<img src="https://tex.z-dn.net/?f=x%20%5E%7B2%7D%20%20%2B%202x%20-%2015" id="TexFormula1" title="x ^{2} + 2x

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natali 33 [55]

3 years ago

+ 2x - 15" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Alja [10]3 years ago

6 0

Answer:

(x-3)(x+5)

Step-by-step explanation:

the only 2 things that can multiply together to get x^2 is x and x

so you already have the first 2 parts of the factoring

(x±?)(x±?)

to get 15 there are a 2 multipliers

1,15

3,5

if you tried 1,15

(x+15)(x-1)

that wouldnt work because 15-1=14

we need to get to 2

Let's trty 5 and 3

(x+5)(x-3)

5x-3=-15

5-3=2

sorry if my explanation doesnt make sense im terrible at explaining things lol

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Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

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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.

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And p is the probability of X happening.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

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

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$

0.0479 = 4.79% probability that fewer than 11 of them will vote

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