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3 years ago

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A building contractor buys 7070% of his cement from supplier A and 3030% from supplier B. A total of 9595% of the bags from A

arrive undamaged, while 7070% of the bags from B arrive undamaged. Find the probability that anan undamagedundamaged bag is from supplier Upper AA.

1 answer:

tatiyna3 years ago

8 0

Title:

<h2>The answer is $\frac{133}{174}$ .</h2>

Step-by-step explanation:

Let the building contractor bought 1000 cement in total.

70% of 1000 = 700 cements were from A and (1000 - 700) = 300 cements were from B.

The number of undamaged bag from A was 95% of 700 = $\frac{95}{100} \times700 = 7\times95 = 665$ and the number of undamaged bags from B was 70% of 300 = $\frac{70}{100} \times300 = 210$ .

Total undamaged bags were (665 + 210) = 870.

The probability that an undamaged bag is from A is $\frac{665}{870} = \frac{133}{174}$ .

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Answer:

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Step-by-step explanation:

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with that template in mind, let's see

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Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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Mean $\mu$ , standard deviation $\sigma$

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