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4 years ago

11

Please help ASAP!

1 answer:

Oliga [24]4 years ago

3 0

The age of the students is about 0.67 years away from the 13.3 years.

The Mean Absolute Deviation is the average distance away from the mean. For the last choice, you should select more variable. Since the average is higher, there is a greater distance between the points and the mean.

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A rectangle has side lengths of (3x+6) inches and (2x-4) inches. Write an expression to represent the perimeter of the rectangle

MaRussiya [10]

If I understand the problem correctly, here's the answer:
2 (3x+6) + 2 (2x+4)=94
6x+12+4x+8=94
10x+20=94
10x=74
X=74/10
X=37/5

8 0

4 years ago

Read 2 more answers

PLEASE HELP WILL MARK BRAINLIEST Problem Page Question In a recent year, 36% of all college students were enrolled part-time. If

blsea [12.9K]

I’m not exactly sure but i think it is around four million

7 0

4 years ago

( −2 x^3) ^ 3 simplify

densk [106]

Answer:

$-8x^9$

Step-by-step explanation:

We can split up the exponent into multiple products: $(-2x^3)^3=(-2)^3*(x^3)^3$ .

We know that $(-2)^3=(-2)*(-2)*(-2)=-1*-1*-1*2*2*2=-1*8=-8$ .

We also know that $(x^3)^3=x^{3*3}=x^9$ .

So, our answer is $\boxed{-8x^9}$ and we're done!

5 0

3 years ago

Read 2 more answers

a chalk line is placed around the perimeter of the football field what is the length of this line in feet

masya89 [10]

50 so thatbis the snswer

3 0

3 years ago

9. Plot the points A (-2,3), B(2,-3), C(2,3) ,D(-2,3) join the points and find the distance between each point.

pav-90 [236]

Answer:

AB=7.21 unit

BC=6 unit

CD=7.21 unit

AD= 6 unit

AC=4 unit

BD=4 unit

Step-by-step explanation:

Coordinates of A =(-2,3)

Coordinates of B = (2,-3)

Coordinates of C = (2,3)

Coordinates of D =(-2,-3)

Distance formula : $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$A=(x_1,y_1)=(-2,3)\\B=(x_2,y_2)=(2,-3)\\AB = \sqrt{(2+2)^2+(-3-3)^2}$

AB=7.21 unit

$B=(x_1,y_1)=(2,-3)\\C=(x_2,y_2)=(2,3)\\BC=\sqrt{(2-2)^2+(3+3)^2}$

BC=6

$C=(x_1,y_1)=(2,3)\\D=(x_2,y_2)=(-2,-3)\\CD=\sqrt{(-2-2)^2+(-3-3)^2}$

CD=7.21

$A=(x_1,y_1)=(-2,3)\\D=(x_2,y_2)=(-2,-3)\\AD=\sqrt{(-2+2)^2+(-3-3)^2}$

AD=6

$A=(x_1,y_1)=(-2,3)\\C=(x_2,y_2)=(2,3)$

$AC=\sqrt{(2+2)^2+(3-3)^2}$

AC=4

$B=(x_1,y_1)=(2,-3)\\D=(x_2,y_2)=(-2,-3)$

BD= $\sqrt{(-2-2)^2+(-3+3)^2}$

BD=4

4 0

3 years ago