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NO LINKS!!!

Three statements were made about each problem. Two are true and one is false. Mark each statement as true or false and rewrite the false statement to make it true. ​

NO LINKS!!! Three statements were made about each problem. Two are true and one is-example-1
User ZaxR
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2 Answers

24 votes
24 votes

Answer:

Explanation:

NO LINKS!!! Three statements were made about each problem. Two are true and one is-example-1
User DrivingInsanee
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2.8k points
18 votes
18 votes

Answer:

1) True, True, False

Rewritten statement: 46,013.86 cm³ of water was used to fill 40 balloons

2) False, True, True

Rewritten statement: The formula
\sf V= \sf \frac43 \pi (2.4)^3 can be used to find the volume of the model of Earth.

3) True, False, True

Rewritten statement: The volume of all the basketballs is 13,467.62 in³

Explanation:

Question 1


\textsf{Volume of a sphere}=\sf \frac43 \pi r^3 \quad \textsf{(where r is the radius)}

Given:

  • r = 6.5 cm

Substituting the given radius into the formula:


\begin{aligned}\implies \sf V & = \sf \frac43 \pi (6.5)^3\\\\& =\sf (2197)/(6) \pi \\\\& = \sf 1150.35\: cm^3 \:(nearest\:hundredth)\end{aligned}


\textsf{Volume of 40 balloons}=\sf 40 * (2197)/(6) \pi=46013.86\:cm^3\:(nearest\:hundredth)

Rewritten statement: 46,013.86 cm³ of water was used to fill 40 balloons

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Question 2


\textsf{Volume of a sphere}=\sf \frac43 \pi r^3 \quad \textsf{(where r is the radius)}

Earth

Given:

  • diameter = 4.8 cm ⇒ r = 2.4 cm

Substituting the given radius into the formula:


\implies \sf V= \sf \frac43 \pi (2.4)^3


\implies \sf V=57.91\:cm^3\:(nearest\:hundredth)

Rewritten statement:

The formula
\sf V= \sf \frac43 \pi (2.4)^3 can be used to find the volume of the model of Earth.

Saturn

Given:

  • diameter = 45.6 cm ⇒ r = 22.8 cm

Substituting the given radius into the formula:


\implies \sf V= \sf \frac43 \pi (22.8)^3


\implies \sf V=49647.02\:cm^3\:(nearest\:hundredth)

Difference between models= 49,647.02 - 57.91 = 49,589.11 cm³

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Question 3


\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

Given:

  • r = 21 in
  • h = 54 in

Substituting the given values into the formula:


\implies \sf V=\pi (21)^2(54)


\implies \sf V=23814\pi


\implies \sf V=74813.89\:in^3\:(nearest\:hundredth)


\textsf{Volume of a sphere}=\sf \frac43 \pi r^3 \quad \textsf{(where r is the radius)}

Given:

  • diameter = 9.5 in ⇒ r = 4.75 in
  • 30 basketballs


\begin{aligned}\textsf{Volume of all 30 basketballs} &=\sf 30 * (4)/(3)\pi (4.75)^3\\ & =\sf 13467.62\:in^3\:(nearest\:hundredth)\end{aligned}

Rewritten statement: The volume of all the basketballs is 13,467.62 in³

Empty space in the container = volume of container - volume of basketballs

⇒ 74813.89 - 13467.62 = 61,346.27 in³

User Sfussenegger
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