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den301095 [7]
3 years ago
12

A container of candy is shaped like a

Mathematics
1 answer:
Simora [160]3 years ago
3 0

Answer:

The radius of the container is 2 centimeter

Step-by-step explanation:

Given that a container of candy is shaped like a cylinder

Given that volume = 125.6 cubic centimeters

Height of conatiner = 10 centimeter

To find: radius of the container

We can use volume of cylinder formula and obatin the radius value

The volume of cylinder is given as:

Where "r" is the radius of cylinder

"h" is the height of cylinder and  is constant has value 3.14

Substituting the values in formula, we get

Taking square root on both sides,

Thus the radius of the container is 2 centimeter

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 So, if 600 is 25%, that means we do 600*4=2400 600 is 1/4 of 2400. 

So the answer is 2400


I hope this helps!
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3 years ago
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3 years ago
3/4 x 2/3 as a fraction
svet-max [94.6K]

Answer:

Hi, There!

Your Answer is \frac{6}{12} ~or~\frac{1}{2}

Step-by-step explanation:

\frac{3}{4} ~x~\frac{2}{3} = \frac{3 ~x~2}{4~x~3}

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\frac{6}{12}

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5 0
3 years ago
The oil prices for 2014, rounded to the nearest dollar, were: 95, 101, 101, 102, 102, 106, 104, 97, 93, 84, 76, 59. what is the
kipiarov [429]
  • Interquartile Range (IQR) = Q_3-Q_1 , with Q_3 as the upper quartile and Q_1 as the lower quartile.

Firstly, rearrange the data so that it's in ascending order: \{59,76,84,93,95,97,101,101,102,102,104,106\}

Next, find the median:

\{59,76,84,93,95,\boxed{97,101,}101,102,102,104,106\}\\\\\frac{97+101}{2}\\\\\frac{198}{2}\\\\99

Now to find the lower quartile, find the "median" of the data set that's to the left of 99:

\{\overbrace{59,76,84,93,95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\{\overbrace{59,76,\boxed{84,93}95,97}^{\textsf{to the left of the median}},101,101,102,102,104,106\}\\\\\frac{84+93}{2}\\\\\frac{177}{2}\\\\88.5

Now to find the upper quartile, it's the similar process as finding the lower quartile, except that you are finding the "median" of the data set to the right of 99:

\{59,76,84,93,95,97,\overbrace{101,101,102,102,104,106}^{\textsf{to the right of the median}}\}\\\\\{59,76,84,93,95,97,\overbrace{101,101,\boxed{102,102} 104,106}^{\textsf{to the right of the median}}\}\\\\\frac{102+102}{2}\\\\\frac{204}{2}\\\\102

Now that we have the upper and lower quartile, subtract them:

102-88.5=13.5

<u>In short, the IQR of this data set is 13.5.</u>

8 0
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Math help, please help thank you ! :)
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Answer:

construct a angle that has the same measure to ABC.

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