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3 years ago

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If A can do a piece of work alone in 7 days and B can do it alone in 4 days, how long will it take the two working together to c

omplete the job?

1 answer:

Sergeeva-Olga [200]3 years ago

3 0

3 days give me brainliest plz

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Evelyn drinks 8 glasses of water,which is 56 glasses of water a week.How many glasses of water does she drink in a year?(1 year

Sav [38]

Evelyn drinks 2,912 glasses of water a year.

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3 years ago

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4.2 divided by 8.19 =

Damm [24]

Well, 4.2 divided by 8.19 = 0.51282051282 But if you switch it to 8.19 divided by 4.2 that would equal 1.95

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4 years ago

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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

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

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

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

The margin of error in this interval is:

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

The formula to compute the sample size is:

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

(1)

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.

8 0

3 years ago

What's The Answer??????????????????

Yuri [45]

$f(x) = \dfrac{x}{2} - 2$

$g(x) = 2x^2 + x - 3$

$(f + g)(x) = \dfrac{x}{2} - 2+ 2x^2 + x - 3$

$(f + g)(x) = 2x^2 + \dfrac{3}{2}x - 5$

7 0

3 years ago

Help me with my math please

DIA [1.3K]

Answer:

your answer would come out to be 16 square inches.

7 0

3 years ago