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3 years ago

Factor 3x^2-19x+20

Mathematics

1 answer:

dusya [7]3 years ago

4 0

Answer:

(3x-4)(x-5)

Step-by-step explanation:

This is in the form

ax²+bx+c.

To factor this, we find factors of a·c that sum to b; this means factors of 3(20) = 60 that sum to -19:

60 = 1(60) or -1(-60); 2(30) or -2(-30); 3(20) or -3(-20); 4(15) or -4(-15); 5(12) or -5(-12); 6(10) or -6(-10). The only of these that sum to -19 are -4 and -15. This means we will split up -19x into -4x and -15x:

3x²-4x-15x+20

Next we group the first two terms and the last two terms:

(3x²-4x)+(-15x+20)

Factor out the GCF of each group. For the first group, this is x:

x(3x-4)

For the second group, this is -5:

-5(3x-4)

The common factor for these two groups is (3x-4):

(3x-4)(x-5)

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Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

3 years ago

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MA_775_DIABLO [31]

Answer:

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56-36=20

20/2= 10

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Hope this helps! Have a nice day :)

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Ghella [55]

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Factor 3x^2-19x+20 ​

Factor 3x^2-19x+20