r3 β β 459 7ab 24b2 β 43r2 + 12r 35 β 24) 16x3 + 44x2 + 44x + 40 32) 42u4 + 76u3v + 17u2v2 18v4 β 25) 12n3 20n2 + 38n 20 β β 4b2 4b 12 β β 26) 8b3 β 27) 36x3 24x2y + 3xy2 + 12y3 β 28) 21m3 + 4m2n 8n3 β 29) 48n4 β 16n3 + 64n2 |
6n + 36 β 30) 14a4 + 30a3 13a2 β 31) 15k4 + 24k3 + 48k2 + 27k + 18 β 12a + 3 33) 18x2 34) 10x2 35) 24x2 36) 16x2 37) 7x2 β 38) 40x2 12 15x β 55x + 60 18x 15 β β β β 44x 12 β β 49x + 70 10x 5 β β 39) 96x2 6 β 40) 36x2 + 108x + 81 5.6 1) x2 2) a2 3) 1 β 4) x2 64 β 16 β 9p2 9 β 49n2 25 64 64 β β 9 5) 1 β 6) 64m2 7) 25n2 8) 4r2 β 9) 16x2 10) b2 β 11) 16y2 12) 49a2 β 13) 16m2 14) 9y2 β 49 β β x2 49b2 64n2 β 9x2 Answers to Multiply Special Products 4y2 15) 36x2 β 16) 1 + 10n + 25n2 17) a2 + 10a + 25 18) v2 + 8v + 16 19) x2 16x + 64 β 12n + 36n2 20) 1 21) p2 + 14p + 49 22) 49k2 β 98k + 49 β 70n + 25n2 40x + 25 23) 49 β 24) 16x2 β 25) 25m2 80m + 64 β 26) 9a2 + 18ab + 9b2 27) 25x2 + 70xy + 49y2 28) 16m2 8mn + n2 β 29) 4x2 + 8xy + 4y2 30) 64x2 + 80xy + 25y2 31) 25 + 20r + 4r2 32) m2 14m + 49 β 33) 4 + 20x + 25x2 49 49 34) 64n2 β 35) 16v2 36) b2 β β 16 37) n2 25 β 38) 49x2 + 98x + 49 39) 16k2 + 16k + 4 40) 9a2 64 β 5.7 1) 5x + 1 4 + 1 2x Answers to Divide Polynomials 3) 2n3 + n2 10 + 4n 2) 5x |
3 9 + 5x2 + 4x 9 4) 3k2 8 + k 2 + 1 4 460 5) 2x3 + 4x2 + x 2 6) 5p3 4 + 4p2 + 4p 19) x 3 + 3 10x β β 20 33) 3n2 9n 10 β β 8 n + 6 β 5 9) x 7) n2 + 5n + 1 5 8) m2 3 + 2m + 3 10 + 9 x + 8 β 10 11) n + 8 9 β 12 13) v + 8 β 14) x 3 β β v 10 β 5 x + 7 15 16) x 6 x β β β 17) 5p + 4 + 3 4 9p + 4 18) 8k 9 β 1 β 3k 1 β 6.1 1) 9 + 8b2 2) x 5 β 3) 5(9x2 β 4) 1 + 2n2 5) 5) 7(8 β 6) 10(5x 7) 7ab(1 5p) 8y) 5a) β β 8) 9x2y2(3y3 8x) β 21) r 1 + 2 4x + 3 β 22) m + 4 + 1 m β 1 23) n + 2 24) x 4 + 4 2x + 3 β 25) 9b + 5 5 3b + 8 β 26) v + 3 3v β 9 5 β 7 5 4x β 3 4n + 5 7 β 27) x 7 β β 28) n 7 β β 29) a2 + 8a 30) 8k2 2k β β 31) x2 32) x2 β β 10 4x β 8x + β 34) k2 35) x2 β β 3k 9 1 β k β β 7x + 3 + 1 x + 7 36) n2 + 9n 37) p2 + 4p 1 + 3 2n + 3 1 + 4 9p + 9 β β 38) m2 8m + 7 β 7 8m + 7 β 39) r2 + 3r β 4 40) x2 + 3x β 8 4 r β β 7 + 5 2x + 6 41) 6n2 3n β β 3 + 5 2n + 3 42) 6b2 + b + 9 + 3 4b β 6v + 6 + 1 43) v2 7 4v + 3 β Answers - Chapter |
6 Answers - Greatest Common Factor 9) 3a2b( 1 + 2ab) β 10) 4x3(2y2 + 1) 11) 5x2(1 + x + 3x2) β 12) 8n5( β 13) 10(2x4 4n4 + 4n + 5) 3x + 3) β 14) 3(7p6 + 10p2 + 9) 15) 4(7m4 + 10m3 + 2) 461 16) 2x( β 5x3 + 10x + 6) 17) 5(6b9 + ab 3a2) β 18) 3y2(9y5 + 4x + 3) 19) β 8a2b (6b + 7a + 7a3) 20) 5(6m6 + 3mn2 5) β 21) 5x3y2z(4x5z + 3x2 + 7y) 22) 3(p + 4q 5q2r2) β 23) 10(5x2y + y2 + 7xz2) 6.2 1) (8r2 2) (5x2 3) (n2 β 4) (2v2 5)(5r β 8)(7x β 3)(3n 1) β 2) β 2) β 1)(7v + 5) β 5) (3b2 7)(5b + 7) β 6) (6x2 + 5)(x 8) β 7) (3x2 + 2)(x + 5) 8) (7p2 + 5)(4p + 3) 9) (7x2 4)(5x 4) β 10)(7n2 5)(n + 3) β β 24) 10y4z3 (3x5 + 5z2 x) β 25) 5q(6pr p + 1) β 26) 7b(4 + 2b + 5b2 + b4) 27) 3( β 6n5 + n3 7n + 1) β 28) 3a2(10a6 + 2a3 + 9a + 7) 29) 10x11( 30) 4x2( β 31) 4mn( 4 β 6x4 β β 2x + 5x2 5x3) β x2 + 3x + 1) 8n7 + m5 + 3n3 + 4 |
) β 5 + 3y3 β 2xy3 4xy) β β 32) 2y7( Answers - Grouping 11) (7x + 5)(y β 12) (7r2 + 3)(6r 7) 21) (4u + 3)(8v 5) β 7) 22) 2(u + 3)(2v + 7u) β 13) (8x + 3)(4y + 5x) 14) (3a + b2)(5b 15) (8x + 1)(2y 2) 7) β β 8) 16) (m + 5)(3n β 17) (2x + 7y2)(y 4x) β 5)(5n + 2) y)(8y + 7) 1)(y + 7) 18) (m 19) (5x 20) (8x β β β 23) (5x + 6)(2y + 5) 24) (4x 25) (3u β β 5y2)(6y 5) β 2u) 7)(v β 26) (7a 2)(8b β 27) (2x + 1)(8y 7) 3x) β β 6.3 1) (p + 9)(p + 8) 2) (x 3) (n β β 8)(x + 9) 8)(n 1) β Answers - Trinomials where a = 1 4) (x 5)(x + 6) β 5) (x + 1)(x 10) β 6) (x + 5)(x + 8) 462 7) (b + 8)(b + 4) 17) (u 5v)(u 3v) 27) 5(a + 10)(a + 2) β β 18) (m + 5n)(m 19) (m + 4n)(m β β 8n) 2n) 10)(b 7) β 7)(x + 10) 8) (b β 9) (x β 10) (x β 3)(x + 6) 20) (x + 8y)(x + 2y) 11) (n 5)(n β 12) (a + 3)(a 3) β 9) β 13) (p + 6)(p + 9) 14) (p + 10)(p 3) β 7) 8)(n β 5n)(m 10n) β 15) (n β 16) (m β 6.4 |
21) (x 22) (u 23) (x β β β 9y)(x 2y) β 7v)(u 2v) β 3y)(x + 4y) 24) (x + 5y)(x + 9y) 25) (x + 6y)(x 2y) β 26) 4(x + 7)(x + 6) 28) 5(n 29) 6(a 30) 5(v β β β 8)(n 1) β 4)(a + 8) 1)(v + 5) 31) 6(x + 2y)(x + y) 32) 5(m2 + 6mn 18n2) β 33) 6(x + 9y)(x + 7y) 34) 6(m β 9n)(m + 3n) 1) (7x 6)(x 6) β β 2) (7n 2)(n β 3) (7b + 1)(b + 2) β 6) 4) (7v + 4)(v 5) (5a + 7)(a 4) 4) β β 6) Prime 7) (2x 1)(x 2) β β 8) (3r + 2)(r 2) β 9) (2x + 5)(x + 7) 10) (7x 11) (2b β β 6)(x + 5) 3)(b + 1) 12) (5k 6)(k β 13) (5k + 3)(k + 2) β 4) Answers - Trinomials where a 1 15) (3x 16) (3u β β 5)(x 4) β 2v)(u + 5v) 17) (3x + 2y)(x + 5y) 18) (7x + 5y)(x y) β 7y)(x + 7y) 19) (5x β 20) (5u 4v)(u + 7v) β 21) 3(2x + 1)(x 7) 6) β β 22) 2(5a + 3)(a 23) 3(7k + 6)(k 5) 24) 3(7n 25) 2(7x β β 26) (r + 1)(4r β 6)(n + 3) 2)(x 4) β 3) β 27) (x + 4)(6x + 5) 29) (k 30) (r β β 4)(4k 1) β 1 |
)(4r + 7) 31) (x + 2y)(4x + y) 32) 2(2m2 + 3mn + 3n2) 3n)(4m + 3n) 33) (m β 34) 2(2x2 3xy + 15y2) β 35) (x + 3y)(4x + y) 36) 3(3u + 4v)(2u 3v) β 37) 2(2x + 7y)(3x + 5y) 38) 4(x + 3y)(4x + 3y) 39) 4(x 2y)(6x y) β β 40) 2(3x + 2y)(2x + 7y) 14) (3r + 7)(r + 3) 28) (3p + 7)(2p 1) β 6.5 1) (r + 4)(r 2) (x + 3)(x 4) β 3) β Answers - Factoring Special Products 3) (v + 5)(v 4) (x + 1)(x 5) 1) β β 463 27) 2(2x 3y)2 β 28) 5(2x + y)2 29) (2 m)(4 + 2m + m2) β 30) (x + 4)(x2 4x + 16) β 31) (x 4)(x2 + 4x + 16) β 32) (x + 2)(x2 2x + 4) β u)(36 + 6u + u2) 33) (6 β 34) (5x β 6)(25x2 + 30x + 36) 35) (5a β 4)(25a2 + 20a + 16) 36) (4x 3)(16x2 + 12x + 9) β 37) (4x + 3y)(16x2 12xy + 9y2) β 38) 4(2m 3n)(4m2 + 6mn + 9n2) β 39) 2(3x + 5y)(9x2 β 40) 3(5m + 6n)(25m2 15xy + 25y2) 30mn + 36n2) β 41) (a2 + 9)(a + 3)(a β 42) (x2 + 16)(x + 4)(x 3) 4) β z) β 5) (p + 2)(p β 6) (2 |
v + 1)(2v 7) (3k + 2)(3k 8) (3a + 1)(3a 2) β β β 1) 2) 1) 3) 9) 3(x + 3)(x β 10) 5(n + 2)(n β 11) 4(2x + 3)(2x 2) 3) β 12) 5(25x2 + 9y2) 13) 2(3a + 5b)(3a 5b) β 14) 4(m2 + 16n2) 15) (a 1)2 β 16) (k + 2)2 17) (x + 3)2 4)2 3)2 2)2 18) (n 19) (x β β 20) (k β 21) (5p β 22) (x + 1)2 23) (5a + 3b)2 24) (x + 4y)2 25) (2a β 26) 2(3m 5b)2 2n)2 β 6.6 β 3) 1) 3(2a + 5y)(4z 2) (2x β 3) (5u 5)(x β 4v)(u β 4) 4(2x + 3y)2 β 1)2 43) (4 + z2 )(2 + z)(2 1) 44) (n2 + 1)(n + 1)(n β 45) (x2 + y2)(x + y)(x β 46) (4a2 + b2)(2a + b)(2a β 47) (m2 + 9b2)(m + 3b)(m y) b) 3b) β 48) (9c2 + 4d2)(3c + 2d)(3c 2d) β Answers - Factoring Strategy 3h) 6) 5(4u 7) n(5n β β 3u2) x)(v β 3)(n + 2) v) 8) x(2x + 3y)(x + y) 5) 2( β x + 4y)(x2 + 4xy + 16y2) 9) 2(3u 10) 2(3 β β 2)(9u2 + 6u + 4) 4x)(9 + 12x + 16x2) 464 11) n(n 1) β 12) (5x + 3)(x 13) (x β |
14) 5(3u 3y)(x 5v)2 β 5) y) β β 15) (3x + 5y)(3x 5y) β 3y)(x2 + 3xy + 9y2) 16) (x β 17) (m + 2n)(m 27) (3x 4)(9x2 + 12x + 16) β 28) (4a + 3b)(4a 29) x(5x + 2) 30) 2(x β 31) 3k(k 2)(x β 5)(k β β 32) 2(4x + 3y)(4x 3b) β 3) 4) 3y) β 4x)(n + 3) 2n) β 33) (m β 18) 3(2a + n)(2b β 19) 4(3b2 + 2x)(3c 3) 2d) β 20) 3m(m + 2n)(m 4n) 21) 2(4 + 3x)(16 β β 12x + 9x2) 22) (4m + 3n)(16m2 12mn + 9n2) 23) 2x(x + 5y)(x β β 2y) 24) (3a + x2)(c + 5d2) 25) n(n + 2)(n + 5) n)(16m2 + 4mn + n2) 26) (4m β 6.7 34) (2k + 5)(k 2) β 35) (4x y)2 β 36) v(v + 1) 37) 3(3m + 4n)(3m 4n) β 38) x2(x + 4) 5y)(x + 4y) 39) 3x(3x β 40) 3n2(3n 41) 2(m β 42) v2(2u 1) β 2n)(m + 5n) 5v)(u 3v) β β 1) 7, 2) β 3) 1, 4) β 5) β 6) 4, 7) 2, 2 β 4, 3 4, 7 β 5 2 5, 5 β 8 7 β 5, 6 8) 9) 10) 11) 12 Answers - Solve by Factoring 13) 4, 0 14) 8, 0 15) 1, 4 16) 4, 2 17) 3 7, 18) β 19) |
4, 7 20) 1 4 21) β 22) 8, β β 4 8 3 2 23) 8, β 24) 4, 0 465, 1 25) 8 3, 26) 27) 28) β β β 29) β 30) 2, 3 β 7, 7 β 6 8 7 4, 5 2 6 5,, β β β β β β 31) 32) 33) 34) 35) 4 5, 6 β 36) 5 3, 2 β Answers - Chapter 7 7.1 1) 3 2) 1 3 3) 1 5 β 4) undeο¬ned 5) 1 2 6) 6 10 7) β 8) 0, 2 5 2 9) β 10) 0, 10 β 11) 0 12) β 13) β 14) 0, 15) β 16) 0, 1 7 10 3 2 1 2 β 8, 4 17) 7x 6 18) 3 n 19) 3 5a 7.2 1) 4x2 2) 14 3 Answers - Reduce Rational Expressions 20) 7 8k 21) 4 x 22) 9x 2 23) 3m β 10 4 24) 10 9n2(9n + 4) 25) 10 2p + 1 26) 1 9 27) 1 x + 7 28) 7m + 3 9 29) 8x 7(x + 1) 30) 7r + 8 8r 31) n + 6 n 5 β 32) b + 6 b + 7 37) 3a 5 β 5a + 2 38) 9 p + 2 39) 2n 1 β 9 40) 3x 5 β 5(x + 2) 41) 2(m + 2) 5m 3 β 42) 9r 5(r + 1) 43) 2(x 3x 4) 4 β β 44) 5b 8 β 5b + 2 45) 7n 4 β 4 46) 5(v + 1) 3v + 1 47) (n 1)2 β 6(n + 1) 48) 49) 7x 6 β (3x + 4)(x + 1) 7a + 9 2(3a 2) β 50) 2(2k + 1) 9(k 1) β 9 33) v 10 β 34) 3(x 3) β 5x + 4 7 7 35) 2x 5x 36) k β β 8 β k + 4 Answers - Multipy and Divide 3) 63 10n 4 |
) 63 10m 466 5) 3x2 2 6) 5p 2 7) 5m 8) 7 10 9) r 6 β r + 10 10) x + 4 11) 2 3 12) 9 β b 5 13) x 14) v 10 10 β 7 1 β 15) x + 1 16) a + 10 a 6 β 17) 5 18) p β p β 10 4 7.3 1) 18 2) a2 3) ay 4) 20xy 5) 6a2c3 6) 12 7) 2x 8) x2 8 β 2x β x 9) x2 β 10) x2 β 3 β 12 β 11x + 30 19) 3 5 20) x + 10 x + 4 21) 4(m β 5m2 5) 22) 7 23) x + 3 4 24) n 9 β n + 7 25) b + 2 8b 26) v 9 β 5 1 β 6 27) n β 28 29) 30) 7 8(k + 3) 31) x 4 β x + 3 32) 9(x + 6) 10 33) 9m2(m + 10) 34) 10 9(n + 6) 35) p + 3 6(p + 8) 36) x 8 β x + 7 37) 5b b + 5 38) n + 3 8 39) r β 40) 18 5 41) 3 2 42) 1 a + 2b 43) 1 x + 2 44) 3(x 2) β 4(x + 2) Answers - Least Common Denominators 467 11) 12a4b5 12) 25x3y5z 13) x (x 3) β 14) 4(x 2) β 15) (x + 2)(x 16) x(x β 4) β 7)(x + 1) 17) (x + 5)(x 5) β 3)2(x + 3) 18) (x β 19) (x + 1)(x + 2)(x + 3) 20) (x 2)(x β 6a4 10a3b2, 2b 10a3b2 5)(x + 3) β 21) 22) 23) 3x2 + 6x (x 4)(x + 2) β x2 + 4x + 4 3)(x + 2) (x β, (x 2x 8 4)(x + 2) β β, x2 ( |
x 6x + 9 β 3)(x + 2) β 24) x(x 5, 2x x(x 6) 3x 12 6), β x(x 6) β β β x2 4x 4)2(x + 4) β, 25) (x β (x 5x + 1 5)(x + 2) β x2 + 7x + 6 26) 27) β 3x2 + 12x 4)2(x + 4) (x β, (x 4x + 8 5) (x + 2) β 6)(x + 6)2, 2x2 (x 9x 18 β β 6)(x + 6)2 β (x β 28) (x 3x2 + 4x + 1 4)(x + 3)(x + 1) β, (x β 2x2 8x 4)(x + 3)(x + 1) β 4x 3)(x + 2), x2 + 4x + 4 3)(x + 2) (x β 3x2 + 15x 4)(x 2)(x + 5) β 29) 30) (x β (x β 7.4 1) 6 a + 3 4 2) x β 3) t + 7 x2 β 4)(x, (x β 4x + 4 2)(x + 5) β, 5x 4)(x β β (x β 20 2)(x + 5) Answers - Add and Subtract 4) a + 4 a + 6 5) x + 6 x 5 β 6) 3x + 4 x2 468 d2 β 27) β (x + 3)(x + 1) 6x2 3xy β β 2x2y2 28) (x 2x + 3 1)(x + 4) β 40) 2 r + s 5 7) 24r 8) 7x + 3y x2y2 9) 15t + 16 18t3 10) 5x + 9 24 11) a + 8 4 12) 5a2 + 7a 9a2 3 β 13 7x β 4x c2 + cd c2d2 13) β 14) β 15) 3y2 4x 16) x2 17) β 1 β z2 + 5z z2 1 18) 11x + 15 4x(x + 5) β 19) 14 x2 20) x2 x2 3x 4 x 25 β β β β 7 |
.5 1) x 2) 1 x β β y 1 y 3) β a a + 2 4) β 6) b3 + 2b b 2 β β 8b 7) 2 5 8) 4 5 9) 1 2 β 10) β 11) x2 1 2 x 1 x2 + x + 1 β β β β 21) 4t 5 4(t 3) 22) 2x + 10 (x + 3)2 23) 24) 20x 6 β 15x(x + 1) 9a 4(a 5) β 25) t2 + 2ty y2 26) 2x2 β x(x y2 β t2 β 10x + 25 5) β 3 x 34) 2x + 7 x2 + 5x + 6 35) 2x β 5x x2 β 8 β 14 36) β 3x2 + 7x + 4 x) 3(x + 2)(2 β 37) a a2 2 9 β β 38) 2 y2 y β 39) z 2z 3 1 β β 29) 30) 31) 32) x 8 β (x + 8)(x + 6) 2) 2x 5 β 3)(x (x β β 5x + 12 x2 + 5x + 6 41) 42) 5(x 1) β (x + 1)(x + 3) 5x + 5 x2 + 2x 15 β 4x + 1 (x + 1)(x β 2) 43) β (x β (x 29) 3)(x + 5) β 33) 2x + 4 x2 + 4x + 3 44) Answers - Complex Fractions 5x 10 β x2 + 5x + 4 3a + 3 2a β β 4a2 12) 2a2 β 13) x 3 14) 3x + 2 b) 15) 4b(a β a 16) x + 2 x 1 β 17) x 5 β x + 9 3)(x + 5) 5x + 4 β 18) 19) (x β 4x2 β 1 3x + 8 20) 1 x + 4 21) x 2 β x + 2 22) x 7 β x + 5 469 23) x 3 β x + 4 24) β 2a 3a 2 β 4 β b 2 β 2b + 3 25) β 26) x + y y x β 27) a 3b β a + 3b 28) 2x |
