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Vesnalui [34]
3 years ago
8

McKenzie has enough paint to paint 108 square feet. She wants to paint her garage door, which has a height of 12 feet. The garag

e door is the shape of a rectangle. If MAcKenzie has just enough paint to cover the garage door, what is the width of the door?
Mathematics
2 answers:
natta225 [31]3 years ago
7 0

Answer:

The width of the door is 9 feet

Step-by-step explanation:

The area of the garage door is

Area= height x width

In this case,  we are looking for the width, and we have the data of the other 2 elements

Area is 108,  because she has enough to paint the area of the garage door

Height=12

Width=Area/height

Width=108/12

Width= 9 feet

almond37 [142]3 years ago
4 0
108/12=9

(You have to divide because you are seeing how many times 12 will go into 108 square feet)

The width of the door is 9 feet!

You can check your answer by multiplying 12 feet times 9 feet.
Your answer will be 108 ft squared! :)

So, the width of the door is 9 feet.

Hope I helped! :)



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The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.
devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

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b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0
3 years ago
Can someone help me
Colt1911 [192]
The answer is B. (-1,3) hope this helped
3 0
2 years ago
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