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irina1246 [14]
2 years ago
9

A homeowner uses a $10 bill to pay a neighbor for cutting his lawn. The

Mathematics
1 answer:
timama [110]2 years ago
7 0

Answer:

c

Step-by-step explanation:

It has a constant value.

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Paul joins a gym that has an initial membership fee and a monthly cost. He pays a total of $295 after three months, and $495 aft
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Answer:

a

Step-by-step explanation:

6 0
1 year ago
Javier writes an integer on a piece of paper. The absolute value of his number is less than 4. Which is true of the possible val
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<span>The possible values are less than 4 but greater than –4.</span>
7 0
2 years ago
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Sameera predicted that she would sell 38 blankets, but she actually sold 28 blankets. Which expression would find the percent er
antoniya [11.8K]

Solution: We are given:

Predicted Sales by Sameera =38

Actual Sales by Sameera =28

Now to find the Percent error, we have to use the below formula:

Percent-Error= \frac{|Predicted-value - Actual-value|}{Actual-value} \times 100 \%

                       =\frac{|38-28|}{28} \times 100 \%

                       =\frac{10}{28}\times 100 \%

                       =35.71 \%

Therefore, the percent error is 35.71 \%      

8 0
2 years ago
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There are two jobs you can apply for. the first job pays $22,000 the first year, with raises of $4,000 each year thereafter. the
jasenka [17]
We let the number of years that the two jobs will have the same payment be denoted as t. Equating the wages of these two jobs after t - 1 years will give us an equation of,
                          22,000 + 4000(t -1) = 26,000 + 2000(t - 1)
The value of t from the generated equation is 3. Therefore, after 3 years the jobs will be paying the same wages.
7 0
2 years ago
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A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12
marissa [1.9K]

Answer:

Step-by-step explanation:

Hello!

You have two random samples obtained from two different normal populations.

Sample 1

n₁= 15

X[bar]₁= 350

S₁= 12

Sample 2

n₂= 17

X[bar]₂= 342

S₂= 15

At α: 0.05 you need to obtain the p-value for testing variances for a one tailed test.

If the statistic hypotheses are:

H₀: σ₁² ≥ σ₂²

H₁: σ₁² < σ₂²

The statistic to test the variances ratio is the Stenecor's-F test. F_{H_0}=(\frac{S^2_1}{Sigma^2_1}) * (\frac{S^2_2}{Sigma^2_2} )~F_{n_1-1;n_2-1}

F_{H_0}= \frac{(12)^2}{(15)^2} * 1= 0.64

The p-value is:

P(F_{14;16}≤0.64)= 0.02

I hope it helps!

7 0
2 years ago
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