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3 years ago

Please help Consider: A (-3,-4), B (5,2), C (11,-6) What is AB? What is BC?

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

7 0

Answer:

AB = 10, BC = 10

Step-by-step explanation:

AB = length of segment AB = sqrt( (-3-5)^2 + (-4-2)^2 ) = sqrt( 64 + 36) = sqrt(100) = 10

BC = sqrt( (5-11)^2 + (2-(-6))^2 ) = sqrt( 36 + 64 ) = sqrt( 100 ) = 10

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3 years ago

Suppose that 60 batteries are shipped to an auto parts store, and that 5 of those are defective. A fleet manager then buys 6 of

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Answer:

In 8910 ways can at least 4 defective batteries be included in the purchase

Step-by-step explanation:

The order in which the batteries are there is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

4 defective:

4 defective from a set of 5 and 2 non-defective from a set of 60 - 5 = 55. So

$C_{5,4}C_{60,2} = \frac{5!}{1!4!} \times \frac{60!}{2!58!} = 8850$

5 defective:

5 defective from a set of 5 and 1 non-defective from a set of 60 - 5 = 55. So

$C_{5,5}C_{60,5} = \frac{5!}{0!5!} \times \frac{60!}{1!59!} = 60$

In how many ways can at least 4 defective batteries be included in the purchase?

$T = 8850 + 60 = 8910$

In 8910 ways can at least 4 defective batteries be included in the purchase

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