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lbvjy [14]
2 years ago
12

The common factors of 4p3mn2 and 2pm2n3 is​

Mathematics
1 answer:
pantera1 [17]2 years ago
7 0

Given:

The given terms are

4p^3mn^2 and 2pm^2n^3

To find:

The common factors of given terms.

Solution:

We have,

4p^3mn^2 and 2pm^2n^3

The factor forms are

4p^3mn^2=2\times 2\times p\times p\times p\times m\times n\times n

2pm^2n^3=2\times p\times m\times m\times n\times n\times n

The common factors are 2, p, m, n, n. So,

Common factors = 2\times p\times m\times n\times n

                            = 2pmn^2

Therefore, the common factor of given terms is 2pmn^2.

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A newspaper editor wants to know which features in the Sunday edition of the newspaper most of its subscribers prefer. Which pop
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Answer:

B) Subscribers to the Sunday edition of the paper

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1. Which equation is the inverse of y = 2x2 + 25 ?​
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Imma go with F just because I think it’s right
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3 years ago
A council of three people is to be chosen from a group of 4 lawyers, one artist, and 3 teachers
padilas [110]

Answer: 0.2857

<u>Step-by-step explanation:</u>

At least one teacher means exactly 1 or exactly 2 or exactly 3 teachers. Find the probability of each and add them up.

\text{One teacher}:\dfrac{3\ teachers}{8\ total\ people}\times\dfrac{5\ others}{7\ people}\times \dfrac{4\ others}{6\ people}=\dfrac{5}{28}=0.1786\\\\\\\text{Two teachers}:\dfrac{3\ teachers}{8\ total\ people}\times\dfrac{2\ teachers}{7\ people}\times\dfrac{5\ others}{6\ people}=\dfrac{5}{56}=0.0893\\\\\\\text{Three teachers}:\dfrac{3\ teachers}{8\ total\ people}\times\dfrac{2\ teachers}{7\ people}\times\dfrac{1\ teacher}{6\ people}=\dfrac{1}{56}=0.0179

One teacher + Two teachers + Three teachers = 0.2857

4 0
3 years ago
A car is traveling at a speed of 30 miles per hour. what is the car's speed in miles per minute? how many miles will the car tra
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5 0
3 years ago
Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos
crimeas [40]

Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c)  P(X = 2) is unusual, P(X = 0) is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And \pi is the probability of X happening.

A number of sucesses x is considered unusually low if P(X \leq x) \leq 0.05 and unusually high if P(X \geq x) \geq 0.05

In this problem, we have that:

Two cans are randomly chosen, so n = 2

Two out of 18 cans are filled with diet coke, so \pi = \frac{2}{18} = 0.11

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that P(X = 2) is unusual, since P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958

There is a 19.58% probability that exactly one is diet and exactly one is regular.

8 0
3 years ago
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