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2 years ago

What's the median and IQR of this graph?

Mathematics

1 answer:

djyliett [7]2 years ago

7 0

Enable to get the median. you have to put the numbers in order from least to greatest. Once you have done that, if there are two numbers in the middle, add them up and divide by 2.
your IQR is: 14.75
&& your Median is: 21.5
I believe. I hope this is right

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What does the expression 2(21x + 75) represent

Alla [95]

Answer:

42x+150

Step-by-step explanation:

=(2)(21x+75)

=(2)(21x)+(2)(75)

Is there a story that goes with the question.

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2 years ago

the average cost of regular gasoline in north carolina increased by 73% from 2006 to 2007. If the average cost of a gallon of ga

mash [69]

You would need to do 2.069+73% which would get you 3.57937

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3 years ago

The range, a measure of variability, Group of answer choices is the difference between the largest (L) and smallest (S) value in

Vlad [161]

The correct answer is all of the above.

Why is the range the most convenient measure of variability?

Your data's spread from the lowest to the greatest value in the distribution is indicated by the range.
The calculation of this variability index is the simplest.
Simply subtract the lowest value from the highest value in the data set to determine the range.

How do you find the range?

The difference between the lowest and highest values in a list or set is known as the range.
Put all the numbers in order before determining the range. The lowest number should then be subtracted from the highest.
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Learn more about range

brainly.com/question/2114456

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1 year ago

Statistics are numerical data. True False

Nina [5.8K]

False
Statistics data can be categorical too

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3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers