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WARRIOR [948]
2 years ago
7

A shipping company claims that 90% of its packages are delivered on time. Jenny noticed that out of the last 10 packages shipped

, 2 were late. What is the probability that 2 out of 10 randomly selected shipments would be late? 0. 04 0. 10 0. 19 0. 43.
Mathematics
1 answer:
nasty-shy [4]2 years ago
3 0

The probability that 2 out of 10 randomly selected shipments would be late is 0.19.

<h2>Given </h2>

A shipping company claims that 90% of its packages are delivered on time.

Jenny noticed that out of the last 10 packages shipped, 2 were late.

<h2>What is probability?</h2>

The probability of success and failure remain the same throughout the trials.

The probability that 2 out of 10 randomly selected shipments would be late is given by;

\rm =^{10}C_2 \times (\dfrac{2}{10})^2 \times (0.8)^8\\\\=45 \times 0.04\times 0.16\\\\ = 0.19

Hence, the probability that 2 out of 10 randomly selected shipments would be late is 0.19.

To know more about Probability click the link given below.

brainly.com/question/5053059

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Answer:

She bought 5 pounds.

Step-by-step explanation:

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Maru [420]

Answer:

Fourth term: a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{3} = \sqrt{3}

Fifth term: a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{4} = 1

Sixth term: a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}

Step-by-step explanation:

The geometric progression is:

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The first term, a, is 9

To find the common ratio, r, all we have to do is divide a term by its preceding term.

Let us divide the second term by the first:

r = \frac{3\sqrt{3}}{9}\\ \\r = \frac{\sqrt{3}}{3}

That is the common ratio.

Geometric progression is given generally as:

a_n = ar^{(n - 1)}

where a = first term

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a_n = nth term

We need to find the 4th, 5th and 6th terms.

Fourth term: a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{3} = \sqrt{3}

Fifth term: a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{4} = 1

Sixth term: a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}

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