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marissa [1.9K]
3 years ago
10

What is the volume of the can below? Use = 3.14 and round your answer to the nearest tenth.

Mathematics
2 answers:
Nadya [2.5K]3 years ago
8 0

Answer: 328,268.2

Step-by-step explanation:

radius=66/2=33

volume=(radius^2)pi)(height)

volume=33^2(3.14)(96)

volume=328,268.2 mm^3

vagabundo [1.1K]3 years ago
6 0

Answer:

volume= 328,268.2 mm^3

Step-by-step explanation:

radius=66/2=33

volume=(radius^2)pi)(height)

volume=33^2(3.14)(96)

volume=328,268.2 mm^3

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Find the probability of obtaining 3 heads when flipping 4 coins

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A gallon of Moo Milk cost $5.12. What is the price, in dollars, of an 8 oz glassof Moo Milk?
MA_775_DIABLO [31]
The cost for a gallon of Moo Milk is $5.12
To find the cost of 8 fl. oz. of Moo Milk, the unit price must be found.

Therefore:

One gallon of liquid equals 128 fluid ounces. 

5.12 ÷ 128 = 0.04

0.04 × 8 = 0.32

Therefore, the cost of 8 fluid ounces of Moo Milk is $0.032 

I hope this helped - Maria S.
5 0
2 years ago
Jacob and his father measured the lengths of scrap wood in their yard. Make a line plot of their data.
drek231 [11]

The line plot is used to show the lengths of the scrap woods using check marks above the line

<h3>How to draw the line plot</h3>

From the question, we have the following frequency table

<u>Lengths      Frequency</u>

4\frac 18                       2

4\frac 14                       3

4\frac 38                       1  

4\frac 12                       6

4\frac 58\\                       1

4\frac 34                      2

4\frac 78                      2

5                         1

Next, we plot the number line.

Such that the lengths of the scrap woods are on the x-axis while their frequencies are on the y-axis

See attachment for the line plot

Read more about line plots at:

brainly.com/question/15853311

3 0
2 years ago
A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft

3 0
3 years ago
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