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2 years ago

Bit confused need help

Mathematics

1 answer:

Triss [41]2 years ago

6 0

Answer:

Step-by-step explanation:

I think they want you to reffer to each column

Institution column holds -descriptive data

grants column holds -quantitative data

aplications columns holds-quatitative data

sucess rate holds -qualitative data

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3 years ago

Solve for x. Round to the nearest tenth of a degree, if necessary.

MissTica

Answer:

x ≈ 54.8°

Step-by-step explanation:

Using the tangent ratio in the right triangle

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2 years ago

A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop

Katyanochek1 [597]

Answer: $(0.367,\ 0.473)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

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

, where $\hat{p}$ is the sample proportion, n is the sample size , $z_{\alpha/2}$ is the critical z-value.

Given : Significance level : $\alpha:1-0.99=0.01$

Sample size : n= 85

Critical value : $z_{\alpha/2}=2.576$

Sample proportion: $\hat{p}=\dfrac{36}{85}\approx0.42$

Now, the 99% confidence level will be :

$\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)$

Hence, the 99% confidence interval estimate of the true proportion of families who own at least one DVD player is $(0.367,\ 0.473)$

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2 years ago