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4 years ago

3(x-7)=9(4x+5) what is the answer????

1 answer:

5 0

PEMDAS

3 ( x - 7 ) = 9 ( 4x + 5 )
3x - 21 = 36x + 45
-36
-33x - 21 = 45
+21

-33x = 66
------ ------
-33 -33

x = -2

Your answer should be -2

I hope this helps!

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3 years ago

2x +4- 3 (x + 5) = -2x + 4 x=-26 X=15 X=-5 x=-15

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3 years ago

Find the value of X to the nearest tenth

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Answer:

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2 years ago

A cylinder and a cone have the same volume. The cylinder has a radius of 2 inches and a hieght of 3 inches. The cone has a radiu

arlik [135]

The height of the cone is inches, if the cylinder and cone have the same volume.The cylinder has a radius of 2 inches and a height of 3 inches. The cone has a radius of 3 inches.

Step-by-step explanation:

The given is,

A cylinder and a cone have the same volume

Cylinder has a radius 2 inches and height of 3 inches.

Cone has a radius of 3 inches

Step:1

For Cylinder'

Formula to calculate the volume of cylinder is,

$V = \pi r^{2} h$ ..................................................(1)

where,

r - 2 inches

h - 3 inches

From the equation (1)

= $\pi$ × $2^{2}$ × 3

= 37.70

V = 37.70 cubic inches

Step:2

For cone,

Formula to calculate the volume of cone is,

$V = \pi r^{2} \frac{h}{3}$ ..................................................(2)

From the statement,

cylinder and a cone have the same volume

$V_{cylinder}$ = $V_{cone}$

37.70 = $\pi$ × $3^{2}$ × $\frac{h}{3}$

37.70 = 9.42478 × h

Height of the cone, h = 4 inches

Result:

Thus the height of the cone is 4 inches, if a cylinder and cone have the same volume.The cylinder has a radius of 2 inches and a height of 3 inches. The cone has a radius of 3 inches.

4 0

3 years ago

A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collecte

allsm [11]

Answer:

The degrees of freedom are given by;

$df =n-1= 5-1=4$

The significance level is 0.1 so then the critical value would be given by:

$F_{cric}= 7.779$

If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays

Step-by-step explanation:

For this case we have the following observed values:

Mon 25 Tue 22 Wed 19 Thu 18 Fri 16 Total 100

For this case the expected values for each day are assumed:

$E_i = \frac{100}{5}= 20$

The statsitic would be given by:

$\chi^2 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}$

Where O represent the observed values and E the expected values

The degrees of freedom are given by;

$df =n-1= 5-1=4$

The significance level is 0.1 so then the critical value would be given by:

$F_{cric}= 7.779$

If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays

5 0

3 years ago