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jek_recluse [69]
3 years ago
13

Show to different ways to use compensation to find the sum. Then solve. 37+16+5

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer:

The answer to the question is 58

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Line AOB is a straight line.
alexandr1967 [171]

Answer:

It is not 1,2,3

Step-by-step explanation:

4 0
3 years ago
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A roofer calculates his bid price using the formula P = 1.85s + 4.2f, where s is the area of the roof in square feet and f is th
solmaris [256]

Answer: 1,811 square foot

Step-by-step explanation:

Hi, to answer this question we have to solve the equation given, by substituting P = 4,148 and f=190 in the equation.

P = 1.85s + 4.2f

4,148 = 1.85s +4.2 (190)

Solving for s:

4,148 = 1.85s +798

4,148-798 =1.85s

3,350 = 1.85s

3,350/1.85 =s

s = 1,810.81 = 1,811 square foot  (rounded)

Feel free to ask for more if needed or if you did not understand something.

8 0
3 years ago
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Unit rate 3 inches of rain in 6 hours
Neporo4naja [7]

Answer:

It is either 1.5 inches per hour or 1 inch in 2/3 hours. I also think 1:3

Step-by-step explanation:

Three inches in six hours is equivalent to 0.5 inches per hour.

4 0
2 years ago
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Which statement is true for the equation 5n − 4 = 5n − 3?
Andrew [12]
It has one solution that is the answer
5 0
3 years ago
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A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali
Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

W = Wednesday

T = Thursday

F = Friday

S = Saturday

Their corresponding probabilities are

P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1

Since Y = number of days beyond Wednesday that it takes for both magazines to arrive(so possible Y values are 0, 1, 2 or 3)

The possible number of outcomes are therefore 4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)

The values associated for each of the outcomes are as follows:

Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3

The probability mass function of Y is,

P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19

7 0
3 years ago
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