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vladimir2022 [97]

1 year ago

5

A mouse traveled a total distance of StartFraction 3 Over 24 EndFraction of a mile in a maze over the past 3 hours. The mouse tr

aveled the same distance each hour. To determine the distance that the mouse traveled each hour, Matt performed the calculations below.
StartFraction 3 Over 24 EndFraction divided by 3 = StartFraction 3 Over (24 divided by 3) EndFraction = StartFraction 3 Over 8 EndFraction
He concluded that the mouse traveled StartFraction 3 Over 8 EndFraction of a mile each hour. What is Matt’s error?

2 answers:

Vlad1618 [11]1 year ago

8 0

Answer:

Matt only divided the denominator by 3 and not the whole fraction

Masteriza [31]1 year ago

5 0

Answer:

The formating is a bit confusing but I think the answer to your question would be, "Matt's error is that he just divided the denominator and not the numerator "

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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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