Algebra 2 Word Problem: Set up using 3 variables & 3 equations:

1 answer:

7 0

A) The answer is
x + y + z = 24
3x + 2y + z = 53
x = y + z

x - the number of swimmers in the first place
y - the number of swimmers in the second place
z - the number of swimmers in the third place

1. The e-mail states that 24 individuals placed: x + y + z = 24
2. First place earned 3 points, second place earned 2 points, and third place earned 1 point, earning a combined total of 53 points: 3x + 2y + z = 53
3. There were as many first-place finishers as second and third-place finishers combined: x = y + z

The system of three equations is:
x + y + z = 24
3x + 2y + z = 53
x = y + z

B) The answer is
12 swimmers in the first place
5 swimmers in the second place
7 swimmers in the third place

(i) x + y + z = 24
(ii) 3x + 2y + z = 53
(iii) x = y + z
______
Substitute y + z from the third equation into the first one:
(i) x + y + z = 24
(iii) x = y + z
______
x + x = 24
2x = 24
x = 24 / 2
x = 12
_____

x = y + z
x = 12
y + z = 12
z = 12 - y

3x + 2y + z = 53
3 * 12 + 2y + (12 - y) = 53
36 + 2y + 12 - y = 53
2y - y + 36 + 12 = 53
y + 48 = 53
y = 53 - 48
y = 5

z = 12 - y
z = 12 - 5
z = 7

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An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

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Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .