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3 years ago

I NEED SOME SERIOUS HELP IM BEING TIMED ON THIS TEST

Mathematics

2 answers:

Katyanochek1 [597]3 years ago

7 0

Answer:

Seventy-five percent of the data values lie between 20 and 50.

Airida [17]3 years ago

4 0

Answer:

Seventy-five percent of the data values lie between 20 and 50.

Hope this helps

Sorry if I am wrong but I saw the answer in the internet

If correct can I get brainliest please

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Sara’s hair was 25 inches long. Three years later it was 43 inches long. What was the percent increase rounded to the nearest hu

Karo-lina-s [1.5K]

Answer:

Increase Percentage =

$\frac{increase}{original} * 100= \frac{43-25}{25} *100= \frac{18}{25}*100 = 72$

Percentage increase in hairlength = 72%

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Which property is shown in the following statement?

dusya [7]

Answer:

Step-by-step explanation:

Identity property of multiplication: Any number times 1 (unity) is the number itself. 7 times 1 is 7, the number with which we started.

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3 years ago

What is the following simplified product? Assume x>0.

tensa zangetsu [6.8K]

Answer:

$24x ^{3} \sqrt{5 - 4x ^{3} } \sqrt{10x}$

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3 years ago

Please help me!!! Geometry question.

tigry1 [53]

Answer:

The volume of the figure is $(\frac{l^{3}}{3})[\frac{\pi }{2}-1]\ units^{3}$

Step-by-step explanation:

we know that

The volume of the figure is equal to the volume of the cone minus the volume of the square pyramid

step 1

Find the volume of the cone

The volume of the cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=\frac{l\sqrt{2}}{2}\ units$ ----> the diagonal of the square base of pyramid is equal to the diameter of the cone

$h=l\ units$

substitute

$V=\frac{1}{3}\pi (\frac{l\sqrt{2}}{2})^{2}(l)$

$V=\frac{1}{6}\pi (l)^{3}\ units^{3}$

step 2

Find the volume of the square pyramid

The volume of the pyramid is equal to

$V=\frac{1}{3}Bh$

where

B is the area of the base

h is the height of the pyramid

we have

$h=l\ units$

$B=l^{2}\ units^{2}$

substitute

$V=\frac{1}{3}l^{2}(l)$

$V=\frac{1}{3}l^{3}\ units^{3}$

step 3

Find the volume of the figure

$\frac{1}{6}\pi (l)^{3}\ units^{3}-\frac{1}{3}l^{3}\ units^{3}=(\frac{l^{3}}{3})[\frac{\pi }{2}-1]\ units^{3}$

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3 years ago