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Elan Coil [88]
3 years ago
13

If the sides of two similar figures have a similarity ratio of 3:2 what is the ratio of their areas

Mathematics
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

I DON'T KN

OW

Step-by-step explanation:

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Simplify the expression -2.5f+0.8f-13-5
Jobisdone [24]

To simplify we can collect like terms so we get -1.7f-18 :)

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3 years ago
Ok Same Thing But Different Questions
siniylev [52]

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7 0
3 years ago
Jack's car used 18 gallons of gasoline to go 468 miles. At that rate, how many gallons would be used to go 754 miles?
Brilliant_brown [7]
29 gallons because if he uses 18 gallons to go 468 miles, he would be getting 26 mpg. 26 goes into 754 29 times, so he uses 29 gallons to go 754 miles.
4 0
3 years ago
How are exponents related to multiplication?
Contact [7]

Answer: Exponents are like the multiplication of a number against itself that many times.

Step-by-step explanation: 2^4=2x2x2x2

3^3=3x3x3

3 0
3 years ago
Read 2 more answers
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
3 years ago
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