x2 + 1 β 2 y 29) β 30) x2 31) y β xy 1 β x 32) x2 β y xy + y2 x β 33) x2 + y2 xy 34) 2x 1 β 2x + 1 7.6 1) 40 3 = a 2) n = 14 3 3) k = 12 7 4) x = 16 5) x = 3 2 6) n = 34 7) m = 21 8) x = 79 8 9) p = 49 10) n = 25 40 3 11) b = β 12) r = 36 5 13) x = 5 2 14) n = 32 5 7.7 1) β 2) β 3) 3 4) β 5) 2 6) 1 3 1, 2 2 3 3, 1 1, 4 7) 1 β 35) 3x)2 (1 β x2(x + 3)(x β 3) 36) x + y xy Answers - Proportions 16 7 15) a = 6 7 16) v = β 17) v = 69 5 18) n = 61 3 19) x = 38 3 20) k = 73 3 21) x = 22) x = 8, 5 7, 5 β β 23) m = 7, 8 β 24) x = 25) p = 26) n = 27) n = 28) n = 3, 9 2 7, β 6, 9 1 4, 1 β β β β β β 7, 1 1, 3 29) x = β 30) x = β 31) S9.31 32) 16 33) 2.5 in 34) 12.1 ft 35) 39.4 ft 36) 3.1 in 37) T: 38, V: 57 38) J: 4 hr, S: 14 hr 39) S8 40) C: 36 min, K: 51 min Answers - Solving Rational Equations 8) 9) 10) 1 3 5 β β β 11) β 12) 5, 10 7 15 5, 0 13) 16 3, 5 14) 2, 13 470 8 15) β 16 17) β 18) β 19) 3 2 20) 10 21) 0, 5 2, 5 3 22) β 23) 4, 7 1 24) β 25) 2 3 7.8 1) 12320 yd 2) 0.0073125 T 3) 0.0112 g 4) 135,000 cm 5) |
6.1 mi 6) 0.5 yd2 7) 0.435 km2 8) 86,067,200 ft2 9) 6,500,000 m3 10) 239.58 cm3 11) 0.0072 yd3 12) 5.13 ft/sec 13) 6.31 mph 14) 104.32 mi/hr 15) 111 m/s 8.1 1) 7 5β 2) 5 5β 26) 1 2 27) 3 10 28) 1 29) 30) 2 3 1 β β 31) 13 4 32) 1 10 33) β 34) 7 4 Answers - Dimensional Analysis 16) 2,623,269,600 km/yr 17) 11.6 lb/in2 18) 63,219.51 km/hr2 19) 32.5 mph; 447 yd/oz 20) 6.608 mi/hr 21)17280 pages/day; 103.4 reams/month 22) 2,365,200,000 beats/lifetime 23) 1.28 g/L 24) S3040 25) 56 mph; 25 m/s 26) 148.15 yd3; 113 m3 27) 3630 ft2, 522,720 in2 28) 350,000 pages 29) 15,603,840,000 ft3/week 30) 621,200 mg; 1.42 lb Answers - Chapter 8 Answers - Square Roots 3) 6 4) 14 471 5) 2 3β 6) 6 2β 7) 6 3β 8) 20 2β 9) 48 2β 10) 56 2β 11) β 112 2β 21 7β 12) β 13) 8 3nβ 14) 7 7bβ 15) 14v 16) 10n nβ 17) 6x 7β 8.2 1) 5 53β 2) 5 33β 3) 5 63β 4) 5 23β 5) 5 73β 6) 2 33β 8 64β 7) β 16 34β 8) β 9) 12 74β 10) 6 34β 2 74β 11) β 12) 15 34β 4β 13) 3 8a2 4β 14) 2 4n3 18) 10a 2aβ 19) 20) 21) 22) 10k2 20p |
2 7β 56x2 16 2nβ β β β β 23) 30 mβ β 24) 32p 7β 25) 3xy 5β 26) 6b2a 2aβ 27) 4xy xyβ 28) 16a2b 2β 29) 8x2y2 5β 30) 16m2n 2nβ Answers - Higher Roots 5β 15) 2 7n3 5β 2 3x3 16) β 17) 2p 75β 18) 2x 46β 19) β 20) β 21) 4v2 7β 6 7r 7β 16b 3b 3β 6v 22) 20a2 23β 23) 24) β β 25) 28n2 53β 8n2 3β 3xy 5x2 β 26) 4uv u2 3β 27) β 3β 2xy 4xy 3β 28) 10ab ab2 4x3β 29) 4xy2 472 31) 24y 5xβ 32) 56 2mnβ 33) 35xy 5yβ 34) 12xy 35) 36) 37) β β β 12u 5uvβ 30y2x 2xβ 48x2z2y 5β 38) 30a2c 2bβ 39) 8j2 5hkβ 40) 41) 42) β β β 4yz 2xzβ 12p 6mnβ 32p2m 2qβ 30) 3xy2 73β 31) 32) β β 21xy2 3β 3y 3 8y2 7x2y2 33) 10v2 3u2v2 3β p 34) β 3β 40 6xy 3β 12 3ab2 35) β 36) 9y 5x3β 37) 38) β β 18m2np2 3 2m2p 12mpq 4 p 5p3 39) 18xy 4 p 8xy3z2 p 18ab2 40) β 41) 14hj2k2 4β 5ac 4β 8h2 42) 8.3 β 18xz 4 4x3yz3 p 1) 6 5β Answers - Adding Radicals 21) β 4 6β + 4 5β 3 6β 5 |
3β β 3 2β + 6 5β 3β β 5 6β 5 6β 3 3β 2) 3) 4) 5) 6) β β β β β 7) 3 6β + 5 5β 8) 9) 10) 11) 12) 5β + 3β 8 2β β β 6 6β + 9 3β 3 6β + 3β 2 5β 6 6β β β β β 2 2β 13) β 14) 8 5β 3β β 15) 5 2β 16) 17) 9 3β 3 6β β β 3β β 18) 3 2β + 3 6β 12 2β + 2 5β 3 2β 19) 20) 8.4 β β 48 5β 1) β 2) 25 6β β 3) 6m 5β 25r2 2rβ 4) β 5β 22) β 23) 8 6β 3 6β β 9 3β + 4 2β β 6β β 24) β 25) 2 23β 10 3β 26) 6 53β 3 33β β 34β 27) β 28) 10 44β 29) 24β 3 34β β 30) 5 64β + 2 44β 31) 6 34β 3 44β β 32) β 6 34β + 2 64β 33) 2 24β + 34β + 6 44β 2 34β 9 54β 3 24β β 34) 35) β 65β β 6 25β β 6 67β + 3 57β 36) 37β β 37) 4 55β 4 65β β 11 27β 4 46β β 2 57β β 6 56β 38) 39) β β 4 26β β Answers - Multiply and Divide Radicals 5) 2x2 x3β 6) 6a2 5a3β 7) 2 3β + 2 6β 8) 5 2β + 2 5β 473 45 5β 9) β β 10 15β 10) 45 5β + 10 15β 11) 25n 10β + 10 5β 12) 5 3 |
β 9 5vβ β 13) β 14) 16 2 4 2β β 9 3β β 15) 15 11 5β β 16) 30 + 8 3β + 5 15β + 4 5β 17) 6a + a 10β + 6a 6β + 2a 15β 18) β 4p 10β + 50 pβ 19) 63 + 32 3β 10 mβ + 25 2β + 2mβ 5 β 20) 21) β 3β 25 25) 26) 15β 3 10β 15 27) 4 3β 9 28) 4 5β 5 29) 5 3xyβ 12y2 30) 4 3xβ 16xy2 31) 6pβ 3 32) 2 5nβ 5 33) 34) 35) 36) 103β 5 153β 4 103β 8 4β 8 8 4β 37) 5 10r2 2 38) 4β 4n2 mn 22) 15β 4 23) 1 20 24) 2 8.5 1) 4 + 2 3β 3 2) β 4 + 3β 12 3) 2 + 3β 5 4) 1 3β β 4 5) 2 13β β 52 5 65β 6) 85β + 4 17β 68 7) 9 6β β 3 8) 30β 2 3β β 18 Answers - Rationalize Denominators 9) 15 5β β 43 5 2β 10) β 5 3β + 20 5β 77 11) 10 2 2β β 23 12) 2 3β + 2β 2 13) β 9 3β 12 β 11 2 2β 4 β 14) β 15) 3 16) 5β 3β β 5β β 2 17) 1 + 2β 474 29) a2 β 2a bβ + b a2 b β bβ 30) aβ β 31) 3 2β + 2 3β 32) a bβ + b aβ a b β 33) a bβ + b aβ ab 34) 24 β 4 6β + 9 2β 15 β 3 3β 35) β 1 + 5β 4 36) 2 5β β 5 |
2β β 30 10 + 5 10β 37) β 5 2β + 10 β 5 3β + 6β 38) 8 + 3 6β 10 18) 16 3β + 4 5β 43 19) 2β 1 β 20) 3 + 2 3β 21) 2β 22) 2β 23) aβ 24) 3 2 2β β aβ 25) 26) 1 3 27) 4 β 28) 2 5β 8.6 2 3β + 2 6β 3 2β β β 2 15β + 3β + 3 2 β 1) ( m5β )3 2) 1 4β )3 ( 10r 3) ( 7xβ )3 4) 1 3β )4 ( 6b 5) (6x)β 3 2 1 2 6) v 7 4 7) nβ 8) (5a) 1 2 Answers - Rational Exponents 15) 1 1 2b 1 2 a 16) 1 17) 1 3a2 3 2 1 4 35 8 7 6 26) a 2b 27) m n 25 12 18) y x 5 6 19) u2v 11 2 20) 1 1 2 21) y 22) v2 7 2 u 28) 1 5 4x 3 2 y 29) 1 n 3 4 1 3 1 4 30) y x 31) xy 4 3 4 3 32) x y 10 3 1 2 3 4 33) u v 475 9) 4 10) 2 11) 8 12) 1 1000 4 3 y 13) x 14) 4 v 1 3 5 2 23) b 3 4 7 4a 3 17 6 7 4 24) 2y x 1 12 25) 3y 2 34) x 15 4 y 17 4 8.7 4 1) 4x2y3 p2) 3xy3 p3) 6 8x2y3z4 p4) 5β 2x 5) 3β 36xy 3y 5 6) x3y4z2 p7) 4 x2y3 p8) 5 8x4y2 p9) 4 x3y2z p10) 5yβ 3 11) 2xy2 p12) 4 3x2y3 6β p13) 5400 14) 12β 300125 6 15) 49x3y2 p16) 15 27 |
y5z5 p17) 6 x3(x 2)2 β 3x(y + 4)2 p18) 4 p19) 10 x9y7 10β p20) 4a9b9 21) 12 x11y10 20β p22) a18b17 23) 20β a18b17c14 Answers - Radicals of Mixed Index 24) 30 x22y11z27 p25) a a4β 26) x xβ 27) b b9 10β 28) a a5 12β 6 29) xy xy5 10β 30) a ab7 p 31) 3a2b ab 4β 32) 2xy2 6 2x5y 33) x 12 p 59049xy11z10 34) a2b2c2 p 6β a2b c2 6 35) 9a2(b + 1) 243a5(b + 1)5 36) 4x(y + z)3 p 6 2x(y + z) p 37) 12β a5 38) 15β x7 39) 12 x2y5 40) p 15β a7b11 b 41) 10β ab9c7 42) yz 10 xy8z3 20 43) p44) 12 p (3x 1)3 β (2 + 5x)5 p45) 15 (2x + 1)4 p46) 12 (5 3x) β p 476 8.8 1) 11 2) 3) 4) β β β 5) β 6) 8) 9) β 10) 13 11) 48 12) 24 13) 40 14) 32 15) β 16) 28 4i β 4i 3 + 9i 6i 13i β β 12i 11i 6i 2i 8i β β β β 49 21i β 17) 11 + 60i 18) β 19) 80 20) 36 32 β β 128i β 10i 36i 21) 27 + 38i 9.1 1) 3 2) 3 3) 1, 5 4) no solution 5) 2 Β± Answers - Complex Numbers 22) 28 + 76i β 23) 44 + 8i 24) 16 25) β 26) 18i β 3 + 11i 1 + 13i β 27) 9i + 5 28) β 2 3i 3 β 9 29) 10 |
i β 6 30) 4i + 2 3 31) 3i 6 β 4 32) 5i + 9 9 33) 10i + 1 2i 34) β 35) β 40i + 4 101 36) 9i 45 β 26 37) 56 + 48i 85 6i 38) 4 β 13 39) 70 + 49i 149 40) β 36 + 27i 50 41) β 5 30i 37 β 42) 48i β 85 56 43) 9i 44) 3i 5β 45) 2 5β β 2 6β 46) β 47) 1 + i 3β 2 48) 2 + i 2β 2 i 49) 2 β 50) 3 + 2i 2β 2 51) i i 52) β 53) 1 54) 1 55) 1 β 56) i 57) 58) 1 i β β Answers - Chapter 9 Answers - Solving with Radicals 6) 3 7) 1 4 8) no solution 9) 5 10) 7 477 11) 6 12) 46 13) 5 14) 21 15) 3 2 β 16) 7 3 β 9.2 1) 2) Β± β 3) Β± 4) 3 5) 6) Β± β 7) β 8) 1, 5 9) β 9.3 5 3β 2 2 2β 2 6β 3, 11 5 β 1 3 5 1) 225; (x 2) 144; (a 15)2 12)2 β β β 17)2 15 β 3) 324; (m 18)2 4) 289; (x 5) 225 1 1 β 4 ; (x 324; (r 4; (y β 4 ; (p 2 )2 18)2 β 1 2)2 2 )2 17 β 6) 7) 1 8) 289 9) 11, 5 10) 4 + 2 7β, 4 11) 4 + i 29β, 4 2 7β β 12) β 13) β β 1 + i 42β, 2 + i 38β 2 2, β 1 β β i 38β β 2 14) 3 + 2i 33β 3, 3 β 2i 33β 3 15) 5 + i 215β 5, 5 β i 215β 5 16) β 4 + 3 2β 4 4, β β 4 3 2β Answers - Solving with Exponents 10) β 3 2β 1 Β± 2 |
11) 65, 12) 5 63 β 13) β 14) β 15) 11, 2 16) 17) β β 7 11 2, 5 2 5 2 β 191 64 3 8, β 5 8 18) 9 8 19) 5 4 20) No Solution 34 3, 10 β 21) β 22) 3 23) 17 2 β 24) No Solutoin Answers - Complete the Square 17) β 5 + 86β, 18) 8 + 2 29β, 8 β β 19) 9, 7 86β 5 β 2 29β β 1 + i 21β, 20) 9, 1 21) β 22) 1, 23) 3 2, 24) 3, β β 3 7 2 1 β 5 + 2i, 25) β β 26) 7 + 85β, 7 5 β 1 β β i 21β 2i β 85β i 29β 27) 7, 3 i 42β 28) 4, 14 β 29) 1 + i 2β, 1 i 2β β 30) 5 + i 105β 5, 5 β i 105β 5 31) 4 + i 110β 2, 4 β i 110β 2 32) 1, 3 β 478 i 39β β 33) 4 + i 39β, 4 1. 34) β 35) 7, 1 7 β 36) 2, 37) β 6 β 6 + i 258β 6, β 6 β i 258β 6 38) β 6 + i 111β 3, β 6 β i 111β 3 39) 5 + i 130β 5, 5 β i 130β 5 40) 2, 41) β 4 β 5 + i 87β 2, β 5 i 87β β 2 42) β 7 + 181β 2, β 7 181β β 2 43) 3 + i 271β 7, 3 β i 271β 7 44) β 1 + 2i 6β 2, β 1 2i 6β β 2 9.4 1) i 6β 2, i 6β 2 β 45) 7 + i 139β 2, 7 β i 139β 2 46) 5 + 3i 7β 2, 5 β 3i 7β 2 47) 12 5, 4 β 48) 1 + i 511β 4, 1 β i 511β 4 49) |
9 + 21β 2, 9 β 21β 2 50) 1 + i 163β 2, 1 β i163 2 51) β 5 + i 415β 8, β 5 β i 415β 8 52) 11 + i 95β 6, 11 β i 95β 6 53) 5 + i 191β 2, 5 β i 191β 2 54) 8, 7 55) 1, 56) 3, 5 2 3 2 β β Answers - Quadratic Formula 3 + 141β 6 15) β, β 3 141β β 6 5β 17) β 16) 3β, 3β β 3 + 401β 14 3, β 401β β 14 3, 2) i 6β i 6β 3 3) 2 + 5β, 2 β β 4) 5) 6β 6, β 6β 6 6β 2, 6β 2 β 1 + i 29β 5 6) β, β 1 i 29β β 5 1 3 7) 1, β 8) 1 + 31β 2 9) 3, 3 β 10) i 2β, 11) 3, 1, 1 β 31β 2 i 2β β 12) β 13) β 1 + i, 1 i β 3 β, β i 55β β 4 3 + i 55β 4 14) β 3 + i 159β 12, β 3 i 159β 12 β 18) β 5 + 137β 8, β 5 137β β 8 19) 2, 20) 5, 21) β 22) 3, 23) 7 2, 24β 3 i 3β 1, β β 2 i 3β 3, β β 3 25) 7 + 3 21β 10, 7 β 3 21β 10 26) β 5 + 337β 12 5, β 337β β 12 27) β 3 + i 247β 16, β 3 i 247β 16 β 479 28) 3 + 33β 6, 3 β 33β 6 29) 1, β β 30) 2 2β, 3 2 2 2β 31) 4, 32) 2, β 4 β β 4 33) 4, 9 34) 2 + 3i 5β β 7 9 2 35) 6, β 36) 5 + i 143β 14, 5 β |
i 143β 14 37) β 3 + 345β 14 3, β 345β β 14 38) 6β 2, 39) 26β 2 β, 6β 2 26β 2 β 1 + 141β 10 1, β 141β β 10, 2 β 3i 5β 7 40) β 9.5 Answers - Build Quadratics from Roots NOTE: There are multiple answers for each problem. Try checking your answers because your answer may also be correct. 14x + 13 = 0 22x + 40 = 0 β β β 7x + 10=0 9x + 18 = 0 1) x2 2) x2 3) x2 4) x2 5) x2 6) x2 7) x2 = 0 8) x2 + 7x + 10 = 0 8x + 16 = 0 9x = 0 β β β 7x 44 = 0 β 2x β 9) x2 β 10) x2 β 11) 16x2 12) 56x2 13) 6x2 14) 6x2 β β 3 = 0 β 16x + 3 = 0 75x + 25 = 0 β 5x + 1 = 0 7x + 2 = 0 31x + 12 = 0 20x + β β 9x 25 = 0 15) 7x2 16) 9x2 β 17) 18x2 β 18) 6x2 7x β 19) 9x2 + 53x β 20) 5x2 + 2x = 0 21) x2 22) x2 β 23) 25x2 24) x2 25) x2 26) x2 β 27) 16x2 3 = 0 28) x2 + 121 = 0 β 7 = 0 12 = 0 11 = 0 1 = 0 1 = 0 β β β β 29) x2 + 13 = 0 30) x2 + 50 = 0 31) x2 4x β β 2 = 0 32) x2 + 6x + 7 = 0 33) x2 β 2x + 10 = 0 34) x2 + 4x + 20 = 0 35) x2 β 12x + 39 = 0 36) x2 + 18x + 86 = 0 37) 4x2 + 4x 5 = 0 β 12x + 29 = 0 38) 9x2 β 39) 64x2 96x + 38 = 0 β 40) 4x2 + 8x + 19 = 0 9. |
