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2 years ago

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Which is a characteristic of any function created by the product of two linear factors?

1 answer:

Afina-wow [57]2 years ago

5 0

The characteristic of any function that is created by the product of two linear factors is B. has two x-intercepts

<h3>What is a function?</h3>

A function is an expression, rule, or law in mathematics that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

The property of any function formed by the product of two linear factors is that it has two x-intercepts. According to the question, we must determine which of the following is a characteristic of any function created by the product of two linear factors.

A quadratic function is the product of two linear factors. Because a quadratic function is a two-degree polynomial, 2. In essence, the result of the product of two linear factors has two x-intercepts.

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In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

<u>1. Start with the first 11 inspected packages:</u>

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

<u>2. The 12th package</u>

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

<u>3. Finally</u>

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

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