What is a array of 50

Find the equation of the inverse.<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%7B2%7D%5E%7Bx%20%2B%205%7D%20%20-%206" id="TexF

Contact [7]

Find the equation of the inverse.

$y = {2}^{x + 5} - 6$

we have

$y=\mleft\{2\mright\}^{\mleft\{x+5\mright\}}-6$

step 1

Exchange the variables

x for y and y for x

$x=\{2\}^{\{y+5\}}-6$

step 2

Isolate the variable y

$(x+6)=2^{(y+5)}$

apply log both sides

$\begin{gathered} \log (x+6)=(y+5)\cdot\log (2) \\ y+5=\frac{\log (x+6)}{\log (2)} \\ \\ y=\frac{\log(x+6)}{\log(2)}-5 \end{gathered}$

step 3

Let

$f^{(-1)}(x)=y$

therefore

the inverse function is

$f^{(-1)}(x)=\frac{\log(x+6)}{\log(2)}-5$

7 0

1 year ago

What’s the error and how can I correct it number 24.

Aloiza [94]

The error is that 9 shouldnt be written as nine it should be written as 3x3 or 3^2. So the answer should be 2^3 x 3^2 because 9 can be divided into even more prime factors of 72.

4 0

3 years ago

Read 2 more answers

Hello, I have a Pi Day Question for you, it's pretty simple.

stira [4]

The value of pi is 3.14

7 0

3 years ago

Can someone awsner this for me honestly I’m sorry but I really need these awsner but I need the answers for these 3 to

zalisa [80]

1. 50 + 20x = 240

2. 85% so more than 75%

3. 25.9% so about 26%

Hope this helps and please consider marking branliest if this helped you

5 0

3 years ago

Read 2 more answers

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

$Z = \frac{X - \mu}{\sigma}$

$1.75 = \frac{X - 3.6}{0.4}$

$X - 3.6 = 0.4*1.75$

$X = 4.3$

They last at least 4.3 minutes

7 0

3 years ago