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3 years ago

10

Don has two jobs for job 1 he earns 7.55 an hour. For job 2 he earns 8.45 an hour last week he worked at the first job for 10 ho

urs and the second job for 15 hours. What were his average earnings per hour

1 answer:

Crazy boy [7]3 years ago

8 0

Answer:

for the first job is $75.05 and for the second job is $126.75

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Olin [163]

Answer:

A

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3 years ago

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

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Answer:

(1) 97

(2) 385

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Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

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The margin of error in this interval is:

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$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

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Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

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andreev551 [17]

Answer:

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Step-by-step explanation:

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