All categories

3 years ago

The two dot plots below show the heights of some sixth graders and some seventh graders:

Mathematics

2 answers:

Kitty [74]3 years ago

8 0

2.0 because if you divide 1.2 by 0.6 it equals 2.0

Ber [7]3 years ago

8 0

Answer: The variability in the heights of the sixth graders is 2 times the variability in the height of the seventh graders.

Step-by-step explanation:

Since we have given tha t

Mean absolute deviation for the first set of data = 1.2

Mean absolute deviation for the second set of data = 0.6

We need to find the number of times the variability in the height of the seventh grade is the variability in the heights of the sixth graders.

So, our expression becomes

$\dfrac{MAD\ of Seventh\ greade}{MAD\ of\ sixth\ grade}\\\\=\dfrac{1.2}{0.6}\\\\=2$

Hence, the variability in the heights of the sixth graders is 2 times the variability in the height of the seventh graders.

You might be interested in

The population of a certain animal species you are studying decreases at a rate of 3.5% per year. Only 80 of the animals in the

kotegsom [21]

The function:
f ( x ) = x * 0.965^t
where x is the initial amount and t is the number of the years
80 = x * 0.965^3
80 = x * 0.89632
x = 80 : 0.89632 ≈ 89
Answer:
The initial amount of animals was <span>89.</span>

4 0

3 years ago

A dilation centered at the origin maps the point (1,3) to the point (4,12). What is the scale factor of the dilation?

Travka [436]

Answer:

Step-by-step explanation:

4,12

1,3

4/12=4

12/3=4

8 0

4 years ago

the expression 2 * L + W equals the perimeter of the rectangle write an equivalent expression using the distributive property

labwork [276]

Answer:

2*L+2*W=Perimeter

5 0

4 years ago

(6/-7)/(3/11)<br> 6/-7 divided by 3/11

beks73 [17]

Answer:

-22/7

Step-by-step explanation:

To divide fractions, we multiply by the reciprocal:

(6/-7)/(3/11)

(6/-7)(11/3)

66/-21

22/-7

-22/7

Therefore, your simplifed answer is -22/7

3 0

3 years ago

Read 2 more answers

(a) What can you say about a solution of the equation y' = - (1/4)y2 just by looking at the differential equation? 1. The functi

Harlamova29_29 [7]

Answer:

Step-by-step explanation:

a) We can state that the value of y' is negative on an interval because y^2 is positive in any interval (except in y(0), in which y'(0)=0). If y' is negative, as it is the slope of the tangent of the function y in any point, we can conclude that y is decreaing or equal to zero on any interval on which is defined.

b) First, we calculate y'

$y=\frac{4}{x+C}\\\\y'=4*\frac{(-1+0)-0}{(x+C)^2}=\frac{-4}{(x+C)^2}$

Then we replace in the differential equation and get the same result

$y'=-\frac{1}{4} y^2=-\frac{1}{4} (\frac{4}{x+C} )^2=-\frac{1}{4}\frac{16}{(x+C)^2}=\frac{-4}{(x+C)^2}$

c) The question is not readable, so I am going to solve this initial-value problem:

$y'=-\frac{1}{4}y^2\\\\y(0)=4$

$dy/dx=-(1/4)y^2\\\\ \int \frac{dy}{y^2}=-(1/4)\int dx\\\\-\frac{1}{y} =-(1/4)(x+C)\\\\y=\frac{4}{x+C}\\\\y(0)=4=\frac{4}{0+C} \\\\C=1\\\\\\y=\frac{4}{x+1}$

6 0

3 years ago