All categories

3 years ago

What is the mean of the following data values? 22, 37, 49, 15, 92

Mathematics

2 answers:

nordsb [41]3 years ago

8 0

Answer:

Step-by-step explanation:

when you add them all up you then have to divide by how many numbers there are.

In this case we have five so all of the numbers added up divided by 5 will give you your answer.

so 215/5 equals 43 and that's your answer

attashe74 [19]3 years ago

7 0

Answer:

Step-by-step explanation:

First, you have to add 22+37+49+15+92 to get 215. To find the mean, you need to divide by how much numbers there are (5). So 215 divided by 5 is 43!

You might be interested in

Any help with this would be greatly appreciated

lesya692 [45]

Answer:

Step-by-step explanation:

using similar triangles

Let x be the boy's height

$\frac{x}{27}$ = $\frac{72}{43.2}$

cross multiply to get 43.2x=1944

then divide by 43.2 both sides

to get x=45

3 0

3 years ago

Answer the 3 screenshotted questions below. 30 points!

Triss [41]

Answer:

13. $36a^2b^4$ π

14. $16x^3y^4$ π

15. $-\frac{c^3}{4a^7b^3}$

Step-by-step explanation:

13.

A = π * $r^2$

r = $6ab^2$

A = π* $(6ab^2)^2$

A = π * $(6ab^2)(6ab^2)$

A = π * $36a^2b^4$

A = $36a^2b^4$ π

14.

V = π * $r^2$ * h

r = $2x$

h = $4xy^4$

V = π * $(2x)^2$ * $4xy^4$

V = π * $4x^2$ * $4xy^4$

V = π * $16x^3y^4$

V = $16x^3y^4$ π

15. $\frac{3a^{-4}b^2}{-12a^3b^5c^{-3}}$ = $-\frac{c^3}{4a^7b^3}$

$\frac{3a^{-4}b^2}{-12a^3b^5c^{-3}}$

$\frac{a^{-4}}{-4a^3b^5c^{-3}}$

$\frac{b^2}{-4^{3+4}b^5c^{-3}}$

$\frac{b^2}{-4^7b^5c^{-3}}$

$\frac{1}{-4a^7b^{5-2}c^{-3}}$

$\frac{1}{-4a^7b^3c^{-3}}$

$\frac{c^3}{-4a^7b^3}$

$-\frac{c^3}{4a^7b^3}$

Hope this helps!

3 0

3 years ago

45(15x+20)−7x=56(12x−24)+6 Enter your answer in the box.

zzz [600]

X=1119/2 hope this helps :)

3 0

3 years ago

Help please 150 x 189

ExtremeBDS [4]

It is 28350 merry Christmas

8 0

3 years ago

Read 2 more answers

What are the common factors between 100 50 an 30

Sonbull [250]

10, 5, 2, and 1 are the common factors.

4 0

3 years ago