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Anna007 [38]
2 years ago
12

Jesper can run 6 miles in 45 minutes. He is thinking about running in two different races, a 9-mile race and a 15-mile race. At

his current rate, how many more minutes will it take Jesper to complete the 15-mile race than the 9-mile race?
Mathematics
1 answer:
marin [14]2 years ago
4 0

Find minutes per mile by dividing time by distance:

45 minutes / 6 miles = 7.5 minutes per mile

Now multiply his speed by each distance:

7.5 x 9 = 67.5 minutes

7.5 x 15 = 112.5 minutes

Now subtract the two:

112.5 - 67.5 = 45 minutes more

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How does making a frequency table or line plot help you see which value occurs most often in a data set?
klemol [59]

Answer:

A frequency table and a line plot show which value occurs most often. this is done by labeling every value. The person can clearly see the organized layedout numbers in a frequency table or line plot.

Step-by-step explanation:

If you have any questions feel free to ask in the comments.

4 0
3 years ago
How to put 63 over 84 in simplest form
Tpy6a [65]
63/84 divided by 21/21
=3/4
 or
63 divided by 84=0.75 which is the same thing as 3/4
8 0
3 years ago
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IF A SUBSCRIPTION IS $499 PLUS 8% TAX FOR 30 DAYS, BUT IS BEING PRORATED FOR 7 DAYS, PLUS THERE IS A $10 OFF COUPON, WHAT'S THE
ELEN [110]

Answer:

Your final answer is $528.92

Step-by-step explanation:

499 + 8\% = 538.92

538.92 - 10 = 528.92

5 0
3 years ago
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Is x = 1 a solution of the equation 2 – 8x = –6?
borishaifa [10]

Let's verify by substituting 1 to x.

2 - 8x = -6

2 - 8(1) = -6

2 - 8 = -6

-6 = -6

It is equal, therefore x = 1 is the solution.

4 0
3 years ago
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Comparing Leaking Pool Problem
Blababa [14]

Answer:

<em>Pool 1 leaks faster than pool 2.</em>

Step-by-step explanation:

<u>Rates of change</u>

The rate of change (ROC) is a measure that compares two quantities, usually to know how fast one variable changes in time.

We are given two rates of change for two pools that are leaking. The first one loses 2/3 gallon in 15 minutes, and the other loses 3/4 gallon in 20 minutes.

To compare them, we are required to express time in hours. Recall one hour has 60 minutes, or equivalently, one minute has 1/60 hours. Converting both times, we have:

15 minutes = 15/60 = 1/4 hours

20 minutes = 20/60 = 1/3 hours

Now compute both rates of change:

Pool 1:

\displaystyle ROC_1=\frac{2/3}{1/4}=\frac{8}{3}\approx 2.67\ gal/h

Pool 2:

\displaystyle ROC_2=\frac{3/4}{1/3}=\frac{9}{4}= 2.25\ gal/h

Comparing both ratios, it's clear pool 1 leaks faster than pool 2.

8 0
3 years ago
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