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2 years ago

7

Kyle and Cooper are driving. Kyle drives 5 mph faster than Cooper. The same amount of time of Kyle drives 87.1 miles Cooper driv

es 80.6 miles how quickly are each driver driving

1 answer:

alexandr1967 [171]2 years ago

4 0

Answer:

Kyle, 67 mph
Cooper, 62 mph

Step-by-step explanation:

Their difference in distance is 87.1 -80.6 = 6.5 miles. Since their difference in speed is 5 mph, it took them (6.5 mi)/(5 mi/h) = 1.3 hours to achieve that separation.

Kyle's speed is 87.1 mi/(1.3 h) = 67 mi/h.

Cooper's speed is 80.6 mi/(1.3 h) = 62 mi/h. (5 mi/h slower)

Kyle is driving 67 mph; Cooper is driving 62 mph.

_____

<em>Alternate solution</em>

The ratio of their speeds is 87.1/80.6 = 67/62. The difference in ratio units happens to equal their difference in miles per hour, so their speeds must be 67 and 62 miles per hour.

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frutty [35]

Answer:

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r =0.934

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Step-by-step explanation:

Previous concepts

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And in order to calculate the correlation coefficient we can use this formula:

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The correlation coeffcient for this case was provided:

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1 year ago

Maria has a cube-shaped box that measures 9 inches along each edge. Can she fit 1,000 1-cubic-inch cubes inside the box?

morpeh [17]

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2 years ago

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) <u>Find the pounds of road salt </u>

$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

therefore

<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

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2 years ago