All categories

3 years ago

Let f(x)=2/3x-6 what is the inverse function?

1 answer:

4 0

Change f(x) to a y and switch places with y and x as below.
$x= \frac{2}{3} y-6$

Solve the equation for y
$x + 6 = \frac{2}{3} y$

Multiply each term by the reciprocal of $\frac{2}{3}$
$\frac{3}{2} (x+6) = y$

Distribute
$\frac{3}{2} x+9=y$

Change y to $f^{-1}(x)$
$f^{-1}(x)= \frac{3}{2} x+9$

You might be interested in

If charged per cut, how much will it cost

melisa1 [442]

Answer:

24 because if your cutting 8 pieces and 4 pieces is 12 then all you have to do is multiply 12 x 2 which equals 24!

Step-by-step explanation:

7 0

3 years ago

The lengths of the sides of a triangle are 3, 3, and the square root 3. Can the triangle be a right triangle?

Readme [11.4K]

First find the decimal equivalent of square root 3: SQRT(3) = 1.732 ( roughly)

If the base and height were each 3, then the hypotenuse would need to be:

3^2 + 3^2 = C^2
9 + 9 = C^2
18 = C^2
C = SQRT(18) = 4.24

This is larger than sqrt(3), so this cannot be a right triangle.

If one leg was 3 and the other leg was sqrt(3) then the hypotenuse would be:
3^2 + 1.73^2 = C^2
9 + 3 = C^2
12 = C^2
C = SQRT(12) = 3.46

This is larger than 3, this cannot be a right triangle.

The answer is b) no.

8 0

4 years ago

Read 2 more answers

Kesha it’s plain to rent a van for a trip to Mt.rainier two of her friends each went to the same type of van from the same car r

Lelechka [254]

Answer:

Step-by-step explanation:

8 0

3 years ago

What is the surface area of the prism represented by the net? Plz zzz hurry I have to turn it in soon

UNO [17]

Answer:

not enough info

Step-by-step explanation:

6 0

3 years ago

A company wants to find out if the average response time to a request differs across its two servers. Say µ1 is the true mean/ex

lorasvet [3.4K]

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

Test statistic t=0.88

The P-value is obtained from a t-table, taking into acount the degrees of freedom (419) and the type of test (two-tailed).

b) A P-value close to 1 means that a sample result have a high probability to be obtained due to chance, given that the null hypothesis is true. It means that there is little evidence in favor of the alternative hypothesis.

c) The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

d) The consequences of the confidence interval containing 0 means that the hypothesis that there is no difference between the response time (d=0) is not a unprobable value for the true difference.

This relate to the previous conclusion as there is not enough evidence to support that there is significant difference between the response time, as the hypothesis that there is no difference is not an unusual value for the true difference.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that there is significant difference in the time response for the two servers.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

$M_d=M_1-M_2=12.5-12.2=0.3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{196}+\dfrac{4^2}{225}}\\\\\\s_{M_d}=\sqrt{0.046+0.071}=\sqrt{0.117}=0.342$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.3-0}{0.342}=\dfrac{0.3}{0.342}=0.88$

The degrees of freedom for this test are:

$df=n_1+n_2-1=196+225-2=419$

This test is a two-tailed test, with 419 degrees of freedom and t=0.88, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>0.88)=0.381$

As the P-value (0.381) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there is significant difference in the time response for the two servers.

<u>Confidence interval </u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

The estimated standard error of the difference is s_Md=0.342.

The critical t-value for a 95% confidence interval and 419 degrees of freedom is t=1.966.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.966 \cdot 0.342=0.672$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 0.3-0.672=-0.372\\\\UL=M_d+t \cdot s_{M_d} = 0.3+0.672=0.972$

The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

7 0

4 years ago