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Vadim26 [7]
3 years ago
9

Ruth has 6 chores to do. If each chore does not depend on the others, in how many different orders could she do the chores?

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0
The answer is above you shoukd mark it brainliest
Mice21 [21]3 years ago
3 0
First chore, 6 to choose from.
Second chore, 5 to choose from.
Third chore, 4 to choose from.
Fourth chore, 3 to choose from.
Fifth chore, 2 to choose from.
Sixth chore, 1 to "choose" from.
The total number of orders possible is the product
6*5*4*3*2*1=6!=720
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Mark A machine has two parts labeled A and B. The probability that part A works for one year is 0.8 and the probability that par
Mrrafil [7]

Answer:

0.625 = 62.5% probability that part B works for one year, given that part A works for one year.

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

The probability that part A works for one year is 0.8 and the probability that part B works for one year is 0.6.

This means that P(A) = 0.8, P(B) = 0.6

The probability that at least one part works for one year is 0.9.

This means that: P(A \cup B) = 0.9

We also have that:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

So

0.9 = 0.8+0.6 - P(A \cap B)

P(A \cap B) = 0.5

Calculate the probability that part B works for one year, given that part A works for one year.

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.5}{0.8} = 0.625

0.625 = 62.5% probability that part B works for one year, given that part A works for one year.

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3 years ago
How do you solve #21?
Inessa05 [86]
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Answer:

The ant moved 10.75 in after three hours

Step-by-step explanation:

18.5-13.5=5+5.75=10.75

46/8=5.75

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