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1 year ago

I don’t know how to do it, please teach me !!!

Mathematics

1 answer:

cricket20 [7]1 year ago

5 0

As described in z-distribution the answers are given below:

a) The 95% confidence interval estimate for the population mean spending by a customer on visiting salon per month is given as follows: (747, 853).

b) The sampling error at 95% confidence level is of: $35.78.

What is a z-distribution ?

The normal distribution with a mean of 0 and a standard deviation of 1 is referred to as the standard normal distribution (also known as the Z distribution) (the green curves in the plots to the right). It is frequently referred to as the bell curve since the probability density graph resembles a bell.

solution:

The bounds of the confidence interval are given as follows:

In which:

is the sample mean.

z is the critical value.

n is the sample size. is the standard deviation for the population.

The parameters for this problem are given as follows:

Hence the lower bound of the interval is of:

800 - 200 x 1.96/square root of 55 = 747.

The upper bound of the interval is of:

800 + 200 x 1.96/square root of 55 = 853.

The sampling error for a sample size of 120 is calculated as follows:

200 x 1.96/square root of 120 = $35.78.

To learn more about the z-distribution from the given link

brainly.com/question/4079902

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3 years ago

How many extraneous solutions does the equation below have? StartFraction 9 Over n squared 1 EndFraction = StartFraction n 3 Ove

BigorU [14]

The equation has one extraneous solution which is n ≈ 2.38450287.

Given that,

The equation;

$\dfrac{9}{n^2+1} =\dfrac{n+3}{4}$

We have to find,

How many extraneous solutions does the equation?

According to the question,

An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.

To solve the equation cross multiplication process is applied following all the steps given below.

$\rm \dfrac{9}{n^2+1} =\dfrac{n+3}{4}\\\\9 (4) = (n+3) (n^2+1)\\\\36 = n(n^2+1) + 3 (n^2+1)\\\\36 = n^3+ n + 3n^2+3\\\\n^3+ n + 3n^2+3 - 36=0\\\\n^3+ 3n^2+n -33=0\\$

The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y

with 0 and solve for x. The graph of the equation is attached.

n ≈ 2.38450287

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