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dsp73
3 years ago
10

The volume of a cube, in cubic centimeters, is given by the function V(x) = x3, where x is the side length of the cube in centim

eters. Write a new function for the volume of the cube in cubic millimeters.
Mathematics
1 answer:
Sav [38]3 years ago
5 0

I think we can all agree that 1 centimeter=10millimeters.

And if x is the side in centimeters then 10x will be the side in millimeters.

We just need to rewrite x^3 into (10x)^3.

Because you need to do the exponent to all part of the ( )...

(10x)^3=1000x^3

V(x)=1000x^3 cubic millimeters is the answer

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The weight distribution of parcels sent in a certain manner is normal with mean of 12 lb and standard deviation of 3 lb. The par
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Answer:

21.16

Step-by-step explanation:

Starting from the theory we have the following equation:

fi*P(x

Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:

P( z \leq \frac{c-1-12}{3.5}) =0.99/fi

solving for "c", knowing that fi is a tabulating value:

\frac{c-13}{3.5}=0.99/fi\\\frac{c-13}{3.5}=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155

therefore the value of c is equal to 21.16

8 0
3 years ago
Marissa used the set of ordered pairs below to graph a relation.
Rudiy27

Answer:

(3,2,4).............

4 0
3 years ago
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Marina CMI [18]

The correct answer I believe you are looking for is D. Elevator 1 is 13 feet above ground level, and Elevator 2 is 10 feet below ground level.

<u>Explaination:</u>

This is because if you start at (0, 0) Elevator one, will go up 13 on the y axis. Putting it at 13 feet above ground level. If you start Elevator 2 at ground level on (0, 0) and go down 10 it will make it -10 AKA 10 feet below ground level.


I hope this helps! :)

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3 years ago
The volume
Sedaia [141]
\bf \begin{array}{cccccclllll}&#10;\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\&#10;\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\&#10;y&=&{{ k}}&\cdot&x&#10;&&  y={{ k }}x&#10;\end{array}\\ \quad \\&#10;

and also

\bf \begin{array}{llllll}&#10;\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\&#10;\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\&#10;y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}&#10;&&y=\cfrac{{{  k}}}{x}&#10;\end{array}&#10;

now, we know that V varies directly to T and inversely to P simultaneously
thus\bf V=T\cdot \cfrac{k}{P}

so     \bf V=T\cdot \cfrac{k}{P}\qquad &#10;\begin{cases}&#10;V=42\\&#10;T=84\\&#10;P=8&#10;\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k&#10;\\\\\\&#10;V=\cfrac{4T}{P}\qquad now\quad &#10;\begin{cases}&#10;V=74\\&#10;P=10&#10;\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T
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3 years ago
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