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Charra [1.4K]
3 years ago
6

Two urns contain white balls and yellow balls the first urn contains 9 white balls and 9 yellow balls and the second urn contain

s 8 white balls and 3 yellow balls. A ball is drawn at random from each urn. What is the probability that both balls are white?
A. 4/11
B. 17/29
C. 1/72
D. 17/198
Mathematics
2 answers:
Kruka [31]3 years ago
8 0
The answer is B 17/29
Lubov Fominskaja [6]3 years ago
5 0

Answer:

B. 17/29 because it's the white balls added together and the total so 17/29

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Write the solution of the inequality in set builder notation: 2(3p-11)>=-16
Dima020 [189]
The answer would be 
p≥1

hope this helps :3
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3 years ago
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It has been estimated that 1000 curies of a radioactive substance introduced at a point on the surface of the open sea would spr
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Find the value of f(9)<br> y = f(x)
goblinko [34]

f(9) is the x value, find where the line is in the y direction at x = 9

The line crosses y = 3 at x = 9

Answer:

f(9) = 3

8 0
3 years ago
​together, steve and tom sold 7777 raffle tickets for their school. steve sold 1414 more than twicetwice as many raffle tickets
storchak [24]
Steve = S , Tom = T
1) S+T = 7777
2) S= 1414 + 2T

put 2) in 1)

1414+2T + T = 7777
1414 +3T = 7777
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so T = 2121
then Steve sold 1414+2(2121) = 1414+4242=5656 tickets
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5 0
4 years ago
(I've been trying to figure this out for 3 days and I really need help)
liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7

2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55

3)

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

4)

pretty much the same thing, we get the volume of the cone and its top, add them up.

\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^

8 0
3 years ago
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