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Ilya [14]
3 years ago
14

Chandresh is helping his father paint the fence in their back yard. The fence is represented by the shaded part of the diagram b

elow. They will paint only the surfaces of the fence that face the yard. They will apply 2 coats of paint. One can of paint covers 500 square feet. What is the area Chandresh and his father will paint, and how many cans of paint must they buy to complete the job?
A. The total area painted is 432 ; they must buy one can of paint.

B. The total area painted is 864 ; they must buy two cans of paint.

C. The total area painted is 1728 ; they must buy four cans of paint.

D. The total area painted is 2080 ; they must buy five cans of paint.
Mathematics
1 answer:
Lyrx [107]3 years ago
7 0

B. The total area painted is 864; they must buy two cans of paint.

Step-by-step explanation:

Step 1:

A rectangle's area can be calculated by multiplying its length and its width. The wall is made up of 5 different types of rectangular walls. All walls are 8 feet tall but the length varies.

Step 2:

The area of the 20 feet long wall = (length)(width)= (20)(8) =160,

The area of the 10 feet long wall = (length)(width)= (10)(8) =80,

The area of the 5 feet long wall = (length)(width)= (5)(8) =40,

The area of the 4 feet long wall = (length)(width)= (4)(8) =32,

The area of the 15 feet long wall = (length)(width)= (15)(8) =120.

The area of all the walls = 432 square feet.

Since there are two sides for every wall, total area = 432(2)=864 square feet.

Step 3:

If one paint can covers 500 square feet,

the number of cans required to paint 864 square feet = \frac{864}{500} = 1.728 cans.

so 2 paint cans are needed to paint 864 square feet which is option B.

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B 0.17 is closest to 0
5 0
3 years ago
Quick guys I need help
Minchanka [31]

Answer:

No

Step-by-step explanation:

You are given three pairs of angle ≅. You can not prove Δ≅Δ by AAA.

you do not have a side that you know are ≅.

The ASA means Angle  - Side - Angles

6 0
2 years ago
Assume that the probability of a defective computer component is 0.02. Components arerandomly selected. Find the probability tha
solmaris [256]

Answer:

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

Step-by-step explanation:

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.

First six not defective, each with 0.98 probability.

7th defective, with 0.02 probability. So

p = (0.98)^6*0.02 = 0.0177

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

Find the expected number and variance of the number of components tested before a defective component is found.

Inverse binomial distribution, with p = 0.02

Expected number before 1 defective(n = 1). So

E = \frac{n}{p} = \frac{1}{0.02} = 50

Variance is:

V = \frac{np}{(1-p)^2} = \frac{0.02}{(1-0.02)^2} = 0.0208

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

5 0
3 years ago
Solve for x using the given diagram
blagie [28]

Answer:

<h2>x = 1</h2>

Step-by-step explanation:

Look a t the picture.

The triangles on the picture are similar.

Therefore the sides are in proportion:

\dfrac{8x}{10}=\dfrac{4}{5x}              <em>cross multiply</em>

(8x)(5x)=(4)(10)

40x^2=40           <em>divide both sides by 40</em>

x^2=1\to x=\sqrt1\\\\x=1

7 0
3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

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:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

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

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

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

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
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