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3 years ago

Factor the polynomials by finding the GCF

1 answer:

3 0

3x + 12

The GCF here is 3.

3x / 3 = x
12 / 3 = 4

So we have 3(x + 4)

7y - 7

The GCF here is 7.

7y / 7 = y
-7 / 7 = -1

So we have 7(y - 1)

5x + 30y

The GCF here is 5.

5x / 5 = x
30y / 5 = 6y

So we have 5(x + 6y)

8m + 36n

The GCF here is 4.

8m / 4 = 2m
36n / 4 = 9n

So we have 4(2m + 9n)

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The Georgia state constitution contains a bill of rights .which of these is Most likely the best entry in this section of Georgi

Pavlova-9 [17]

That would be the freedom of speech and the freedom of the press.

Both of these are without a doubt a center of political to the political system existing in the United States. They state that both the freedom of speech and the press is to be expressed freely. These rights are the key to the political discourse as they allow an honest conversation and exchange of opinion that is of paramount importance in democracy. People are free to express their opinions and through the press they gain the necessary information to make educated decisions about the burning political questions.

Hope this helps!

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3 years ago

An item is regularly priced at $35. It is now priced at a discount of 80% off the regular price. What is the price now?

lys-0071 [83]

Answer:

Price of item now = $35×20%

= $7

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3 years ago

Lauren had 2 2/3 yards of ribbon. She used 4/5 of the ribbon. How many yards of ribbon did Lauren use?

strojnjashka [21]

Answer:

2 2/15 or 2.13333333

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3 years ago

Read 2 more answers

Four rats are selected at random from a cage of 5 male (M) and 6 female (F) rats. If the random variable Y is concerned with the

Ludmilka [50]

Using the hypergeometric distribution, it is found that the distribution is:

P(X = 0) = 0.0152

P(X = 1) = 0.1818

P(X = 2) = 0.4545

P(X = 3) = 0.3030

P(X = 4) = 0.0455

Hence:

A) There is a 0.0455 = 4.55% probability that all selected rats are female.

B) There is a 0.1818 = 18.18% probability that only one of the selected rats is female.

C) There is a 0.803 = 80.3% probability that at least two of the selected rats is female.

D) The mean of the probability distribution is of 2.18.

E) The standard deviation of the probability distribution is of 0.833.

The rats are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

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

The parameters are:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this problem:

There is a total of 11 rats, hence N = 11.

6 of the rats are female, hence k = 6.

4 rats are taken, hence n = 4.

The distribution is the probability of each outcome, hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,11,4,6) = \frac{C_{6,0}C_{5,4}}{C_{11,4}} = 0.0152$

$P(X = 1) = h(1,11,4,6) = \frac{C_{6,1}C_{5,3}}{C_{11,4}} = 0.1818$

$P(X = 2) = h(2,11,4,6) = \frac{C_{6,2}C_{5,2}}{C_{11,4}} = 0.4545$

$P(X = 3) = h(3,11,4,6) = \frac{C_{6,3}C_{5,1}}{C_{11,4}} = 0.3030$

$P(X = 4) = h(4,11,4,6) = \frac{C_{6,4}C_{5,0}}{C_{11,4}} = 0.0455$

Item a:

P(X = 4) = 0.0455, hence:

There is a 0.0455 = 4.55% probability that all selected rats are female.

Item b:

P(X = 1) = 0.1818, hence:

There is a 0.1818 = 18.18% probability that only one of the selected rats is female.

Item c:

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.4545 + 0.3030 + 0.0455 = 0.803$

There is a 0.803 = 80.3% probability that at least two of the selected rats is female.

Item d:

The mean of the hypergeometric distribution is:

$\mu = \frac{nk}{N}$

Hence:

$\mu = \frac{4(6)}{11} = 2.18$

The mean of the probability distribution is of 2.18.

Item e:

The standard deviation of the hypergeometric distribution is:

$\sigma = \sqrt{\frac{nk(N-k)(N-n)}{N^2(N-1)}}$

Hence:

$\sigma = \sqrt{\frac{4(6)(11-6)(11-4)}{11^2(11-1)}} = 0.833$

The standard deviation of the probability distribution is of 0.833.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

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2 years ago

Please help me on this math problem!

creativ13 [48]

Answer:

427 feet.

Step-by-step explanation:

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3 years ago