PLS HELP WILL MARK BRAINLIEST!1

13+26 [(2.8 x 5) divided by 7]<br> HELPPP

Katen [24]

65
1. you work in the parentheses and multiply 2.8 by 5 which is 14
2. you divide 14 by 7 which is 2
3. you follow PEMDAS and multiply 26 by what is in the parentheses (2) which is 52
4. you add 13 with 52 which is 65
hope this helped!!

4 0

2 years ago

What is the inverse of f(x)=2^x

GuDViN [60]

Answer:

Step-by-step explanation:

$f(x)=2^x$

$y = {2}^{x}$

$x = {2}^{y}$

${2}^{y} = x$

$ln( {2}^{y} )=ln(x)$

$yln(2)=ln(x)$

<h3> $\frac{yln(2)}{ln(2)} = \frac{ln(x)}{ln(2)}$ </h3><h3> $y = \frac{ln(x)}{ln(2)}$ </h3><h3> ${f}^{ - 1} (x) = \frac{ln(x)}{ln(2)}$ </h3><h3>Hope it is helpful...</h3>

3 0

3 years ago

Can somebody please answer this asap

erastovalidia [21]

Answer:

you have to add all of them up it will get you some of your awner

5 0

2 years ago

Read 2 more answers

How did you identify the line of sight, angle of elevation and angle of depression?

Arturiano [62]

Answer:

The term angle of elevation denotes the angle from the horizontal upward to an object. An observer's line of sight would be above the horizontal. The term angle of depression denotes the angle from the horizontal downward to an object. An observer's line of sight would be below the horizontal.

Step-by-step explanation:

ヾ(＾-＾)ノ

7 0

1 year ago

A recent study found that the average length of caterpillars was 2.8 centimeters with a

pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 2.8, \sigma = 0.7$ .

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

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

$Z = \frac{4 - 2.8}{0.7}$

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0

1 year ago