6 1) 2) 3) 4) 1, 2, i, 5β 2 2β 2 Answers - Quadratic in Form 5) 6) 7) 8) 1, 3, 3, 6, 7 1 4 2 Β± Β± Β± Β± Β± Β± Β± Β± 480 9) 2, 4 Β± Β± 10) 2, 3, 1 i 3β, β 3 3i 3β Β± 2 Β± i 3β, β i 3β 3 Β± 2 β 2, 3, 1 6β, Β± 2i 6β 2 Β±, Β± 2i 3β 3 1 3 β 125, 343 5 4, 1 5 11) β 12) Β± 13) Β± 14) 1 4, 15) β 16) β 17) 1, Β± Β± Β± Β± Β± 18) 19) 20) 21) 22) 23), 1 1 2 Β± i 3β 4 1, β Β± 2 i 3β 3β Β± β 2, 3β i, Β± i 5β, i 2β Β± 2β 2 2β, i, Β± 2 Β± 6 1, 2 2β Β± 24) 2, Β± 23β, 25) 1β i 3β, β 6β i 108 23β Β± 2 i 3β 1, β Β± 2 26) 1 2, β 1 1, β Β± 4 i 3β, 1 Β± i 3β 2 9.7 1) 6 m x 10 m 2) 5 3) 40 yd x 60 yd 4) 10 ft x 18 ft 5) 6 x 10 6) 20 ft x 35 ft 7) 6β x 6β 8) 6 yd x 7 yd 9) 4 ft x 12 ft 27) 1, 2, i, Β± Β± 2i Β± Β± 28) 6, 0 (b + 3), 7 b β 4 4, 6 29) 30) β β 31) β 32) 8, 33) β 34) 2, 35) β 36) 5, 0 2 1 β 2, 10 6 β 1, 11 4 3 β 6β, 37) 4, 38) Β± 1, 39) Β± 40) 0, Β± 41) 511, 3 β 1, β 2β, 5 3 Β± 1 3 2 β 1339 24 42) 43) 44) 3, 1, 3, β Β± β Β± β β 45) 1, |
β Β± 46) 1, 2, 1, 3 1 2, β 2, 1 3 1 β Answers - Rectangles 10) 1.54 in 11) 3 in 12) 10 ft 13) 1.5 yd 14) 6 m x 8 m 15) 7 x 9 16) 1 in 17) 10 rods 18) 2 in 481 19) 15 ft 20) 20 ft 21) 1.25 in 22) 23.16 ft 23) 17.5 ft 24) 25 ft 25) 3 ft 26) 1.145 in 9.8 1) 4 and 6 2) 6 hours 3) 2 and 3 4) 2.4 5) C = 4, J = 12 6) 1.28 days 7) 1 1 3 days 8) 12 min 9) 8 days 10) 15 days 9.9 Answers - Teamwork 11) 2 days 12) 4 4 9 days 13) 9 hours 14) 12 hours 15) 16 hours 16) 7 1 2 min 17) 15 hours 18) 18 min 19) 5 1 4 min 20) 3.6 hours 21) 24 min 22) 180 min or 3 hrs 23) Su = 6, Sa = 12 24) 3 hrs and 12 hrs 25) P = 7, S = 17 1 2 26) 15 and 22.5 min 27) A = 21, B = 15 28) 12 and 36 min Answers - Simultaneous Product 1) (2, 36), ( 2) ( 9, β 3) (10, 15), ( β 4) (8, 15), ( 18, β 20), ( 4) β 40, β 5 3) β 12) β 90, β 10, β β 5) (5, 9), (18, 2.5) 6) (13, 5), ( 20, β β 13 4 ) 9.10 1) 12 2) S4 3) 24 4) 55 5) 20 6) 30 7) 25 @ S18 8) 12 @ S6 9.11 9 2 ) 7) (45, 2), ( 8) (16, 3), ( β β 10, 9) β 8) 4) β 6, 3, 9) (1, 12), ( β 10) (20, 3), (5, 12) β 11) (45, 1), ( 12) (8, 10), ( β β, 5 3 β 10, 27) 8) β Answers - Revenue and Distance 9) 60 mph, 80 mph 17 |
) r = 5 10) 60, 80 11) 6 km/hr 12) 200 km/hr 13) 56, 76 14) 3.033 km/hr 15) 12 mph, 24 mph 18) 36 mph 19) 45 mph 20) 40 mph, 60 mph 21) 20 mph 16) 30 mph, 40 mph 22) 4 mph 482 1) (-2,0) (4,0) (0,-8) (1,-9) 2) (-1,0) (3, 0) (0, -3) (1, -4) 3) (0,10) (1,0) (5,0) (3,-8) 4) 5) 6) (0,16) (2,0) (0,-18) (1,0) (-10,0) (4,0) (3, -2) (3,0) (3,8) (5,0) Answers - Graphs of Quadratics 13) (4,8) (2,0) (6,0) (0,-24) 14) (-3,0) (-1,-8) (1,0) (0,-6) (3,0) (4,3) (5,0) (0,-45) (1,0) (2,3) (3,0) 7) 8) 9) (-9,0) (0,5) (-1,0) (2,9) 15) (5,0) (-3,0) (-1,0) (0,9) (-2,-3) (0,45) (-3,0) 16) 17) (0,75) (3,0) (5,0) (4,-5) (0,15) (-1,0) 18) (-3,0) (-2,-5) (1,0) (2,1) (3,0) (0,-3) (3,4) (5,0) (1,0) (0,-5) 10) 11) 12) (3,0) (4,2) (5,0) (0,-30) 483 19) (-6,5) (-7,0) (-5,0) 20) (1,0) (2,5) (3,0) (0,-175) (-15,0) Answers - Chapter 10 10 |
.1 1) a. yes b. yes c. no d. no e. yes f. no g. yes h. no 2) all real numbers 3) x 6 5 4 4) t 0 5) all real numbers 6) all real numbers 1, 4 7) x > 16 8) x β 9) x > 4, x 5 10) x Β± 4 11) β 5 3 25 12) β 13) 2 10.2 1) 82 2) 20 3) 46 4) 2 5) 5 Answers - Function Notation 14) 85 7 15) β 16) 7 17 9 6 21 17) β 18) β 19) 13 20) 5 21) 11 22) β 23) 1 24) 4 β 21 25) β 26) 2 27) 28) 60 32 β β 29) 2 30) 31 32 31) 64x3 + 2 β 32) 4n + 10 33) β 1 + 3x 34) β 35) 2 12+a 4 3 2 Β· 3n2 1 + 2 β β 36) 1 + 1 16 x2 37) 3x + 1 38) t4 + t2 39) 5β x 3 β β2+n 40) 5 2 + 1 Answers - Operations on Functions 6) 30 β 3 7) β 8) 140 9) 1 10) 43 β 484 11) 100 74 9 26 β x3 12) β 13) 1 5 14) 27 15) β 16) n2 17) 18) β β x2 19) β 20) 2t2 β β 8t 2n 4x 2 β β x3 + 2x2 3 β 8x + 2 21) 4x3 + 25x2 + 25x 22) β 23) x2 2t3 15t2 β 4x + 5 25t β β 24) 3x2 + 4x 25) n2 + 5 3n + 5 9 β 2x + 9 26) β 27) β 2a + 5 3a + 5 28) t3 + 3t2 3t + 5 β 29) n3 + 8n + 5 30) 4x + 2 x2 + 2x 31) n6 β 32) 18n2 9n4 + 20n2 15n 25 β β 33) x + 3 34) 2 3 β 35) t4 + 8t2 + 2 10.3 1) Yes 36) 3n n2 β 6 4n β β x3 |
2x β 3x + 4 37) β β 38) x4 39) β 4x2 3 β β n2 β 3 2n 40) 32 + 23n 8 β n3 155 41) β 42) 5 43) 21 44) 4 45) 103 46) 12 47) 50 β 48) 112 49) 176 50) 147 51) 16x2 + 12x 4 β 8a + 14 8a + 2 52) 53) β β 54) t 55) 4x3 12n 16 β 2n2 β 2x + 8 56) β 57) β 58) 27t3 β 16t 5 59) β β 60) 3x3 + 6x2 4 β 108t2 + 141t 60 β Answers - Inverse Functions 2) No 485 3 β 2)3 + 2 1 β 3) Yes 4) Yes 5) No 6) Yes 7) No 8) Yes 9) Yes 10) No 11) f β 1(x) = x 5β 12) gβ 1(x) = (x 13) gβ 1(x) = 4 β 2x β x 14) f β 1(x) = β 3 + 3x x 15) f β 1(x) = β 2x β x + 2 2 16) gβ 1(x) = 3x 9 β 10.4 1) 0 1 2) β 3) 0 5) 6) β β 7) β 8) 0 9) β 10) 0 11) 5 6 12) 0 13) 14) 2 5 6 β β 17) f β 18) f β 1(x) = β 1(x) = 15 + 2x 5x + 10 5 19) gβ 20) f β 1(x) = β 1(x) = β x3β + 1 4x + 12 3 21) f β 1(x) = x3β + 3 2x5 + 2 22) gβ 23) gβ 24) f β 1(x) = β 1(x) = x x β 3x 1(x β 2x 1 25) f β 1(x) = β x 26) hβ 27) gβ 28) gβ 29) gβ 1(x) = β x β 4x + 8 5 3x + 2 1(x) = β 1(x) = β 1(x) = β x + 1 5 30) f |
β 1(x) = 5 + 4x 5 31) gβ 3β 1(x) = x + 1 32) f β 1(x) = β 5 x + 3 2 33) hβ 34) gβ q β 1(x) = ( 2x + 4)3 4 2 + 1 3β 1(x) = x β 2x + 1 1(x) = β x 1 β 36) f β 1(x) = β x 1 β x 37) f β 1(x) = 2x + 7 x + 3 4x 3 x 38) f β 1(x) = β 39) gβ 40) gβ 1(x) = β 1(x) = β 3x + 1 2 35) f β 3 Answers - Exponential Functions 15) 1 16) 1 β 29) 0 30) No solution 17) No solution 4 3 18) β 19) 1 4 β 20) 3 4 21) No solution β 22) 0 23) β 24) 2 5 25) β 26) 1 4 27) β 28) 1 3 3 2 1 1 2 486 31) 1 32) 3 33) 1 3 34) 2 3 35) 0 36) 0 37) 3 8 38) 39) 1 3 β β 40) No solution 10.5 1) 92 = 81 2) bβ 16 = a 2 = 1 49 4) 162 = 256 3) 7β 5) 132 = 169 6) 110 = 1 7) log8 1 = 0 2 1 289 = 8) log17 β 9) log15 225 = 2 10) log144 12 = 1 2 11) log64 2 = 1 6 12) log19 361 = 2 13) 1 3 14) 3 10.6 1) Answers - Logarithmic Functions 1 3 15) β 16) 0 17) 2 3 18) β 19) 2 20) 1 2 21) 6 22) 5 23) 5 24) 512 25) 1 4 26) 1000 27) 121 28) 256 29) 6552 30) 45 11 125 3 1 4 54 11 2401 3 1 2 1 11 621 10 31) 32) 33) 34) 35) 36) β β β β β β 37) β 38) 283 243 39) 2 5 40) 3 Answers - Interest Rate Problems a. 740.12; 745.91 b. 851.11; 859.99 c. 950.08 |
; 953.44 e. 1209.52; 1214.87 f. 1528.02; 1535.27 g. 2694.70; 2699.72 d. 1979.22; 1984.69 h. 3219.23; 3224.99 i. 7152.17; 7190.52 2) 1640.70 3) 2868.41 4) 2227.41 5) 1726.16 6) 1507.08 7) 2001.60 8) 2009.66 9) 2288.98 10) 6386.12 11) 13742.19 487 12) 28240.43 13) 12.02; 3.96 14) 3823.98 15) 101.68 10.7 1) 0.3256 2) 0.9205 3) 0.9659 4) 0.7660 5) 7 25 6) 8 15 7) 7 16 8) 3 5 9) 2β 2 10) 4 5 11) 16.1 12) 2.8 13) 32 10.8 1) 29β¦ 2) 39β¦ 3) 41β¦ 4) 52β¦ 5) 24β¦ 6) 32β¦ 7) 15β¦ 8) 18β¦ 9) 27β¦ 10) 35β¦ Answers - Trigonometric Functions 14) 8.2 15) 26.1 16) 16.8 17) 2.2 18) 9.8 19) 17.8 20) 10.3 21) 3.9 22) 10.6 23) 10.2 24) 8.9 25) 9.5 26) 24.4 27) 4.7 28) 14.6 29) 1 30) 8 31) 1.5 32) 7.2 33) 5.5 34)2 35) 41.1 36) 3.2 37) 18.2 38) 3.3 39) 17.1 40) 22.2 Answers - Inverse Trigonometric Functions 11) 36β¦ 12) 61.7β¦ 13) 54β¦ 14) 46.2β¦ 15) 55.2β¦ 16) 42.7β¦ 17) 58β¦ 18) 20.1β¦ 19) 45.2β¦ 20) 73.4β¦ 488 21) 51.3β¦ 22) 45β¦ 23) 56.4β¦ 24) 48.2β¦ 25) |
55β¦ 26) 30.5β¦ 27) 47β¦ 28) 15.5β¦ 29) 30β¦ 30) 59β¦ 32) mβ B = 22.8β¦, mβ A = 67.2β¦, c = 16.3 33) mβ B = 22.5β¦, mβ A = 67.5β¦, c = 7.6 34) mβ A = 39β¦, b = 7.2, a = 5.9 35) mβ B = 64.6β¦, mβ A = 25.4β¦, b = 6.3 36) mβ A = 69β¦, b = 2.5, a = 6.5 37) mβ B = 38β¦, b = 9.9, a = 12.6 38) mβ B = 42β¦, b = 9.4, c = 14 39) mβ A = 45β¦, b = 8, c = 11.3 40) mβ B = 29.1β¦, mβ A = 60.9β¦, 31) mβ B = 28β¦, b = 15.1, c = 32.2 a = 12.2 48920 From this table we can see that the ratio of the column areas, 68 : 92 : 40, is exactly the same as the ratio of the frequencies, 34 : 46 : 20. In a histogram, the area of a column represents the frequency of the corresponding class, so that the area must be proportional to the frequency. We may see this written as βarea β frequencyβ. This also means that in every histogram, just as in the example above, the ratio of column areas is the same as the ratio of the frequencies, even if the classes do not have equal widths. Also, there can be no gaps between the columns in a histogram because the upper boundary of one class is equal to the lower boundary of the neighbouring class. A gap can appear only when a class has zero frequency. The axis showing the measurements is labelled as a continuous number line, and the width of each column is equal to the width of the class that it represents. When we construct a histogram, since the classes may not have equal widths, the height of each column is no longer determined by the frequency alone, but must be |
calculated so that area β frequency. The vertical axis of the histogram is labelled frequency density, which measures frequency per standard interval. The simplest and most commonly used standard interval is 1 unit of measurement. For example, a column representing 85 objects with masses from 50 to 60 kg has a frequency density of per gram and so on. 85objects β (60 50) kg = KEY POINT 1.3 8.5 objects per kilogram or 0.0085 objects For a standard interval of 1 unit of measurement, rearranged to give Frequency density = class frequency class width, which can be Class frequency = class width Γ frequency density In a histogram, we can see the relative frequencies of classes by comparing column areas, and we can make estimates by assuming that the values in each class are spread evenly over the whole class interval. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 1.2 The masses, m kg, of 100 children are grouped into two classes, as shown in the table. Mass m( kg) No. children f( ) ΓΈ <m40 50 40 ΓΈ <m50 70 60 a Illustrate the data in a histogram. b Estimate the number of children with masses between 45 and 63kg. Answer a Mass m( kg) ΓΈ <m40 50 ΓΈ <m50 70 No. children f( ) Class width (kg) 40 10 60 20 Frequency density Γ· 40 10 = 4 Γ· 60 20 = 3 Frequency density is calculated for the unequal-width intervals in the table. The masses are represented in the histogram, where frequency density measures number of children per 1kg or simply children per kg 40 45 50 55 Mass (m kg) 60 63 65 70 b There are children with masses from 45 to 63kg in both classes, so we must split this interval into two parts: 45 β50 and 50 β 63. Γ 40 = 20 children 1 2 Frequency width = Γ frequency density = (63 50) kg β Γ 3 children 1kg = = 13 Γ 3 children 39 children Our estimate is + 20 39 = 59 children. The class 40β50 kg has a mid-value of 45kg and a frequency of 40. |
