Please help quick on quiz

Mathematics

1 answer:

emmainna [20.7K]3 years ago

3 0

Answer:

D. $y=\frac{1}{x-5} +2$

Step-by-step explanation:

Graph

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Given sec(theta)= 5... find cot(90-theta)

Mumz [18]

$\sec(\theta)=\frac{1}{\cos(\theta)}
\\
\\\cos(\theta)=\frac{1}{\sec(\theta)}=\frac{1}{5}
\\
\\ \sin^2\theta+\cos^2\theta=1
\\
\\\sin\theta= \sqrt{1-\cos^2\theta}= \sqrt{1-( \frac{1}{5})^2 } = \sqrt{1- \frac{1}{25} } = \sqrt{ \frac{25}{25} -\frac{1}{25} }=\frac { \sqrt{24}}{5} 
\\
\\ \cot(90^o-\theta)=\tan{\theta}= \frac{\sin\theta}{\cos\theta} = \frac{\frac { \sqrt{24}}{5} }{ \frac{1}{5} } = \sqrt{24} = \sqrt{4\times6}=2 \sqrt{6} 
\\
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3 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of