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Harrizon [31]
3 years ago
12

Can one of y’all help me with this pls

Mathematics
2 answers:
andrew11 [14]3 years ago
8 0

Answer:

it’s A

Step-by-step explanation:

youre supposed to find the total earnings. to do that, you need to multiply 9.25 by the hours.

Mazyrski [523]3 years ago
3 0

Answer:

your answer is A

Step-by-step explanation:

d is the total

h is the hours

9.25 is the money

So your need to multiply hours with money

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Determine the intercepts of the line.
Lemur [1.5K]

Hello there!

The x-intercept and y-intercept of the line shown on the graph are...

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Please answer fast this is a test question
pishuonlain [190]

Answer:

4

Step-by-step explanation:

We can actually make the two equal each other like so:

3d-4=12d/6

Multiply both sides by 6:

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18d=12d+24

Subtract 12d from both sides:

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3 years ago
Hank estimated the width of the door to his classroom in feet. What is a reasonable estimate?
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4 years ago
Assume that the heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.2 inches. The
anygoal [31]

Answer:

The number of women expected to meet the height requirements is 149 women.

Step-by-step explanation:

To solve the question, we note that

The population mean = 63.5 inches

The population standard deviation = 2.2

The range of required height = 58 to 80

We are required to find the proportion of women between the height range

The z value for 58 is given by

z = \frac{x- \mu}{\sigma}

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x = Required statistic

μ = Population mean

σ = Standard deviation

= (58 - 63.5)/2.2 = -2.5 and

From the z-score table, we have P(z= -2.5) = 0.0062

The z value for 80 is given by

= (80 - 63.5)/2.2 = 7.5

From the z-score table, we have P(z= 7.5) ≈ 1

Therefore the proportion of women that satisfy the given range is between the two given probabilities, that is P(z= 7.5) - P(z= -2.5) or

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3 0
3 years ago
Deonadre runs 17 1/2 times around the reservoir in one week. What is the unit rate per day?
Elena-2011 [213]

Answer:

2 7/5 or 2.5 times per day

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2 7/5 times per day

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