Our estimate for the interval 50β63kg is equal to the area corresponding to this section of the second column. We add together the estimates for the two intervals. Copyright Material - Review Only - Not for Redistribution TIP Column areas are equal to class frequencies. For example, the area of the first column is 4 children 1kg (50 40) kg β Γ = 40 children. TIP If we drew column heights of 8 and 6 instead of 4 and 3, then frequency density would measure children per 2 kg. The area of the first column would be (50 40) kg β Γ 8 children 2 kg = 40 children. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data Consider the times taken, to the nearest minute, for 36 athletes to complete a race, as given in the table below. TIP Time taken (min) No. athletes f( ) 13 4 14 15β 14 16 18β 18 Gaps of 1 minute appear between classes because the times are rounded. Frequency densities are calculated in the following table. Time taken (t min) 12.5 ΓΈ <t 13.5 13.5 ΓΈ <t 15.5 15.5 ΓΈ <t 18.5 No. athletes f( ) Class width (min) Frequency density 4 1 14 2 18 3 4 Γ· = 1 4 14 Γ· = 2 7 18 Γ· = 3 6 This histogram represents the race times, where frequency density measures athletes per minute 12.5 13.5 15.5 Time taken (t min) 18.5 WORKED EXAMPLE 1.3 Use the histogram of race times shown previously to estimate: a the number of athletes who took less than 13.0 minutes b the number of athletes who took between 14.5 and 17.5 minutes c the time taken to run the race by the slowest three athletes. Answer Use class boundaries (rather than rounded values) to find class widths, otherwise incorrect frequency densities will be obtained. TIP The class with the highest frequency does not necessarily have the highest frequency density. TIP Do think carefully about the scales you use when constructing a histogram or any other type of diagram. Sensible scales, such as 1cm for 1, 5, 10, 20 or 50 units, allow you to read |
values with much greater accuracy than scales such as 1cm for 3, 7 or 23 units. For similar reasons, try to use as much of the sheet of graph paper as possible, ensuring that the whole diagram will fit before you start to draw it. 9 We can see that two blocks represent one athlete in the histogram. So, instead of calculating with frequency densities, we can simply count the number of blocks and divide by 2 to estimate the number of athletes involved. a b Γ· = 4 2 2 athletes There are four blocks to the left of 13.0 minutes. 38 2 19 athletes Γ· = There are + 14 24 = 38 blocks between 14.5 and 17.5 minutes. c Between 18.0 and 18.5 minutes. The slowest three athletes are represented by the six blocks to the right of 18.0 minutes. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 1.1 Refer back to the table in Section 1.3 that shows the percentage scores of 100 students who took an examination. Discuss what adjustments must be made so that the data can be represented in a histogram. How could we make these adjustments and is there more than one way of doing this? TIP It is not acceptable to draw the axes or the columns of a histogram freehand. Always use a ruler! EXERCISE 1B 1 In a particular city there are 51 buildings of historical interest. The following table presents the ages of these buildings, given to the nearest 50 years. Age (years) No. buildings f( ) 50β150 15 200β300 18 350β450 12 500β600 6 a Write down the lower and upper boundary values of the class containing the greatest number of buildings. b State the widths of the four class intervals. c Illustrate the data in a histogram. d Estimate the number of buildings that are between 250 and 400 years old. 10 2 The masses, m grams, of 690 medical samples are given in the following table. Mass (m grams) No. medical samples f( ) ΓΈ <m4 12 224 ΓΈ <m12 24 396 ΓΈ <m24 28 p a Find the value of p that appears in |
the table. b On graph paper, draw a histogram to represent the data. c Calculate an estimate of the number of samples with masses between 8 and 18 grams. 3 The table below shows the heights, in metres, of 50 boys and of 50 girls. Height (m) No. boys f( ) No. girls f( ) 1.2β 7 10 1.3β 11 22 1.6β 26 16 1.8β1.9 6 2 a How many children are between 1.3 and 1.6 metres tall? b Draw a histogram to represent the heights of all the boys and girls together. c Estimate the number of children whose heights are 1.7 metres or more. 4 The heights of 600 saplings are shown in the following table. Height (cm) No. saplings f( ) 0β 64 5β 232 15β 240 30β u 64 a Suggest a suitable value for u, the upper boundary of the data. b Illustrate the data in a histogram. c Calculate an estimate of the number of saplings with heights that are: i less than 25cm ii between 7.5 and 19.5cm. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data 5 Each of the 70 trainees at a secretarial college was asked to type a copy of a particular document. The times taken are shown, correct to the nearest 0.1 minutes, in the following table. Time taken (min) No. trainees f( ) 2.6β2.8 15 2.9β3.0 25 3.1β3.2 20 3.3β3.7 10 a Explain why the interval for the first class has a width of 0.3 minutes. b Represent the times taken in a histogram. c Estimate, to the nearest second, the upper boundary of the times taken by the fastest 10 typists. d It is given that 15 trainees took between 3.15 and b minutes. Calculate an estimate for the value of b when: i >b 3.15 ii <b 3.15. 6 A railway line monitored 15% of its August train journeys to find their departure delay times. The results are shown below. It is |
given that 24 of these journeys were delayed by less than 2 minutes 20 16 12 8 4 0 2 4 12 Time (min) 18 20 11 a How many journeys were monitored? b Calculate an estimate of the number of these journeys that were delayed by: i 1 to 3 minutes ii 10 to 15 minutes. c Show that a total of 2160 journeys were provided in August. d Calculate an estimate of the number of August journeys that were delayed by 3 to 7 minutes. State any assumptions that you make in your calculations. 7 A university investigated how much space on its computersβ hard drives is used for data storage. The results are shown below. It is given that 40 hard drives use less than 20GB for data storage 20 60 120 200 Storage space (GB) a Find the total number of hard drives represented. b Calculate an estimate of the number of hard drives that use less than 50GB. c Estimate the value of k, if 25% of the hard drives use k GB or more. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 8 The lengths of the 575 items in a candle makerβs workshop are represented in the histogram. a What proportion of the items are less than 25cm long? b Estimate the number of items that are between 12.4 and 36.8cm long. c The shortest 20% of the workshopβs items are to be recycled. Calculate an estimate of the length of the shortest item that will not be recycled. 9 The thicknesses, k mm, of some steel sheets are represented in the histogram. It is given that 0.4<k for 180 sheets. a Find the ratio between the frequencies of the three classes. Give your answer in simplified form. b Find the value of n, given that frequency density measures sheets per n mm. c Calculate an estimate of the number of sheets for which: i <k 0.5 ii 0.75 kΓΈ < 0.94. 25 20 15 10 12 10 20 30 40 50 Length (l cm) 0.2 0.4 0.6 0.8 Thickness (k mm) 1.0 1.2 d The sheets are classified as thin |
, medium or thick in the ratio 1: 3 : 1. Estimate the thickness of a medium sheet, giving your answer in the form < < your values for a and b? a k b. How accurate are PS 10 The masses, in kilograms, of the animals treated at a veterinary clinic in the past year are illustrated in a histogram. The histogram has four columns of equal height. The following table shows the class intervals and the number of animals in two of the classes. Mass (kg) No. animals f( ) 3β5 a 6β12 371 13β32 1060 33β44 b a Find the value of a and of b, and show that a total of 2226 animals were treated at the clinic. b Calculate an estimate of the lower boundary of the masses of the heaviest 50% of these animals. PS 11 The minimum daily temperature at a mountain village was recorded to the nearest days. The results are grouped into a frequency table and a histogram is drawn. Β°0.5 C on 200 consecutive The temperatures ranged from Β°0.5 C to Β°2 C on n days, and this class is represented by a column of height h cm. The temperatures ranged from β2.5 to column that represents these temperatures. Β° β0.5 C on d days. Find, in terms of n h, and d, the height of the PS 12 The frequency densities of the four classes in a histogram are in the ratio 4 : 3 : 2 : 1. The frequencies of these classes are in the ratio 10 : 15 : 24 : 8. Find the total width of the histogram, given that the narrowest class interval is represented by a column of width 3cm. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data PS 13 The percentage examination scores of 747 students are given in the following table. Score %%( ) No. students f( ) p β50 165 51β70 240 71β80 195 81βq 147 Given that the frequency densities of the four classes of percentage scores are in the ratio 5 : 8 : 13 : 7, find the value of p and of q. DID YOU KNOW? Bar charts first appeared in a book by the Scottish political economist William Playfair, entitled |
The Commercial and Political Atlas (London, 1786). His invention was adopted by many in the following years, including Florence Nightingale, who used bar charts in 1859 to compare mortality in the peacetime army with the mortality of civilians. This helped to convince the government to improve army hygiene. In the past few decades histograms have played a very important role in image processing and computer vision. An image histogram acts as a graphical representation of the distribution of colour tones in a digital image. By making adjustments to the histogram, an image can be greatly enhanced. This has had great benefits in medicine, where scanned images are used to diagnose injury and illness. FAST FORWARD In Chapter 2, Section 2.3 and in Chapter 8, Section 8.1, we will see how a histogram or bar chart can be used to show the shape of a set of data, and how that shape provides information on average values. 1.4 Representation of continuous data: cumulative frequency graphs A cumulative frequency graph can be used to represent continuous data. Cumulative frequency is the total frequency of all values less than a given value. If we are given grouped data, we can construct the cumulative frequency diagram by plotting cumulative frequencies (abbreviated to cf ) against upper class boundaries for all intervals. We can join the points consecutively with straight-line segments to give a cumulative frequency polygon or with a smooth curve to give a cumulative frequency curve. For example, a set of data that includes 100 values below 7.5 and 200 values below 9.5 will have two of its points plotted at (7.5, 100) and at (9.5, 200). We plot points at upper boundaries because we know the total frequencies up to these points are precise. From a cumulative frequency graph we can estimate the number or proportion of values that lie above or below a given value, or between two values. There is no rule for deciding whether a polygon or curve is the best type of graph to draw. It is often difficult to fit a smooth curve through a set of plotted points. Also, it is unlikely that any two people will draw exactly the same curve. A polygon, however, is not subject to this uncertainty, as we know exactly where the line segments must be drawn. 13 TIP A common mistake is to plot points at class mid-values but the total frequency up to each mid-value is not precise β it is an estimate. Copyright Material - Review Only - Not for Redistribution Review |
Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 By drawing a cumulative frequency polygon, we are making exactly the same assumptions that we made when we used a histogram to calculate estimates. This means that estimates from a polygon should match exactly with estimates from a histogram. WORKED EXAMPLE 1.4 The following table shows the lengths of 80 leaves from a particular tree, given to the nearest centimetre. Lengths (cm) No. leaves f( ) 1β2 8 3β4 20 5β7 38 8β9 10 10β11 4 Draw a cumulative frequency curve and a cumulative frequency polygon. Use each of these to estimate: a the number of leaves that are less than 3.7 cm long b the lower boundary of the lengths of the longest 22 leaves. Answer Lengths l( cm) l < 0.5 l < 2..5 7.5 9.5 11.5 14 Addition of frequencies No. leaves cf( ) 0 0 8+ + + 0 8 20 + + 0 8 20 38 + 0 8 20 38 10 + + + + + 0 8 20 38 10 4 + + + + 0 8 28 66 76 80 Before the diagrams can be drawn, we must organise the given data to show upper class boundaries and cumulative frequencies. 76 can also be calculated as 66 + 10, using the previous cf value 80 60 40 20 0 2 4 6 8 10 12 Length (l cm) a The polygon gives an estimate of 20 leaves. The curve gives an estimate of 18 leaves. b The polygon gives an estimate of 6.9cm. The curve gives an estimate of 6.7 cm. We plot points at (0.5,0), (2.5,8), (4.5,28), (7.5,66), (9.5,76) and (11.5, 80). We then join them in order with ruled lines and also with a smooth curve, to give the two types of graph. Copyright Material - Review Only - Not for Redistribution REWIND In Worked example 1.2, we made estimates from a histogram by assuming that the values in each class are spread evenly over the whole class interval. TIP β’ Do include the lowest class boundary |
, which has a cumulative frequency of 0, and plot this point on your graph. β’ The dotted lines show the workings for parts a and b. β’ When constructing a cumulative frequency graph, you are advised to use sensible scales that allow you to plot and read values accurately. β’ Estimates from a polygon and a curve will not be the same, as they coincide only at the plotted points. β’ Note that all of these answers are only estimates, as we do not know the exact shape of the cumulative frequency graph between the plotted points. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data EXPLORE 1.2 Four histograms (1β 4) that represent four different sets of data with equal-width intervals are shown. 1 3 fd 0 fd 0 2 4 fd 0 fd 0 a Which of the following cumulative frequency graphs (AβD) could represent the same set of data as each of the histograms (1β 4) above? A C cf 0 cf 0 B D cf 0 cf 0 How do column heights affect the shape of a cumulative frequency graph? b Why could a cumulative frequency graph never look like the sketch shown below? cf 0 Copyright Material - Review Only - Not for Redistribution 15 FAST FORWARD We will use cumulative frequency graphs to estimate the median, quartiles and percentiles of a set of data in Chapter 2, Section 2.3 and in Chapter 3, Section 3.2. WEB LINK You can investigate the relationship between histograms and cumulative frequency graphs by visiting the Cumulative Frequency Properties resource on the Geogebra website (www.geogebra.org). Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXERCISE 1C 1 The reaction times, t seconds, of 66 participants were measured in an experiment and presented below. Time (t seconds) No. participants cf( ) t < 1.5 t < 3.0 t < t < 4.5 6.5 t < 8.5 t < 11.0 t < 13.0 0 3 8 32 54 62 66 a |
Draw a cumulative frequency polygon to represent the data. b Use your graph to estimate: i the number of participants with reaction times between 5.5 and 7.5 seconds ii the lower boundary of the slowest 20 reaction times. 2 The following table shows the widths of the 70 books in one section of a library, given to the nearest 16 centimetre. Width (cm) No. books f( ) 10β14 3 15β19 13 20β29 25 30β39 24 40β44 5 a Given that the upper boundary of the first class is 14.5cm, write down the upper boundary of the second class. b Draw up a cumulative frequency table for the data and construct a cumulative frequency graph. c Use your graph to estimate: i the number of books that have widths of less than 27 cm ii the widths of the widest 20 books. 3 Measurements of the distances, x mm, between two moving parts inside car engines were recorded and are summarised in the following table. There were 156 engines of type A and 156 engines of type B. Distance x( mm) x < 0.10 x < 0.35 x < 0.60 x < 0.85 x < 1.20 Engine A cf( ) Engine B cf( ) 0 0 16 8 84 52 134 120 156 156 a Draw and label two cumulative frequency curves on the same axes. b Use your graphs to estimate: i the number of engines of each type with measurements between 0.30 and 0.70 mm ii the total number of engines with measurements that were less than 0.55 mm. c Both types of engine must be repaired if the distance between these moving parts is more than a certain fixed amount. Given that 16 type A engines need repairing, estimate the number of type B engines that need repairing. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data 4 The diameters, d cm, of 60 cylindrical electronic components are represented in the following cumulative frequency graph ( 60 40 20 0 0.1 0.2 0.3 0.4 0.5 0.6 Diameter (d cm) a Find the number of components such that 0.2 ΓΈ d < 0.4, and explain why your answer is not |
an estimate. b Estimate the number of components that have: i a diameter of less than 0.15cm ii a radius of 0.16cm or more. c Estimate the value of k, given that 20% of the components have diameters of k mm or more. d Give the reason why 0.1β0.2 cm is the modal class. 17 5 The following cumulative frequency graph shows the masses, m grams, of 152 uncut diamonds 160 120 80 40 0 P 4 8 12 16 20 24 Mass (m grams) a Estimate the number of uncut diamonds with masses such that: i ΓΈ < 17m 9 ii 7.2 ΓΈ < m. 15.6 b The lightest 40 diamonds are classified as small. The heaviest 40 diamonds are classified as large. Estimate the difference between the mass of the heaviest small diamond and the lightest large diamond. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 c The point marked at P(24, 152) on the graph indicates that the 152 uncut diamonds all have masses of less than 24 grams. Each diamond is now cut into two parts of equal mass. Assuming that there is no wastage of material, write down the coordinates of the point corresponding to P on a cumulative frequency graph representing the masses of these cut diamonds. 6 The densities, d g/cm3, of 125 chemical compounds are given in the following table. Density d( g/cm )3 No. compounds cf( ) < d 1.30 32 < d 1.38 77 < d 1.42 92 < d 1.58 107 < d 1.70 125 Find the frequencies a b c,, and d given in the table below. Density d( g/cm )3 No. compounds f( ) 0β a 1.30β b 1.38β c 1.42β1.70 d PS 7 The daily journey times for 80 bank staff to get to work are given in the following table. Time (t min) No. staff cf( ) t 10< 3 t 15< 11 20< t 24 25< t 56 t 30< 68 45< t 76 t 60< 80 18 a How |
many staff take between 15 and 45 minutes to get to work? b Find the exact number of staff who take + x y 2 minutes or more to get to work, given that 85% of the staff take less than x minutes and that 70% of the staff take y minutes or more. M 8 A fashion company selected 100 12-year-old boys and 100 12-year-old girls to audition as models. The heights, h cm, of the selected children are represented in the following graph 100 75 50 25 0 Boys Girls 140 150 160 170 180 Height (h cm) a What features of the data suggest that the children were not selected at random? b Estimate the number of girls who are taller than the shortest 50 boys. c What is the significance of the value of h where the graphs intersect? d The shortest 75 boys and tallest 75 girls were recalled for a second audition. On a cumulative frequency graph, show the heights of the children who were not recalled. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data 9 The following table shows the ages of the students currently at a university, given by percentage. Ages are rounded down to the number of whole years. Age (years) Students %%( ) 18< 0 18β19 20 β 21 22 β 24 25β 28 29β 35 36β 44 27 51 11 5 4 2 a Represent the data in a percentage cumulative frequency polygon. b The oldest 8% of these students qualify as βmatureβ. Use your polygon to estimate the minimum age requirement for a student to be considered mature. Give your answer to the nearest month. c Of the 324 students who are 18β19 years old, 54 are not expected to find employment within 3 months of finishing their course. i Calculate an estimate of the number of current students who are expected to find employment within 3 months of finishing their course. ii What assumptions must be made to justify your calculations in part c i? Are these assumptions reasonable? Do you expect your estimate to be an overestimate or an underestimate? PS 10 The distances, in km, that 80 new cars can travel on 1 litre of fuel are shown in the table. Distance (km) No. cars f( ) 4.4β 5 6. |
