Please answer CORRECTLY !!!!!! Will mark brainliest !!!!!!!!!

What is the inverse of the function f(x) = x – 12?

Luda [366]

Answer:a

Step-by-step explanation:

h(x)=x-12

h(x)=4x-48

h(x)=-1(4x-48)

h(x)=-4x+48

h(x)=48-4x

7 0

3 years ago

How do I do number 4?

Galina-37 [17]

What are you trying to find?

4 0

3 years ago

Read 2 more answers

Find the difference. write the difference in lower term 8 ⅓-5

sergejj [24]

Given data:

The given expression is 8 ⅓-5.

The given expression can be written as,

$\begin{gathered} 8\frac{1}{3}-5=\frac{25}{3}-5 \\ =\frac{25-15}{3} \\ =\frac{10}{3} \end{gathered}$

Thus, the difference of the given expression is 10/3.

5 0

1 year ago

Raleigh went to a gaming arcade for his birthday. His parents gave him $55 to spend. He purchased $14.25 in food and drinks, and

xenn [34]

Answer:

Kindly check explanation

Step-by-step explanation:

Given that :

Total amount to spend = $55

Amount of food and drinks purchased = $14.25

Amount put on gaming card = $(55 - 14.25) = $40.75

Cost per game = $1.25

Number of games (g) he can play

Number of games g:

Amount put on gaming card / cost per game

= $40.75 / $1.25

= 32.6 games

g ≤ 32

6 0

2 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

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

3 years ago