Ask question

Ask question

All categories

3 years ago

11

the length of a rectangle is 4 units less than 3 times the width. The perimeter is 22 units more than twice of the width. Find t

he Length and the width of the rectangle.

1 answer:

mr_godi [17]3 years ago

8 0

L=Length W=Width P=Perimeter

L=3w-4
P=2w+22
P=2L+2w=2w+22 Answer: W=5
2L=22 L=11
L=11
11=3w-4
3w=15
W=5

You might be interested in

Harry purchased 16 hinges for his kitchen project. They were each on sale for $3 off the regular price.

Zielflug [23.3K]

THE ANSWER IS 8 DOLLARS BCUZ 80/16 EQUALS 5, AND 5+3 EQUALS 8!!!!

3 0

3 years ago

Read 2 more answers

Helpppp what can i use to submit this

-Dominant- [34]

-15 and -2 don’t make +17

5 0

3 years ago

Calculate 41% 0f 51kg

vlabodo [156]

20.91

51 x 0.41 = 20.91

5 0

3 years ago

Read 2 more answers

Question 2 (3 points)

Sauron [17]

The answer is D! This shape is a pentagon and has 3 180° angles. 3x180 is 540.

6 0

2 years ago

Social Sciences Alcohol Abstinence The Harvard School of Public Health completed a study on alcohol consumption on college campu

Scilla [17]

Answer:

a) There is a 6.69% probability that a randomly selected female student abstains from alcohol.

b) If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

Step-by-step explanation:

This is a probability problem:

We have these following probabilities:

-20.7% of a woman attending an all-women college abstaining from alcohol.

-6% of a woman attending a coeducational college abstaining from alcohol.

-4.7% of a woman attending an all-women college

- 100%-4.7% = 95.3% of a woman attending a coeducational college.

(a) What is the probability that a randomly selected female student abstains from alcohol?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability of a woman attending an all-women college being chosen and abstaining from alcohol. There is a 0.047 probability of a woman attending an all-women college being chosen and a 0.207 probability that she abstain from alcohol. So:

$P_{1} = 0.047*0.207 = 0.009729$

$P_{2}$ is the probability of a woman attending a coeducational college being chosen and abstaining from alcohol. There is a 0.953 probability of a woman attending a coeducational college being chosen and a 0.06 probability that she abstain from alcohol. So:

$P_{2} = 0.953*0.06 = 0.05718$

So, the probability of a randomly selected female student abstaining from alcohol is:

$P = P_{1} + P_{2} = 0.009729 + 0.05718 = 0.0669$

There is a 6.69% probability that a randomly selected female student abstains from alcohol.

(b) If a randomly selected female student abstains from alcohol, what is the probability she attends a coedücational colege?

<em>This can be formulated as the following problem:</em>

<em>What is the probability of B happening, knowing that A has happened.</em>

Here:

<em>What is the probability of a woman attending a coeducational college, knowing that she abstains from alcohol.</em>

It can be calculated by the following formula:

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

We have the following probabilities:

$P(B)$ is the probability of a woman from a coeducational college being chosen. So $P(B) = 0.953$

$P(A/B)$ is the probability of a woman abstaining from alcohol, given that she attends a coeducational college. So $P(A/B) = 0.06$

$P(A)$ is the probability of a woman abstaining from alcohol. From a), $P(A) = 0.0669$

So, the probability that a randomly selected female student attends a coeducational college, given that she abstains from alcohol is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.953)*(0.06)}{(0.0669)} = 0.8287$

If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

4 0

3 years ago