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2 years ago

Please help asap! Will give brainliest!

Mathematics

1 answer:

love history [14]2 years ago

7 0

Answer:

A. 38

Step-by-step explanation:

F(6)=7(6)-4

Plug the 6 in for x

7*6=42

42-4 = 38

& thats it....

You might be interested in

A professor at a university wants to estimate the average number of hours of sleep students get during exam week. The professor

kotegsom [21]

Answer:

$n=(\frac{1.64(6.995)}{1.939})^2 =35.003 \approx 36$

So the answer for this case would be n=36 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma =6.995$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =1.939 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

If we replace we have this:

$n=(\frac{1.64(6.995)}{1.939})^2 =35.003 \approx 36$

So the answer for this case would be n=36 rounded up to the nearest integer

4 0

3 years ago

If sine ⁡theta equals negative 8/17 and the terminal side of theta lies in quadrant IV, find cosine ⁡ theta .

Lelechka [254]

$\sin^2\theta+\cos^2\theta=1\implies \cos\theta=\pm\sqrt{1-\sin^2\theta}$

Since $\sin\theta=-\dfrac8{17}$ , it follows that

$\cos\theta=\pm\sqrt{1-\dfrac{64}{289}}=\pm\sqrt{\dfrac{225}{289}}=\pm\dfrac{15}{17}$

where the sign depends on the location of the terminal side of the angle. Since the terminal side lies in the fourth quadrant, that means the cosine of the angle is positive, so

$\cos\theta=\dfrac{15}{17}$