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2 years ago

Pls help i need help

Mathematics

1 answer:

Thepotemich [5.8K]2 years ago

4 0

$(5^6)^0 \\\\=5^0\\\\=1$

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A factory received a shipment of 11 hammers, and the vendor who sold the items knows there are 2 hammers in the shipment that ar

zhenek [66]

Question:

If a sample of 2 hammer is selected

(a) find the probability that all in the sample are defective.

(b) find the probability that none in the sample are defective.

Answer:

a $Pr = \frac{2}{110}$

b $Pr = \frac{72}{110}$

Step-by-step explanation:

Given

$n = 11$ --- hammers

$r = 2$ --- selection

This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities

Solving (a): Probability that both selection are defective.

For two selections, the probability that all are defective is:

$Pr = P(D) * P(D)$

$Pr = \frac{2}{11} * \frac{2-1}{11-1}$

$Pr = \frac{2}{11} * \frac{1}{10}$

$Pr = \frac{2}{110}$

Solving (b): Probability that none are defective.

The probability that a selection is not defective is:

$P(D') = \frac{9}{11}$

For two selections, the probability that all are not defective is:

$Pr = P(D') * P(D')$

$Pr = \frac{9}{11} * \frac{9-1}{11-1}$

$Pr = \frac{9}{11} * \frac{8}{10}$

$Pr = \frac{72}{110}$

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3 years ago

Use Pythagorean theorem to find right triangle side lengths 8 and 10

sp2606 [1]

Answer:

Step-by-step explanation:

$a^{2}+b^{2} =c^{2} \\\\8^{2}+b^{2} =10^{2} \\64+b^{2}=100\\b^{2}=36\\\sqrt{36} =6\\b=6$

4 0

3 years ago

A recipe that makes 4 servings calls for Two-thirds cup of flour. How much flour is required to make 20 servings?

Karo-lina-s [1.5K]

Answer:

8/3

Step-by-step explanation:

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3 years ago

An employee worked 36 hours last week. This week, she worked 48 hours. Find the percent increase. Round to the nearest percent.

ioda

Answer:

There was a 25% increase.

Step-by-step explanation:

6 0

3 years ago

PLEASE HELP

stepan [7]

Answer:

to big but i will slove it for u ok so see solution c one is correct

5 0

3 years ago

Read 2 more answers

Pls help i need help​

Pls help i need help