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3 years ago

A polynomial function, f(x), has x-intercepts at (-6,0) and (2,0). What is one possible factor of f(x)? PLEASE GIVE STEPS

1 answer:

8 0

Set
f
(
x
)
=
0
f(x)=0.
If the polynomial function is not given in factored form:
Factor out any common monomial factors.
Factor any factorable binomials or trinomials.
Set each factor equal to zero and solve to find the
x
x- intercepts.

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-3(6) + 2(3) - 18 = 6<br><br>Is this incorrect or correct and why ?

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horsena [70]

Answer:

$P(A) = \frac{1}{9}$

$P(B) = \frac{1}{3}$

$P(C) = \frac{1}{729}$

Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

$P(M) = \frac{1}{3}$ .

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

$P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}$

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is $P(A) = \frac{1}{9}$ . We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

$P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}$ .

C. If the probability that two people have been hired on a specific day of the week is $3(\frac{1}{3}) ^ 2$ , then the probability that 7 people have been hired on the same day is:

$P(C) = (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7\\\\P(C) = \frac{1}{729}$

D. The probability is $\frac{1}{729}$ . This number is quite close to zero. Therefore it is an unlikely bastate event.

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3 years ago

28 divided by negative 7

monitta

28 divded by -7 is -4

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3 years ago

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