Datasets:
NationalTutoringObservatory
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MathEd-PII

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student
What jobs pay
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volunteer
And then try to explain if you are prehighschool graduate
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student
Do u know
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volunteer
Employment based you would want to look out for ones that pay back.
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volunteer
Like TutoringWorks
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[ { "pii_type": "SCHOOL", "surrogate": "TutoringWorks", "start": 5, "end": 18 } ]
student
Is that a app
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volunteer
Emma that is a good one.
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[ { "pii_type": "PERSON", "surrogate": "Emma", "start": 0, "end": 4 } ]
volunteer
That is onsite.
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volunteer
You would have to take the test and pass.
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student
What do I type in
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volunteer
But that is one example.
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volunteer
Just google search and try to search for the job of your interest.
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volunteer
Does this make sense are we good here?
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student
Mr. Lee good bye
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[ { "pii_type": "PERSON", "surrogate": "Mr. Lee", "start": 0, "end": 7 } ]
student
hi
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student
how are you
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volunteer
Hi Chloe
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[ { "pii_type": "PERSON", "surrogate": "Chloe", "start": 3, "end": 8 } ]
volunteer
I'm well, thanks! And you?
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student
im well
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student
im taking a screen shot of the promblem now
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volunteer
Which question do you need help with?
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student
c
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student
the funtion one
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volunteer
OK. Can you give me 5 minutes to solve, and then I'll explain?
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student
ok
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volunteer
We can start. Have you learned about geometric series?
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student
yeah so long ago
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volunteer
We can solve it in two steps. The first step is to see the "pattern" of the series. I'll write on the board.
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volunteer
I'm writing below your upload. Do you see it?
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student
yes
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volunteer
After the first dose, we have Q_1 = 500 mg, correct?
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student
yes
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volunteer
And after the 2nd dose, we have Q_2 = 500 + 500(0.2).
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student
right
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volunteer
The pattern continues. As you can see the "500" is multiplied by increasing powers of 0.2 as "n" increases.
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student
right
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volunteer
Using Summation notation (Σ), we can write the geometric series as I've shown.
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volunteer
The second step is to find the "formula" for the series.
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volunteer
Can you find the formula for Qn ?
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student
i thought it was just qn=500(0.2)^k
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volunteer
What you wrote is the "kth" term of the series, but not the sum.
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volunteer
I can show you the general formula, and you can apply it to your problem if you like.
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student
ok
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volunteer
Looking at the series, can you find what are a = ? and r = ?
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student
is r for rate
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volunteer
Yes. r = 0.2
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volunteer
The formula is correct. For example, if n=3, we have Q_3 = 500(0.2^3-1)/(0.2-1) = ?
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student
620?
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volunteer
Exactly.
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volunteer
So your formula works.
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student
whats this formula called?
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volunteer
It's the formula for a "geometric series". You could also simplify it to read: Qn = 625*(1-0.2^n)
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student
so long term the amount of anitbotic that remains as n approaches infinity
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volunteer
No. For question (d), to find the remaining amount in the long run, you calculate the limit of Qn as n→∞. Can you find it?
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student
infinty
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volunteer
Use the formula: Qn = 625*(1-0.2^n). As n→∞, 0.2^n → 0, correct?
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student
oh i just plugged in infiny for n
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volunteer
I can show you on the board if you wish.
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student
yes please
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volunteer
Did you learn about "limits" ?
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student
yes
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student
anything to the infinty power
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student
is o
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student
?
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volunteer
Because 0.2 is less than 1, 0.2^n goes to zero as "n" goes to infinity. So (1-0.2^n) goes to "1".
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volunteer
For example, if n=1, then 0.2^n = 0.2. If n = 10, then 0,2^10 = 0.0000001024.
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student
its going closer to zero
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volunteer
Yes. And the final limit is: 625*(1-0) = ?
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student
625
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volunteer
Right.
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student
thanks
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volunteer
You're welcome. Is that all for today?
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student
i have one more question
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volunteer
OK.
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volunteer
I need to step away for 3 minutes, and I'll be right back, OK?
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student
ok
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volunteer
I'm back. Can you give me 5 minutes to solve, and then I'll walk you through the solution?
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student
ok
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volunteer
We can start.
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student
ok
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volunteer
Let's say the original area is A_0 = 1, correct?
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student
right
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volunteer
I'll write below your upload.
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volunteer
Now let's find the next area in the sequence, A_1. How much area is removed from A_0 ?
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student
1/9
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volunteer
Correct. 1/9 of the area is removed. So A_1 = ?
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student
0.8889
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volunteer
Yes. We'll write this as 8/9.
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volunteer
Now look at the second carpet, circled in yellow. How much more area is removed from the previous carpet?
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student
8 small squares
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student
the center of
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volunteer
Yes. We have 8 remaining squared (boxed in dark blue), and the center is 1/9 of it.
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volunteer
So the area removed from A_1 is (8/9)*(1/9), correct?
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student
right
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student
8/81
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volunteer
Now let's simplify a bit.
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volunteer
We have 8/9 - 8/9^2. If we add the two fractions, the common denominator is 9^2, correct?
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student
right
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volunteer
So we can simplify A_2 = (8/9)^2.
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volunteer
We have A_0 = (8/9)^0, A_1 = (8/9)^1 and A_2 = (8/9)^2. Do you see the pattern?
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