6β 7 8.8β 52 12.1β 12 15.4β18.7 4 These distances are 10% greater than the distances the cars will be able to travel after they have covered more than 100 000 km. Estimate how many of the cars can travel 10.5km or more on 1 litre of fuel when new, but not after they have covered more than 100 000 km. 19 PSM 11 A small company produces cylindrical wooden pegs for making garden chairs. The lengths and diameters of the 242 pegs produced yesterday have been measured independently by two employees, and their results are given in the following table. Length ( cm) l No. pegs ( )cf 1.0< l 2.0< l 2.5< l 3.0< l 3.5< l 4.0< l 4.5< l 0 0 8 40 110 216 242 Diameter ( cm) d No. pegs ( )cf d 1.0< d 1.5< d 2.0< d 2.5< d 3.0< 0 60 182 222 242 a On the same axes, draw two cumulative frequency graphs: one for lengths and one for diameters. b Correct to the nearest millimetre, the lengths and diameters of n of these pegs are equal. Find the least and greatest possible value of n. c A peg is acceptable for use when it satisfies both l ΓΉ and <d 2.8 2.2. Explain why you cannot obtain from your graphs an accurate estimate of the number of these 242 pegs that are acceptable. Suggest what the company could do differently so that an accurate estimate of the proportion of acceptable pegs could be obtained. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXPLORE 1.3 The following table shows data on the masses, m grams, of 150 objects. Mass ( g)m m 0< m 12< m 30< m 53< m 70< m 80< No. objects cf( ) 0 24 60 106 138 150 By drawing a cumulative frequency polygon, the following estimates will be obtained: a Number of objects with masses less than 20 g = 40 objects. TIP The |
six plotted points, whose coordinates you know, are joined by straight lines. b Arranged in ascending order, the mass of the 100th object 50 g. = However, we can calculate these estimates from the information given in the table without drawing the polygon. Investigate the possible methods that we can use to calculate these two estimates. 1.5 Comparing different data representations Pictograms, vertical line graphs, bar charts and pie charts are useful ways of displaying qualitative data and ungrouped quantitative data, and people generally find them easy to understand. Nevertheless, it may be of benefit to group a set of raw data so that we can see how the values are distributed. Knowing the proportion of small, medium and large values, for example, may prove to be useful. For small datasets we can do this by constructing stem-and-leaf diagrams, which have the advantage that raw values can still be seen after grouping. 20 In large datasets individual values lose their significance and a picture of the whole is more informative. We can use frequency tables to make a compact summary by grouping but most people find the information easier to grasp when it is shown in a graphical format, which allows absolute, relative or cumulative frequencies to be seen. Although some data are lost by grouping, histograms and cumulative frequency graphs have the advantage that data can be grouped into classes of any and varied widths. The choice of which representation to use will depend on the type and quantity of data, the audience and the objectives behind making the representation. Most importantly, the representation must show the data clearly and should not be misleading in any way. The following chart is a guide to some of the most commonly used methods of data representation. FAST FORWARD In Chapter 2 and Chapter 3, we will see how grouping effects the methods we use to find measures of central tendency and measures of variation. Qualitative data Discrete quantitative data Continuous data Ungrouped Grouped Pictogram, vertical line graph, bar chart, pie chart, sectional bar chart Small amount Stem-and-leaf diagram Large amount Histogram Cumulative frequency graph Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data EXERCISE 1D 1 Jamila noted each studentβs answer when her year group was asked to name |
their favourite colour. a List the methods of representation that would be suitable for displaying Jamilaβs data. b Jamila wishes to emphasise that the favourite colour of exactly three-quarters of the students is blue. Which type of representation from your list do you think would be the most effective for Jamila to use? Explain why you have chosen this particular type of representation. 2 A large number of chickensβ eggs are individually weighed. The masses are grouped into nine classes, each of width 2 grams, from 48 to 66g. Name a type of representation in which the fact could be seen that the majority of the eggs have masses from 54 to 60 g. Explain how the representation would show this. 3 Boxes of floor tiles are to be offered for sale at a special price of $75. The boxes claim to contain at least 100 tiles each. a Why would it be preferable to use a stem-and-leaf diagram rather than a bar chart to represent the numbers of tiles, which are 112, 116, 107, 108, 121, 104, 111, 106, 105 and 110? b How may the seller benefit if the numbers 12, 16, 7, 8, 21, 4, 11, 6, 5 and 10 are used to draw the stem-and- leaf diagram instead of the actual numbers of tiles? 4 A charity groupβs target is to raise a certain amount of money in a year. At the end of the first month the group raised 36% of the target amount, and at the end of each subsequent month they manage to raise exactly half of the amount outstanding. 21 a How many months will it take the group to raise 99% of the money? b Name a type of representation that will show that the group fails to reach its target by the end of the year. Explain how this fact would be shown in the representation. 5 University students measured the heights of the 54 trees in the grounds of a primary school. As part of a talk on conservation at a school assembly, the students have decided to present their data using one of the following diagrams 12 Heights of trees (m) Heights of 54 trees tall short medium 2 to 3 m 3 to 5 m 5 to 8 m a Give one disadvantage of using each of the representations shown. b Name and describe a different type of representation that would be appropriate for the audience, and that has none of the disadvantages given in part a. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy |
Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 6 The percentage scores of 40 candidates who took a Health and Safety test are given: 77 44 65 84 52 60 35 83 68 66 50 68 65 57 60 50 93 38 46 55 45 69 61 64 40 66 91 59 61 74 70 75 42 65 85 63 73 84 68 30 a Construct a frequency table by grouping the data into seven classes with equal-width intervals, where the first class is 30β39. Label this as Table 1. b It is proposed that each of the 40 candidates is awarded one of three grades, A, B or C. Construct a new frequency table that matches with this proposal. Label this as Table 2. c A student plans to display all three versions of the data (i.e. the raw data, the data in Table 1 and the data in Table 2) in separate stem-and-leaf diagrams. For which version(s) of the data would this not be appropriate? Suggest an alternative type of representation in each case. M 7 Last year Tom renovated an old building during which he worked for at least 9 hours each week. By plotting four points in a graph, he has represented the time he spent working 52 51 50 49 22 0 34 36 38 40 42 44 Time spent working (hours) a What can you say about the time that Tom spent working on the basis of this graph? b Explain why Tomβs graph might be considered to be misleading. c Name the different types of representation that are suitable for displaying the amount of time that Tom worked each week throughout the year. Consider the benefits of each type of representation and then fully describe (but do not draw) the one you believe to be the most suitable. PS 8 The following table shows the focal lengths, l mm, of the 84 zoom lenses sold by a shop. For example, there are 18 zoom lenses that can be set to any focal length between 24 and 50 mm. Focal length l( mm) No. lenses f( ) 24β50 18 50β108 100β200 150β300 250β400 30 18 12 6 a What feature of the data does not allow them to be displayed in a histogram? b What type of diagram could you use to illustrate the data? Explain clearly how you would do this. Copyright Material - Review Only - |
Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data M 9 The following table shows estimates, in hundred thousands, of the number of people living in poverty and the populations, in millions, of the countries where they live. (Source: World Bank 2011/12) Country No. living in poverty ΓΓ( 10 )5 Population ΓΓ( 10 )6 Chile Sri Lanka Malaysia Georgia Mongolia 24.8 17.25 18.4 20.65 10.8 28.33 7.91 4.47 7.50 2.74 Represent, in a single diagram, the actual numbers and the relative poverty that exists in these countries. In what way do the two sets of data in your representation give very different pictures of the poverty levels that exist? Which is the better representation to use and why? EXPLORE 1.4 Past, present and predicted world population figures by age group, sex and other categories can be found on government census websites. You may be interested, for example, in the population changes for your own age group during your lifetime. This is something that can be represented in a diagram, either manually, using spreadsheet software or an application such as GeoGebra, and for which you may like to try making predictions by looking for trends shown in the raw data or in any diagrams you create. Checklist of learning and understanding β Non-numerical data are called qualitative or categorical data. β Numerical data are called quantitative data, and are either discrete or continuous. β Discrete data can take only certain values. β Continuous data can take any value, possibly within a limited range. β Data in a stem-and-leaf diagram are ordered in rows with intervals of equal width. β In a histogram, column area β frequency, and the vertical axis is labelled frequency density. β Frequency density = class frequency class width and Class frequency = class width frequency density Γ. β In a cumulative frequency graph, points are plotted at class upper boundaries. 23 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics |
