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leva [86]
3 years ago
11

In how many ways can six people sit in a six-passenger car?

Mathematics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

number of ways = 720

Step-by-step explanation:

The number of ways  six people sit in a six-passenger car is given by the number of permutations of 6 elements in 6 different positions ( seats), then

number of ways = number of permutations of 6 elements = 6! = 6 * 5 * 4 *3 * 2 * 1 = 720

Since the first person that sits can be on any of the seats , but then the second person that sits can choose any of 5 seats (since the first person had already occupied one) , the third can choose 4 ... and so on.

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Answer:

Your answer is -3

Step-by-step explanation:

We know your answer is going to be the answer because 8 is a bigger number than 5, so you take 5 away from (-8)

Which leaves you with -3

Hope this helps!

-Payshence

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The following data points represent the number of players on the Russian Bears volleyball team that were injured in each match t
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Can someone help me with this problem.
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Step-by-step explanation:

y-5x is the rate

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3 years ago
Statistics question about random probability Cheese pastureized or Raw MilkA cheese can be classified as either raw-milk or past
blsea [12.9K]

Answer:

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) Or Pr(RRPP) Or Pr(RRRP) Or Pr(RRRR)

= 0.1269 to 4 decimal places

It would not be unusual that at least one of four randomly selected cheeses is raw-milk, because the probability have a value between 0 and 1

Step-by-step explanation:

If is given that 80% of the cheese is classified as pasteurized.

It then implies that 20% of the cheese is classified as Raw-milk

Probability of pasteurized cheese is 0.82(Denoted by Pr(P))

Probability of raw-milk cheese is 0.18(Denoted as Pr(R))

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) + Pr(RRPP) + Pr(RRRP) + Pr(RRRR)

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3 years ago
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