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Lelu [443]
3 years ago
10

Hey! please help i’ll give brainliest

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
6 0

Answer:

D

Step-by-step explanation:

the others are positive

trasher [3.6K]3 years ago
4 0

Answer:

I think it's D

hope this helps

have a good day :)

Step-by-step explanation:

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"A gumball machine is in the shape of a sphere with a radius of 6 inches. A store manager wants to fill up the machine with mini
krek1111 [17]
So Volume (V) of a sphere is:
V =  \frac{4}{3} \pi {r}^{3}
so essentially we need to divide the machine's Volume (V) the gumball volume (v)
V =  \frac{4}{3} \pi {r}^{3}  = \frac{4}{3} \pi {(6)}^{3}  \\ V =  \frac{4}{3}(216) \pi  = 288\pi
v =  \frac{4}{3} \pi {( \frac{1}{3} )}^{3}  =  \frac{4}{3} \times  \frac{1}{27}  \pi  \\ v =  \frac{4}{81} \pi
now we divide V by v:
V/v =  \frac{288\pi}{ \frac{4\pi}{81} }  =  \frac{288}{1}  \times  \frac{81}{4}  = 5832
so b) 5,832 gumballs can fit in the machine

4 0
2 years ago
Answer this pls<br>pls answer​
Helga [31]

Answer:

\dfrac{1}{12}

Step-by-step explanation:

Rewrite the denominators as products of factors, then cancel the common factors.

\begin{aligned}\dfrac{5}{14} \times \dfrac{7}{30} & =\dfrac{5}{2 \cdot 7} \times \dfrac{7}{6 \cdot5}\\\\ & = \dfrac{5 \cdot 7}{2 \cdot 7 \cdot 6 \cdot 5}\\\\ & = \dfrac{1}{2 \cdot 6}\\\\ & = \dfrac{1}{12}{\end{aligned}

5 0
2 years ago
Read 2 more answers
I need help with three problems!!!!!
Marrrta [24]
I hope this helps you

7 0
3 years ago
Please help!!!!!!!!!!!!!!!
Andreas93 [3]

Answer:

a)  12/91

b) about 67 times;  (66 2/3)

Step-by-step explanation:

a) total of frequencies is 90 and frequency for a 3 is 11

however, the question said if the dice is rolled once more, then that raises the 11 to 12 and the frequency total from 90 to 91

let 'x' = # of sixes rolled out of 500 times

b) 12/90 = x/500

cross-multiply:

90x = 6000

x = 66 2/3

7 0
3 years ago
What is the arithmetic mean between 27 and -3
Gekata [30.6K]
Answer: 35.3

Explanation: (-3-2-1+0+1+2+....27)/30 = 35.3
6 0
3 years ago
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