: Probability & Statistics 1 END-OF-CHAPTER REVIEW EXERCISE 1 1 The weights of 220 sausages are summarised in the following table. Weight (grams) Cumulative frequency 20< 0 30< 20 40< 50 45< 100 50< 160 60< 210 70< 220 i State how many sausages weighed between 50 g and 60 g. ii On graph paper, draw a histogram to represent the weights of the sausages. [1] [4] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q4 November 2011[Adapted] 2 The lengths of childrenβs feet, in centimetres, are classified as 14β16, 17β19, 20β22 and so on. State the lower class boundary, the class width and the class mid-value for the lengths given as 17β19. 3 The capacities of ten engines, in litres, are given rounded to 2 decimal places, as follows: 1.86, 2.07, 1.74, 1.08, 1.99, 1.96, 1.83, 1.45, 1.28 and 2.19. These capacities are to be grouped in three classes as 1.0β1.4, 1.5β 2.0 and 2.1β 2.2. a Find the frequency of the class 1.5β2.0 litres. b Write down two words that describe the type of data given about the engines. 4 Over a 10-day period, Alina recorded the number of text messages she received each day. The following stem-and-leaf diagram shows her results Key: 1 5 represents 15 messages 24 messages? a On how many days did she receive more than 10 but not more than 15 b How many more rows would need to be added to the stem-and-leaf diagram if Alina included data for two extra days on which she received 4 and 36 messages? Explain your answer. 5 The following stem-and-leaf diagram shows eight randomly selected numbers between 2 and 4 represents 2.1 Key: 2 1 Given that a bβ find the value of a and of b. = 7 and that the sum of the eight numbers correct to the nearest integer is 24, [2] [1] [2] [1] [2] [3] 6 Eighty people downloaded a particular application and recorded the time taken for the download to complete. The times are given in the following table. Download time ( |
min) No. downloads cf( ) 3< 6 5< 18 6< 66 10< 80 a Find the number of downloads that completed in 5 to 6 minutes. b On a histogram, the download times from 5 to 6 minutes are represented by a column of height 9.6cm. Find the height of the column that represents the download times of 6 to 10 minutes. [1] [2] PS 7 A histogram is drawn with three columns whose widths are in the ratio 1: 2 : 4. The frequency densities of these classes are in the ratio 16 : 12 : 3, respectively. a Given that the total frequency of the data is 390, find the frequency of each class. b The classes with the two highest frequencies are to be merged and a new histogram drawn. Given that the height of the column representing the merged classes is to be 30 cm, find the correct height for the remaining column. c Explain what problems you would encounter if asked to construct a histogram in which the classes with the two lowest frequencies are to be merged. [3] [3] [1] Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 1: Representation of data P 8 The histograms below illustrate the number of hours of sunshine during August in two regions, A and B. Neither region had more than 8 hours of sunshine per day. Region A Region Hours of sunshine Hours of sunshine a Explain how you know that some information for one of the regions has been omitted. [2] b After studying the histograms, two students make the following statements. β Bindu: There was more sunshine in region A than in region B during the first 2 weeks of August. β Janet: In August there was less sunshine in region A than in region B. 25 Discuss these statements and decide whether or not you agree with each of them. In each case, explain your reasoning. [3] 9 A hotel has 90 rooms. The table summarises information about the number of rooms occupied each day for a period of 200 days. Number of rooms occupied 1β20 21β 40 41β50 51β60 61β70 71β90 Frequency 10 32 62 50 28 18 i Draw a cumulative frequency graph on graph paper to illustrate this information. ii Estimate the number of days |
when more than 30 rooms were occupied. iii On 75% of the days at most n rooms were occupied. Estimate the value of n. [4] [2] [2] Cambridge International AS & A Level Mathematics 9709 Paper 62 Q5 June 2011 Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy 26 Chapter 2 Measures of central tendency In this chapter you will learn how to: β find and use different measures of central tendency β calculate and use the mean of a set of data (including grouped data) either from the data itself or β and use such totals in solving problems that may x b ) from a given total xΞ£ or a coded total ( Ξ£ involve up to two datasets. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills IGCSE / O Level Mathematics Calculate the mean, median and mode for individual and discrete data. Use a calculator efficiently and apply appropriate checks of accuracy. 1 Find the mean, median and mode of the numbers 7.3, 3.9, 1.3, 6.6, 9.2, 4.7, 3.9 and 3.1. 2 Use a calculator to evaluate 6 1.7 Γ 8 1.9 11 2.1 + + Γ 6 8 11 + + Γ and then check that your answer is reasonable. 27 Three types of average There are three measures of central tendency that are commonly used to describe the average value of a set of data. These are the mode, the mean and the median. β The mode is the most commonly occurring value. β The mean is calculated by dividing the sum of the values by the number of values. β The median is the value in the middle of an ordered set of data. We use an average to summarise the values in a set of data. As a representative value, it should be fairly central to, and typical of, the values that it represents. If we investigate the annual incomes |
of all the people in a region, then a single value (i.e. an average income) would be a convenient number to represent our findings. However, choosing which average to use is something that needs to be thought about, as one measure may be more appropriate to use than the others. Deciding which measure to use depends on many factors. Although the mean is the most familiar average, a shoemaker would prefer to know which shoe size is the most popular (i.e. the mode). A farmer may find the median number of eggs laid by their chickens to be the most useful because they could use it to identify which chickens are profitable and which are not. As for the average income in our chosen region, we must also consider whether to calculate an average for the workers and managers together or separately; and, if separately, then we need to decide who fits into which category. EXPLORE 2.1 Various sources tell us that the average person: β laughs 10 times a day β falls asleep in 7 minutes β sheds 0.7 kg of skin each year β grows 944km of hair in a lifetime β produces a sneeze that travels at 160 km/h β has over 97 000 km of blood vessels in their body β has a vocabulary between 5000 and 6000 words. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 The average adult male is 172.5cm tall and weighs 80 kg. The average adult female is 159cm tall and weighs 68kg. What might each of these statements mean and how might they have been determined? Have you ever met such an average person? How could this information be useful? You can find a variety of continuously updated figures that yield interesting averages at http://www.worldometers.info/. 2.1 The mode and the modal class As you will recall, a set of data may have more than one mode or no mode at all. The following table shows the scores on 25 rolls of a die, where 2 is the mode because it has the highest frequency. Score on die Frequency ( ) In a set of grouped data in which raw values cannot be seen, we can find the modal class, which is the class with the highest frequency density. WORKED EXAM |
PLE 2.1 28 Find the modal class of the 270 pencil lengths, given to the nearest centimetre in the following table. Length x( cm) No. pencils ( )f 4β7 8β10 11β12 100 90 80 Answer Length ( cm) x No. pencils ( )f 3.5 < <x 7.5 7.5 < <x 10.5 10.5 < <x 12.5 100 90 80 Width (cm) Frequency density 4 3 2 100 Γ· = 4 25 90 3 30 Γ· = 80 2 Γ· = 40 ( 40 30 20 10 0 3.5 7.5 10.5 12.5 Length (cm) The modal class is 11β12 cm (or, more accurately, 10.5 12.5cm). << x Class boundaries, class widths and frequency density calculations are shown in the table. Although the histogram shown is not needed to answer this question, it is useful to see that, in this case, the modal class is the one with the tallest column and the greatest frequency density, even though it has the lowest frequency. Copyright Material - Review Only - Not for Redistribution REWIND We saw how to calculate frequency density in Chapter 1, Section 1.3. KEY POINT 2.1 In histograms, the modal class has the greatest column height. If there is no modal class then all classes have the same frequency density. TIP The modal class does not contain the most pencils but it does contain the greatest number of pencils per centimetre. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency WORKED EXAMPLE 2.2 Two classes of data have interval widths in the ratio 3:2. Given that there is no modal class and that the frequency of the first class is 48, find the frequency of the second class. Answer Let the frequency of the second class be x. 48: 3:2 x = x 48 x 2 3 32 = = The second class has a frequency of 32. Or, let the frequency of the second class be x. x 2 x 48 3 32 = = EXERCISE 2A βNo modal classβ means that the frequency densities of the two classes are |
equal, so class frequencies are in the same ratio as interval widths. Alternatively, frequencies are proportional to interval widths. TIP In the special case where all classes have equal widths, frequency densities are proportional to frequencies, so the modal class is the class with the highest frequency. 1 Find the mode(s) of the following sets of numbers. 29 a 12, 15, 11, 7, 4, 10, 32, 14, 6, 13, 19, 3 b 19, 21, 23, 16, 35, 8, 21, 16, 13, 17, 12, 19, 14, 9 2 Which of the eleven words in this sentence is the mode? 3 Identify the mode of x and of y in the following tables4 27 β3 28 β2 29 β1 27 0 25 4 Find the modal class for x and for y in the following tables. x f 0β 5 4β 9 14β20 8 y f 3β6 66 7β11 12β20 80 134 5 A small company sells glass, which it cuts to size to fit into window frames. How could the company benefit from knowing the modal size of glass its customers purchase? 6 Four classes of continuous data are recorded as 1β7, 8β16, 17β20 and 21β25. The class 1β7 has a frequency of 84 and there is no modal class. Find the total frequency of the other three classes. PS 7 Data about the times, in seconds, taken to run 100 metres by n adults are given in the following table. Time x( s) 13.6 <<x 15.4 15.4 <<x 17.4 17.4 <<x 19.8 No. adults ( )f a b 27 By first investigating the possible values of a and of b, find the largest possible value of n, given that the modal class contains the slowest runners. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 PS 8 Three classes of continuous data are given as 0β4, 4β10 and 10β18. The frequency densities of the classes 0β 4 and 10β18 are in the ratio 4 : 3 and the total frequency of these two |
