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-BARSIC- [3]
3 years ago
12

Joe's Painting charges $100 plus $20 per hour to paint the exterior of your home. Steve's Painting charges $120 plus $15 per hou

r to paint the exterior of your home. How many hours do Joe and Steve have to work for their models to be equal to each other? Which of the following systems is correct for this situation if "x" = hours worked and "y" = total income.
Mathematics
1 answer:
Tanzania [10]3 years ago
5 0

Joe's Painting: 20x + 100 = y

Steve's Painting: 15x + 120 = y

x = hours worked

y = total income

We can find when the two equations intersect by making them equal to each other. That means we put an equal sign in the middle. So, it would look something like this:

20x + 100 = 15x + 120

First, we have to move the 100 by subtracting it from both sides.

20x = 15x + 120 - (100)

20x = 15x + 20

Then, we need to move the 15x by subtracting it from both sides.

20 - (15x) = 20

5x = 20

Lastly, we need to divide 5 from both sides.

5x = 20/5

x = 4

Therefore, Joe and Steve would have to work for 4 hours in order for their models to be equal to each other.

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Answer:

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Step-by-step explanation:

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2 years ago
Do the set of points in the figure represents a function? Yes No
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3 years ago
Read 2 more answers
What is the least common denominator for the fractions 9/14and 2\21
Diano4ka-milaya [45]
Hello : 
14 = 2×7
21 = 3×7
<span> the least common denominator for the fractions 9/14and 2\21  is : 6×7=42</span>
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3 years ago
How do you find surface area
Oksana_A [137]

Answer:

look at my explanation

Step-by-step explanation:

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2 years ago
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
2 years ago
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