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gladu [14]
3 years ago
12

If there are 35 total votes and 5% of people voted for one thing, how many people would be that 5%?

Mathematics
2 answers:
astraxan [27]3 years ago
5 0

Answer:2

Step-by-step explanation:

1/100 * 35

=35/100 or .35

--> .35 x 5(%)

=1.75

So... There would be 2 people because there's no such thing as 1.75 of a person.

hram777 [196]3 years ago
3 0

Answer:

2

Step-by-step explanation:

5% = 0.05

0.05 * 35 = 1.75

So, roughly 2 people voted for it.

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Suppose a sample of 972972 tenth graders is drawn. Of the students sampled, 700700 read above the eighth grade level. Using the
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Answer:

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

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

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For this problem, we have that:

972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so n = 972, \pi = \frac{272}{972} = 0.28

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.259

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.301

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

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