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3 years ago

Which of the following best describes the relationship between (x-3) and the polynomial x^3 + 4x^2 + 2?

Mathematics

2 answers:

sladkih [1.3K]3 years ago

3 0

Answer:

A) (x-3) is not a factor of x^3+4x^2+2

Step-by-step explanation:

(x-3) is a factor of f(x)=x^3+4x^2+2 if f(3)=0. This is by factor theorem.

So let's check it.

f(x)=x^3+4x^2+2

f(3)=3^3+4(3)^2+2

f(3)=27+4(9)+2

f(3)=27+36+2

f(3)=63+2

f(3)=65

Since f(3) doesn't equal 0, then x-3 is not a factor.

mario62 [17]3 years ago

3 0

Answer:

A. (x-3) is not a factor

Step-by-step explanation:

You can find if (x-3) is a factor of the polynomial by dividing the polynomial by (x-3) by using long division or synthetic division.

Long division:

x^2+x+3

(x-3)/x^3+4x^2+0x+2

-(x^3-3x^2)

x^2+0x

-(x^2-3x)

3x+2

-(3x-9)

-7

Here you can see that (x-3) is not a factor of the polynomial because when you divide x^3 + 4x^2 + 2 by (x-3), there is a remainder of -7

Synthetic Division (A shortcut version of long division just to see if there is a remainder and if the supposed factor is really a factor) :

3 1 4 0 2

- 3 21 63

1 7 21 65

As seen before (x-3) is not a factor of the polynomial because there is a remainder. If 65 were 0, the (x-3) would be a factor of the polynomial.

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zhuklara [117]

Answer:

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c) 0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .