What is the median of the data represented in the box plot below 20 30 40 50

Mathematics

2 answers:

PtichkaEL [24]3 years ago

5 0

40 is the answer!!!!!

mr_godi [17]3 years ago

4 0

The median of the data is 40.

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Two expressions that equal 19

aev [14]

There are many like

1(10)+10-1=19

18+1=19

30-11=19

15+4=19

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3 years ago

A certain chain of ice cream stores sells 28 different flavors, and a customer can order a single-, double- , or triple-scoop co

astra-53 [7]

Think about the cones with only 1 scoop of ice cream. Isn't it clear that there are 28 of those?
Now think about the two scoop cones. You have 28 choices for the first scoop and 28 choices for the second scoop.
So the number of possibilities is 28(28).
Now think of the 3 scoop cones.
You have 28 choices for the first scoop, 28 choices for the second scoop, and 28 choices for the third scoop or 28(28)(28) possibilities.
Add them all together and you have the total.

So it will be like this:
28^3+28^2+28

I hope my answer helped you.

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3 years ago

A coin was flipped 150 times. The results of the experiment are shown in the following table:

Wewaii [24]

It is 10% higher than the theoretical probability.

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3 years ago

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tia_tia [17]

Answer:

15.41

Step-by-step explanation:

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2 years ago

Read 2 more answers

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$