All categories

3 years ago

The two dot plots below compare the forearm lengths of two groups of schoolchildren:

2 answers:

7 0

B.

as a, none is longer than 11 inches
b, 2 are longer than 11 inches
c, 3 have 9 inches (which is the shortest)
s, one has 9 inches

Solnce55 [7]3 years ago

3 0

Two dot plots are shown one below the other. The title for the top dot plot is Online Customers and the title for the bottom plot is Walk-In Customers. Below the line for each dot plot is written Number of Packs. There are markings from 1 to 5 on each line at intervals of one. For the top line there are 0, 0, 1, 5, 4 dots respectively for marks from 1 to 5. For the bottom line there are 6, 3, 1, 0, 0 dots respectively for marks from 1 to 5.

You might be interested in

On the first math test, Keiko answered 19 out of 25 questions correctly. On the second test she got 17 out of 20 correct. On whi

jenyasd209 [6]

Answer:

shegot a better score on the second test

50 is double of 25

75is 3 times 25

100 is 4 times 25

Step-by-step explanation:

19 times 4= 76

17 times 5= 85

7 0

4 years ago

Name each polynomial by degree and number of terms of -7

denis-greek [22]

Answer: i couldn’t find the answer to this one either

Step-by-step explanation:

4 0

3 years ago

the graph shows the total number of inches of snow that fell in a town in february over ten year period. what is the approximate

VladimirAG [237]

Answer:

0.6 because its 58% chance which is 0.58 which rounds to 0.6

Step-by-step explanation:

6 0

3 years ago

Use the disk method or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs

JulsSmile [24]

Answer:

1) V = 12 π ㏑ 3

2) $\mathbf{V = \dfrac{328 \pi}{9}}$

Step-by-step explanation:

Given that:

the graphs of the equations about each given line is:

$y = \dfrac{6}{x^2}, y =0 , x=1 , x=3$

Using Shell method to determine the required volume,

where;

shell radius = x; &

height of the shell = $\dfrac{6}{x^2}$

∴

Volume V = $\int ^b_{x-1} \ 2 \pi ( x) ( \dfrac{6}{x^2}) \ dx$

$V = \int ^3_{x-1} \ 2 \pi ( x) ( \dfrac{6}{x^2}) \ dx$

$V = 12 \pi \int ^3_{x-1} \dfrac{1}{x} \ dx$

$V = 12 \pi ( In \ x ) ^3_{x-1}$

V = 12 π ( ㏑ 3 - ㏑ 1)

V = 12 π ( ㏑ 3 - 0)

V = 12 π ㏑ 3

2) Find the line y=6

Using the disk method here;

where,

Inner radius $r(x) = 6 - \dfrac{6}{x^2}$

outer radius R(x) = 6

Thus, the volume of the solid is as follows:

$V = \int ^3_{x-1} \begin {bmatrix} \pi (6)^2 - \pi ( 6 - \dfrac{6}{x^2})^2 \end {bmatrix} \ dx$

$V = \pi (6)^2 \int ^3_{x-1} \begin {bmatrix} 1 - \pi ( 1 - \dfrac{1}{x^2})^2 \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} 1 - ( 1 + \dfrac{1}{x^4}- \dfrac{2}{x^2}) \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} - \dfrac{1}{x^4}+ \dfrac{2}{x^2} \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} {-x^{-4}}+ 2x^{-2} \end {bmatrix} \ dx$

Recall that:

$\int x^n dx = \dfrac{x^n +1}{n+1}$

Then:

$V = 36 \pi ( -\dfrac{x^{-3}}{-3}+ \dfrac{2x^{-1}}{-1})^3_{x-1}$

$V = 36 \pi ( \dfrac{1}{3x^3}- \dfrac{2}{x})^3_{x-1}$

$V = 36 \pi \begin {bmatrix} ( \dfrac{1}{3(3)^3}- \dfrac{2}{3}) - ( \dfrac{1}{3(1)^3}- \dfrac{2}{1}) \end {bmatrix}$

$V = 36 \pi (\dfrac{82}{81})$

$\mathbf{V = \dfrac{328 \pi}{9}}$

The graph of equation for 1 and 2 is also attached in the file below.

5 0

4 years ago

4x-y=2 <br>what is the solution

Rudiy27

Answer= y=-_

Step-by-step explanation:

y= -2+4x

4 0

3 years ago