Factor the GCF: 9x4y − 6x3y2 + 3x2y3.

1 answer:

7 0

For this case we have the following expression:
$9x^4y - 6x^3y^2 + 3x^2y^3
$
For this case what you should do is observe the terms that are repeated in the given expression.
We have then:
$3x^2y
$
It is the term that is present in the whole expression.
Therefore, by doing common factor we have:
$3x^2y (3x^2 - 2xy + y^2)
$
Answer:
the GCF is:
$3x^2y (3x^2 - 2xy + y^2)$
option A: a 3 is missing at the beginning of the option.

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Refer to the accompanying technology display. The probabilities in the display were obtained using the values of n equals n=5 an

zubka84 [21]

Answer:

Yes, it is reasonable to expect that more than one subject will experience headaches

Explanation:

Notice that where it says "assume that 55 subjects are randomly selected ..." there is a typo. The correct statement is "assume that 5 subjects are randomly selected ..."

You are given the table with the probability distribution, assuming, correctly, the binomial distribution with n = 5 and p = 0.732.

p = 0.732 is the probability of success (an individual experiences headaches).
n = 5 is the number of trials (number of subjects in the sample).

The meaning of the table of the distribution probability is:

The probability that 0 subjects experience headaches is 0.0014; the probability that 1 subject experience headaches is 0.0189, and so on.

To answer whether it is reasonable to expect that more than one subject will experience headaches, you must find the probability that:

X = 2 or X = 3 or X = 4 or X = 5

That is:

P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5).

That is also the complement of P(X = 0) or P(X = 1)

1 - P(X = 0) - P(X = 1)

From the table:

P(X = 0) = 0.0014
P(X = 1) = 0.0189

Hence:

1 - P(X = 0) - P(X = 1) = 1 - 0.0014 - 0.0189 = 0.9797

That is very close to 1; thus, it is highly likely that more than 1 subject will experience headaches.

In conclusion, yes, it is reasonable to expect that more than one subject will experience headaches

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10. In a survey of 212 people at the local track and field championship, 72% favored the home team

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Answer:

a. The margin of error for the survey is of 0.0308 = 3.08%.

b. The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.