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3 years ago

If PQ=7 and QR=10, then PR= what ?

Mathematics

1 answer:

Yuliya22 [10]3 years ago

6 0

If PQ=7 and QR=10,

P=(7/Q) and R=(10/Q)

Therefore:

PR=(7/Q)(10/Q)

=(70)/Q²

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A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects

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The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let X denote the number of green candies.

The probability of green candies is, p = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, n = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable X follows a binomial distribution with parameter n = 30 and p = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

$P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328$

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

= 0.17228 + 0.17946 + 0.15328

= 0.50502

Thus, the probability of Steve agreeing with the company’s claim is 0.50502.

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3 years ago