All categories

2 years ago

I need help!!! I’ll mark you brainliest This is timed!!

Mathematics

1 answer:

devlian [24]2 years ago

3 0

Answer:

Using the inverse composition rule you know that if (fog)=x and (gof)=x then the functions are inverses.

( g o f ) = ( 7 x + 1 ) − 1 7 this will result in (gof)=x ( f o g) = 7 x − 1 7 +1 = x − 1 + 1 =x

graphically you can check this if it seems that the functions are reflected over the line x=y.

Step-by-step explanation:

You might be interested in

-\frac{4\left(x+3\right)}{5}=4x-12

Keith_Richards [23]

Answer:

huh

Step-by-step explanation:

6 0

3 years ago

Two groups of students were asked how many pets they had. The table below shows the numbers for each group:

Charra [1.4K]

Whats the question tho

6 0

3 years ago

What's (7 x 100) + (4 x /100) + (8 x 1/1000) in standard form?

IceJOKER [234]

Answer:

(7*100) can be written as 7*10^2

4 * / 100 i don’t know

if it was 4 * 100 then it’s 4*10^2

if it was divided then it’s 4*10^-2

8* 1/1000 = 8/1000 = 8*10^-3

add together to get 1.100008 * 10^3 (if it was 4*100)

or 7.00048 * 10^2 ( if it was 4/100)

7 0

3 years ago

what is the hypothesis and conclusion in the sentence you will receive the trophy if you win the championship

lana [24]

The hypothesis is if you win the championship the conclusion is you will get the trophy

3 0

3 years ago

Find the P-value for a test of the claim that more than 60% of the people following a particular diet will experience increased

lina2011 [118]

Answer:

The p-value of the test is 0.0207.

Step-by-step explanation:

Test of the claim that more than 60% of the people following a particular diet will experience increased energy (H1: p > 0.6).

The null hypothesis is:

$H_0: p = 0.6$

The alternate hypothesis is:

$H_1: p > 0.6$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.6 is tested at the null hypothesis:

This means that $\mu = 0.6, \sigma = \sqrt{0.6*0.4}$

Of 100 randomly selected subjects who followed the diet, 70 noticed an increase in their energy level.

This means that $n = 100, X = \frac{70}{100} = 0.7$

Test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.7 - 0.6}{\frac{\\sqrt{0.6*0.4}}{\sqrt{100}}}$

$z = 2.04$

P-value of the test:

The p-value of the test is the probability of finding a sample proportion of at least 0.7, which is 1 subtracted by the p-value of z = 2.04.

Looking at the z-table, z = 2.04 has a p-value of 0.9793

1 - 0.9793 = 0.0207

The p-value of the test is 0.0207.

4 0

2 years ago