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3 years ago

If 2m−6=8m then 3m= A. 3 B. -1 C. -3 D. -6 E. I don't know.

Mathematics

1 answer:

zmey [24]3 years ago

6 0

For this case we have to:

$2m-6 = 8m$

If we subtract 8m on both sides of the equation, we have:

$2m-6-8m = 8m-8m\\-6m-6 = 0$

Adding 6 to both sides of the equation:

$-6m-6 + 6 = + 6\\-6m = 6$

Dividing between -6 on both sides of the equation:

$\frac {-6m} {6} = \frac {6} {6}\\-m = 1$

So:

$m = -1$

If we want to know the value of 3m, then:

$3m = 3 (-1) = - 3$

Answer:

Option C

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3 years ago

In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 11% of voters a

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Answer:

a) 0.0214 = 2.14% probability that none of the people are Independent.

b) 0.8516 = 85.16% probability that fewer than 6 are Independent.

c) 0.8914 = 89.14% probability that more than 2 people are Independent.

Step-by-step explanation:

For each people, there are only two possible outcomes. Either they are independent, or they are not. For each person asked, the probability of them being Independent voters is the same. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

It is believed that 11% of voters are Independent.

This means that $p = 0.11$

A survey asked 33 people to identify themselves as Democrat, Republican, or Independent.

This means that $n = 33$

A. What is the probability that none of the people are Independent?

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

0.0214 = 2.14% probability that none of the people are Independent.

B. What is the probability that fewer than 6 are Independent?

This is

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X = 2) = C_{33,2}.(0.11)^{2}.(0.89)^{31} = 0.1724$

$P(X = 3) = C_{33,3}.(0.11)^{3}.(0.89)^{30} = 0.2202$

$P(X = 4) = C_{33,4}.(0.11)^{4}.(0.89)^{29} = 0.2041$

$P(X = 5) = C_{33,5}.(0.11)^{5}.(0.89)^{28} = 0.1463$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0214 + 0.0872 + 0.1724 + 0.2202 + 0.2041 + 0.1463 = 0.8516$

0.8516 = 85.16% probability that fewer than 6 are Independent.

C. What is the probability that more than 2 people are Independent?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X < 2) = 0.0214 + 0.0872 = 0.1086$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1086 = 0.8914$

0.8914 = 89.14% probability that more than 2 people are Independent.

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