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3 years ago

Find the sum of 12a + 7b, −6a − 9, and 14a − 12b.

2 answers:

Ivan3 years ago

8 0

The answer is: [A]: " 20a − 5b − 9 " .
______________________________________
Explanation:
_______________________________________________
(12a + 7b) + (−6a − 9) + (14a − 12b) =

12a + 7b + 1(−6a − 9) + 1(14a − 12b) =

12a + 7b + (1*-6a) + (1*-9) + (1*14a) + (1* -12b) =

12a + 7b − 6a − 9 + 14a − 12b = ?

Combine the "like terms:

12a − 6a + 14a = 20a ;

7b − 12b = - 5b ;

and then we have "-9" ;
_____________________________________________________
So, write as: " 20a − 5b − 9 " ; which is: Answer choice: [A].
______________________________________________________

Elden [556K]3 years ago

7 0

Answer:

The answer is (A)

Step-by-step explanation:

(12a + 7b) + (−6a - 9) + (14a − 12b)

Hence,

20a - 5b - 9

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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.

8 0

3 years ago