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valentinak56 [21]
2 years ago
15

What ratios are equivalent to 5/9

Mathematics
1 answer:
NNADVOKAT [17]2 years ago
5 0

Answer:

10/18 and 20/36

Step-by-step explanation:

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Angle 2 is equal to angle 8.

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A sample of 1000 college students at NC State University were randomly selected for a survey. Among the survey participants, 102
cupoosta [38]

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The upper endpoint of the 99% confidence interval for population proportion is 0.13.

Step-by-step explanation:

The confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

<u>Given:</u>

<em>n</em> = 1000

\hat p = 0.102

z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58

*Use the standard normal table for the critical value.

Compute the 99% confidence interval for population proportion as follows:

CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.102\pm 2.58\times\sqrt{\frac{0.102(1-0.102)}{1000}}\\=0.102\pm0.0248\\=(0.0772, 0.1268)\\\approx (0.08, 0.13)

Thus, the upper limit of the 99% confidence interval for population proportion is 0.13.

6 0
2 years ago
Consider the statement: For all integers n, if n3 is even then n is even. a. Explicitly write out what you are supposing and wha
Fittoniya [83]

Answer:

It is proved that if n^3 is even the n is even.

Step-by-step explanation:

Given n is any integer.

To show n^3 is even then n is even.

Proving by contrapositive suppose n^3 is odd. Then we need to show n is odd.

Then, letting k is a ny integer,

n^3=2k+1\implies n=(2k+1)^{\frac{1}{3}}

Now since (2k+1) is odd therefore n is odd.

Conversly let n is odd, then,

n=2k+1\implies n^3=(2k+1)^3

since 2k+1 is odd so n^3 is odd.

This proves, if n is even then n^3 is even.

4 0
3 years ago
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