5. Let f be a continuous and differentiable function for all real numbers. The only critical points for f are located

Set the equations equal to each other and solve for the variable x. then use the answer from x and plug it in to the equations and you should get an answer

3x=63

63÷3=x

21=x

x=21

that is the value of x

73 is the answer to the top equation and the bottom equation in the triangle. Meaning the equations are equal to each other when you have 21 plugged in for the variable x.
make sure to put your label if it asks for it. it means the book or the teacher. go by what the teacher says if it is different from what the book says

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2 years ago

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The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check

Verdich [7]

Answer:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 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)

$\sigma =150$ represent the population standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

$\bar X = \frac{740+920}{2} = 830$

and the margin of error is given by :

$ME = \frac{920-740}{2}= 90$

The margin of error is given by this formula:

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

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

5 0

2 years ago