classes is 120. Find the least possible frequency of the modal class, given that the modal class is 4β10. 2.2 The mean The mean is referred to more precisely as the arithmetic mean and it is the most commonly known average. The sum of a set of data values can be found from the mean. Suppose, for example, that 12 values have a mean of 7.5 : = Mean sum of values number of values sum of values 12 You will soon be performing calculations involving the mean, so here we introduce notation that is used in place of the word definition used above. and sum of values 7.5 12 Γ, so 7.5 = = =. 90 We use the upper case Greek letter βsigmaβ, written Ξ£, to represent βsumβ and x to represent the mean, where x represents our data values. The notation used for ungrouped and for grouped data are shown on separate rows in the following table. Sum Data values Frequency of Number of Sum of data Mean data values data values values Ungrouped Grouped Ξ£ Ξ£ x x β f 30 n Ξ£f Ξ£x x = xfΞ£ or Ξ£fx x = x Ξ£ n xf Ξ£ f Ξ£ WORKED EXAMPLE 2.3 Five labourers, whose mean mass is 70.2 kg, wish to go to the top of a building in a lift with some cement. Find the greatest mass of cement they can take if the lift has a maximum weight allowance of 500 kg. TIP f for grouped = Ξ£n data. TIP fx xf = Ξ£ indicates the Ξ£ sum of the products of each value and its frequency. For example, the sum of five 10s and six 20s is (20 6) (10 5) + Γ Γ = (6 20) 170. (5 10) = Γ + Γ Γ = Answer Ξ£ = Γ x x n = = 70.2 5 Γ 351kg 351 y + = y = 500 149 We first rearrange the formula x to find the sum of the labourersβ masses. = x Ξ£ n We now form an equation using y to represent the greatest possible mass of cement. KEY POINT 2.2 For ungrouped data, the mean is = x Ξ£ n For grouped data, x. The greatest mass of cement they can take is 149 kg. x = xf Ξ£ f Ξ£ or fx οΏ½ |
οΏ½ f Ξ£. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency WORKED EXAMPLE 2.4 Find the mean of the 40 values of x, given in the following table. x f Answer x f xf 31 5 31 5 32 7 32 7 33 9 33 9 34 8 34 8 35 11 35 11 fΞ£ = 40 155 224 297 272 385 xfΞ£ = 1333 The mean, x = xf Ξ£ f Ξ£ = 1333 40 = 33.325. We find the sum of the 40 values by adding together the products of each value of x and its frequency. This is done in the row headed xf in the table. The 40 values of x have a sum of 1333. Combined sets of data There are many different ways to combine sets of data. However, here we do this by simply considering all of their values together. To find the mean of two combined sets, we divide the sum of all their values by the total number of values in the two sets. For example, by combining the dataset 1, 2, 3, 4 with the dataset 4, 5, 6, we obtain a new set of data that has seven values in it: 1, 2, 3, 4, 4, 5, 6. Note that the value 4 appears twice. 1 2 3 4 + + + 4 Individually, the sets have means of 5. The combined sets 4 5 6 + + 3 2.5 and = = have a mean of. WORKED EXAMPLE 2.5 A large bag of sweets claims to contain 72 sweets, having a total mass of 852.4g. A small bag of sweets claims to contain 24 sweets, having a total mass of 282.8g. What is the mean mass of all the sweets together? Answer Total number of sweets = 72 24 + = 96. Total mass of We find the total number of sweets and their total mass. sweets = 852.4 282.8 1135.2 g + = Mean mass = 1135.2 96 = 11.825g Copyright Material - Review Only - Not for Redistribution TIP 31 Note that the mean of the combined sets is not +2.5 5 2. TIP Our |
answer assumes that the masses given are accurate to 1 decimal place; that the numbers of sweets given are accurate; and that the masses of the bags are not included in the given totals. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 WORKED EXAMPLE 2.6 A family has 38 films on DVD with a mean playing time of 1 hour 32 minutes. They also have 26 films on video cassette, with a mean playing time of 2 hours 4 minutes. Find the mean playing time of all the films in their collection. Answer 64films 38 26 + = (1h 32 min 38) Γ + (2 h 4 min 26) Γ = = (92 38) Γ 6720 min + (124 26) Γ We find the total number of films and their total playing time. Mean playing time = = 6720 64 105 min or 1h 45 min. The 64 films have a total playing time of 6720 minutes. P EXPLORE 2.2 In Worked example 2.6, the mean playing time of 105 minutes is not equal to +92 124 2. The mean of A and B β but this is not always the case. mean of A mean of B + 2 32 Suppose two sets of data, A and B, have m and n values with means respectively. In what situations will the mean of A and B together be equal to mean of mean of A B + 2? A Ξ£ m and B Ξ£ n, TIP The symbol β means βis not equal toβ. Means from grouped frequency tables When data are presented in a grouped frequency table or illustrated in a histogram or cumulative frequency graph, we lose information about the raw values. For this reason we cannot determine the mean exactly but we can calculate an estimate of the mean. We do this by using mid-values to represent the values in each class. We use the formula x = xf Ξ£ f Ξ£, given in Key point 2.2, to calculate an estimate of the mean, where x now represents the class mid-values. WORKED EXAMPLE 2.7 Coconuts are packed into 75 crates, with 40 of a similar size in each crate. 46 crates contain coconuts with a total mass from 20 up to but |
not including 25kg. 22 crates contain coconuts with a total mass from 25 up to but not including 40 kg. 7 crates contain coconuts with a total mass from 40 up to but not including 54kg. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency a Calculate an estimate of the mean mass of a crate of coconuts. b Use your answer to part a to estimate the mean mass of a coconut. Answer a Mass (kg) No. crates ( )f 20β 46 Mid-value ( )x 22.5 xf 1035 25β 22 32.5 715 40β54 7 47.0 329 Ξ£ =f 75 Ξ£ =xf 2079 Estimate for mean, x = xf Ξ£ f Ξ£ = 2079 75 = 27.72 kg b Estimate for the mean mass of a coconut 27.72 40 0.693kg = = We tabulate the data to include class mid-values, x, and the products xf. We estimate that the 75 crates have a total mass of 2079 kg. We divide our answer to part a by 40 because each crate contains 40 coconuts. When gaps appear between classes of grouped data, class boundaries should be used to find class mid-values. The following example shows a situation in which using incorrect boundaries leads to an incorrect estimate of the mean. 33 WORKED EXAMPLE 2.8 Calculate an estimate of the mean age of a group of 50 students, where there are sixteen 18-year-olds, twenty 19-year-olds and fourteen who are either 20 or 21 years old. Answer Age ( years ) A Mid-value No. students xf ( )x ( )f < 18 <A 19 < 19 <A 20 < 20 <A 22 18.5 19.5 21.0 16 20 14 296 390 294 =Ξ£f 50 =Ξ£xf 980 Estimate of the mean age is xf Ξ£ f Ξ£ 980 50 19.6years = = The 18-year-olds are all aged from 18 up to but not including 19 years. The 19-year-olds are all aged from 19 up to but not including 20 years. The 20- and 21- |
yearolds are all aged from 20 up to but not including 22 years. The age groups and necessary totals are shown in the table. Copyright Material - Review Only - Not for Redistribution TIP Incorrect mid-values of 18, 19 and 20.5 give an incorrect estimated mean of 19.1 years. FAST FORWARD We will see how the mean is used to calculate the variance and standard deviation of a set of data in Chapter 3, Section 3.3. Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Cambridge International AS & A Level Mathematics: Probability & Statistics 1 EXERCISE 2B 1 Calculate the mean of the following sets of numbers: a 28, 16, 83, 72, 105, 55, 6 and 35 b 7.3, 8.6, 11.7, 9.1, 1.7 and 4. and 7 3 8. 2 a The mean of 15, 31, 47, 83, 97, 119 and p2 is 63. Find the possible values of p. b The mean of 6, 29, 3, 14,, ( q q + 8), q 2 and (10 β ) is 20. Find the possible values of q. q 3 Given that: a n = 14 and x =Ξ£ 325.5, find x. b n = 45 and y = 23.6, find the value of yΞ£. c d e z =Ξ£ 4598 and z = 52.25, find the number of values in the set of data. Ξ£ xf = 86 and x 7 1 6=, find the value of fΞ£. f =Ξ£ 135 and x = 0.842, find the value of xfΞ£. 4 Find the mean of x and of y given in the following tables. a x f 18.0 18.5 19.0 19.5 20.0 8 10 17 24 1 34 5 For the data given in the following table, it is given that 3.62 3.65 3.68 3.71 3.74 127 209 322 291 251 13 9 a 10 11 Calculate the value of a. 6 Calculate an estimate of the mean of x and of y given in the following tables. a b x f y f 0 << x 2 2 << x 4 4 << |
x 8 8 << x 14 8 9 11 2 13 << y 16 16 << y 21 21 << y 28 28 << y 33 33 << y 36 7 17 29 16 11 7 An examination was taken by 50 students. The 22 boys scored a mean of 71% and the girls scored a mean of 76%. Find the mean score of all the students. 8 A company employs 12 drivers. Their mean monthly salary is $1950. A new driver is employed and the mean monthly salary falls by $8. Find the monthly salary of the new driver. MP 9 The mean age of the 16 members of a karate club is 26 years and 3 months. One member leaves the club and the mean age of those remaining is 26 years. Find the age of the member who left the club. Give a reason why your answer might not be very accurate. Copyright Material - Review Only - Not for Redistribution Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Review Copy - Cambridge University Press - Review Copy Chapter 2: Measures of central tendency M 10 The following table shows the hourly rates of pay, in dollars, of a companyβs employees. Hourly rate ($) No. employees ( )f 6 8 7 11 8 17 109 1 a Is the mean a good average to use here? Give a reason for your answer. b Find the mean rate of pay for the majority of the employees. PS P 11 A train makes a non-stop journey from one city to another and back again each day. Over a period of 30 days, the mean number of passengers per journey is exactly 61.5. Exact one-way ticket prices paid by these passengers are given by percentage in the following table. Price ($) Passengers ( )% 34 30 38 41 45 29 a Calculate the total revenue from ticket sales, and explain why your answer is an approximation. b The minimum and maximum possible revenues differ by k$. Find the value of k. 12 The heights, in centimetres, of 54 children are represented in the following diagram. The children are split into two equal-sized groups: a βtall halfβ and a βshort halfβ. Calculate an estimate of the difference between the mean heights of these two groups of children. 0 140 144 150 156 159 35 Height (cm) M 13 The following table summarises the number of tomatoes produced by the plants in the plots |
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