Ask question

Ask question

All categories

3 years ago

7

A clothing store owner tracks the amount of money the last 10 customers spend in her store. The owner concludes that the typical

amount spent by any customer is $28.80. The owner's conclusion

1 answer:

Dima020 [189]3 years ago

4 0

Is not a good one...because there is not enough people...it doesn't represent all of the customers

You might be interested in

I quite don’t get this question and I have 4 minutes remaining

gregori [183]

The length of one of the sides of the larger pentagon is 24 cm. The length of one of the sides of the smaller pentagon is 8 cm.

$24$ ÷ $8=3$

So, if the larger pentagon is 3x the size of the smaller pentagon, we can find the area of the smaller pentagon by dividing the area of the larger pentagon by 3.

$135$ ÷ $3=45$

The area of the smaller pentagon is 45in²

<em>Hope this helps! Pwease mark me brainliest! Have a great day!</em>

<em />

<em>−xXheyoXx</em>

7 0

2 years ago

Please help (: !!

Yuki888 [10]

Add more information <span />

4 0

3 years ago

I do not know this answer

Ivenika [448]

Answer:

The answer is -19

Step-by-step explanation:

Given information -19y^2 , y=-1

-19 * y^2

-19 * (-1)^2

-19 * 1 = -19

7 0

2 years ago

Work out the volume of the shape

pashok25 [27]

Answer:

$\large\boxed{V=\dfrac{1,421\pi}{3}\ cm^3}$

Step-by-step explanation:

We have the cone and the half-sphere.

The formula of a volume of a cone:

$V_c=\dfrac{1}{3}\pi r^2H$

r - radius

H - height

We have r = 7cm and H = (22-7)cm=15cm. Substitute:

$V_c=\dfrac{1}{3}\pi(7^2)(15)=\dfrac{1}{3}\pi(49)(15)=\dfrac{735\pi}{3}\ cm^3$

The formula of a volume of a sphere:

$V_s=\dfrac{4}{3}\pi R^3$

R - radius

Therefore the formula of a volume of a half-sphere:

$V_{hs}=\dfrac{1}{2}\cdot\dfrac{4}{3}\pi R^3=\dfrac{2}{3}\pi R^3$

We have R = 7cm. Substitute:

$V_{hs}=\dfrac{2}{3}\pi(7^3)=\dfrac{2}{3}\pi(343)=\dfrac{686\pi}{3}\ cm^3$

The volume of the given shape:

$V=V_c+V_{hs}$

Substitute:

$V=\dfrac{735\pi}{3}+\dfrac{686\pi}{3}=\dfrac{1,421\pi}{3}\ cm^3$

7 0

2 years ago

Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank i

shepuryov [24]

Answer:D) 1 hr

Step-by-step explanation:

Given

Pump A and B can fill tank in $\frac{6}{5} hr$

Pump A and C can fill tank in $\frac{3}{2} hr$

Pump B and C can fill tank in $2 hr$

Let A be the total hr taken A therefore rate of $A=\frac{1}{A} /hr$

Let B be the total hr taken B therefore rate of $B=\frac{1}{B} /hr$

Let C be the total hr taken C therefore rate of $C=\frac{1}{C} /hr$

$\frac{1}{A}+\frac{1}{B}=\frac{5}{6}$ -----1

$\frac{1}{B}+\frac{1}{C}=\frac{1}{2}$ -----2

$\frac{1}{A}+\frac{1}{C}=\frac{2}{3}$ -----3

Adding 1,2 & 3

$2(\frac{1}{A}+\frac{1}{B}+\frac{1}{C})=\frac{5}{6}+\frac{1}{2}+\frac{2}{3}$

$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{2}$

$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1$

thus time taken by A,B and C combined is 1 hr

7 0

2 years ago