Solve each of the following equations for the variable "a" 1. a^2+ b^2 = c^2

Green Frog convenience store supplies two types of protein bars. One protein bar has 320 calories in the 2-ounce size. How many

Zepler [3.9K]

Answer:

I think it is C) 480

Step-by-step explanation:

320÷2=160

160×3=480

Can I please have Brainliest?

6 0

3 years ago

Find the midpoint of A and B where A has coordinates (2,7) and B has coordinates (6,3).

Veronika [31]

Given that the coordinates of the point A is (2,7) and the coordinates of the point B is (6,3)

We need to determine the midpoint of A and B

Midpoint of A and B:

The midpoint of A and B can be determined using the formula,

$Midpoint =(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substituting the points (2,7) and (6,3) in the above formula, we get;

$Midpoint =(\frac{2+6}{2}, \frac{7+3}{2})$

Adding the numerator, we have;

$Midpoint =(\frac{8}{2}, \frac{10}{2})$

Dividing the terms, we get;

$Midpoint =(4, 5)$

Thus, the midpoint of the points A and B is (4,5)

7 0

3 years ago

A recipe for dessert calls for 2/4 cup of powdered sugar and 1/6 cup of brown sugar . What is the total amount of sugar needed f

Neporo4naja [7]

They used 2/3 cup of sugar mark brainest

4 0

4 years ago

Read 2 more answers

The graph below shows the solution set of which inequality?

GrogVix [38]

Answer:

The answer is the option D

$x^{2} > 4$

Step-by-step explanation:

we know that

The inequality

$x^{2} > 4$

has two solutions

First solution

$x > 2$

The solution is the interval-------> (2,∞)

All real numbers greater than $2$

Second solution

$-x > 2$ ------> $x < -2$ -

The solution is the interval-------> (-∞,-2)

All real numbers less than $-2$

using a graphing tool

see the attached figure

3 0

3 years ago

Read 2 more answers

Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h

Usimov [2.4K]

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 200, \sigma = \sqrt{400} = 20$

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

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

$Z = \frac{150 - 200}{20}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So $Z = 1.28$

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

$1.28 = \frac{X - 200}{20}$

$X - 200 = 20*1.28$

$X = 225.6$

225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

6 0

4 years ago