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NNADVOKAT [17]
2 years ago
11

Two urns contain white balls and yellow balls. The first urn contains 3 white balls and 6 yellow balls and the second urn contai

ns 3 white balls and 8 yellow balls. A ball is drawn at random from each urn. What is the probability that both balls are white?
Mathematics
1 answer:
yuradex [85]2 years ago
5 0
We have 3 white balls in the first urn out of 9. That means we have a 1 in 3 chance at picking the white ball in the first urn.

Now, we have a 3 in 11 chance at picking the white ball in the second urn.

Since, we want them simultaneously, we need to multiply them.

1/3 × 3/11 = 1/11 chance
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Hi.
andreev551 [17]

Answer:

65

Step-by-step explanation:

40 + 5 × 5

40 + 25

65

<h3><em>Good</em><em> </em><em>Luck</em><em>!</em></h3>

Note: I also come from another country •-•

7 0
3 years ago
Read 2 more answers
A painter is painting a wall with an area of 150 ft2. He decides to paint half of the wall and then take a break. After his brea
fgiga [73]
First break= 1/2
second break = 1/2 of half= 1/4
third break = half of 1/4 = 1/8
forth break = 1/2 of 1/8 = 1/16
fifth break = 1/2 of 1/16 =1/32
fraction painted = 1 -1/32 =31/32
portion painted = 150*31/32 = 145.3125 ft^2
3 0
3 years ago
When comparing answers on a group assignment, Jenny and Jeremy found that their answers did not match. Part 1: Who do you believ
vitfil [10]
PART 1:

Jeremy gives the correct answer. 

The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....

Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value. 

PART 2:

The long division method is shown in the picture below

PART 3:

As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct



8 0
2 years ago
Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))
Marrrta [24]

Answer:

\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Step-by-step explanation:

\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine \frac{1}{c} + \frac{1}{h}

\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine the bottom, too.

=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}

Apply the fraction rule

=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}

Cancel

=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Therefore, \frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

5 0
2 years ago
HELP ME I NEED HELP!!!!
eduard

Answer:

D

Step-by-step explanation:

Since they are the same variables

7 0
3 years ago
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