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3 years ago

Number 8 fill in the blanks

Mathematics

2 answers:

atroni [7]3 years ago

3 0

The answer would be prime

erik [133]3 years ago

3 0

Prime is the answer

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Karen wants to buy a $22.50 jacket that is marked 20% off.

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I think the answer is b 0.2(20)

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There are 8 brooms and 6 mops in a janitor's closet. What is the ratio of the number of mops to the total number of brooms and m

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B 3:7

If there are 6 mops, and 14 mops and brooms, you get 6:14, which simplifies to 3:7.

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Yankees wins to mets losses. Yankees wins 15. mets losses 12. use this and write the ratio.

Delvig [45]

Answer:

5:4

Step-by-step explanation:

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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

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Vika [28.1K]

Hope it benefit u !!

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