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3 years ago

9 squared ? 11 ? 8 ? 4 ? 1 = 60 The ? Can be +,-,x or d division

1 answer:

5 0

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Arada [10]

Answer: C. “BEC”

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2 years ago

Maya spent her allowance on playing an arcade game a few times and riding the Ferris wheel more than once.

NISA [10]

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To write a two-variable equation, I would first need to know how much Maya’s allowance was. Then, I would need the cost of playing the arcade game and of riding the Ferris wheel. I could let the equation be cost of playing the arcade games plus cost of riding the Ferris wheel equals the total allowance. My variables would represent the number of times Maya played the arcade game and the number of times she rode the Ferris wheel. With this equation I could solve for how many times she rode the Ferris wheel given the number of times she played the arcade game.

Step-by-step explanation:

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3 years ago

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University Theater sold 479 tickets for a play. Tickets cost $22 per adult and $14 per senior citizen. If total receipts were $7

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Answer:

341 senior tickets were sold

Step-by-step explanation:

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3 years ago

A rectangle has vertices P(-7, -5) , Q(-7, 8), R(3, 8), and S(3, -5). What is the length, in units, of RS¯¯¯¯¯¯¯? 3 units

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3 years ago

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

butalik [34]

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 $\pi = 0.12, n = 5615$ .

We have to find M.

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

$M = 2.81\sqrt{\frac{0.12*0.88}{5615}}$

$M = 0.012$

7 0

3 years ago