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3 years ago

How many solutions does this equation have? –10q = –10q − 7

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

None (no solutions)

Step-by-step explanation:

-10q=-10q-7 (add 10q to both sides to get the costant 7 by itself)

-10q+10q=-10q+10q-7

0=-7 ( zero will never equal -7 hence there are no solutions to this equation)

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3 years ago

Which represents the solution(s) of the system of equations, y = x^2 - 6x + 8 and y = -x + 4? Determine the solution set by grap

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Answer:

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Step-by-step explanation:

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3 years ago

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

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Answer:

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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

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3 years ago