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erastovalidia [21]
3 years ago
14

Carmen is printing an essay that is 15 pages long, and each page contains 52 lines of text. It takes her 5 minutes to print the

essay. At what rate did the essay print?
A.
13.4 lines per minute
B.
161 lines per minute
C.
156 lines per minute
D.
780 lines per minute
Mathematics
1 answer:
Maru [420]3 years ago
8 0
Find number of lines
lines=lines per page times pages
lines=52 times 15
lines=780 lines

780 lines per 5 minutes
156 lines per 1 minute


C is answer
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2 years ago
A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he
UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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}}

For this problem, we have that:

We dont know the true proportion, so we use \pi = 0.5, which is when we are are going to need the largest sample size.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2

n = 600.25

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.02}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2

n = 2401

The minimum sample size is 2401.

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Evaluate the following: 6+ (5 + 1).
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6+(5+1)
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A. Divide by 46,656

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