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4 years ago

Multiply.(x^4 + 1)(3x^2 + 9x + 2)

Mathematics

1 answer:

barxatty [35]4 years ago

7 0

Answer: (x4 + 1) • (3x2 + 9x + 2)

Step-by-step explanation: Factoring 3x2+9x+2

The first term is, 3x2 its coefficient is 3 .

The middle term is, +9x its coefficient is 9 .

The last term, "the constant", is +2

Step-1 : Multiply the coefficient of the first term by the constant 3 • 2 = 6

Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is 9 .

-6 + -1 = -7

-3 + -2 = -5

-2 + -3 = -5

-1 + -6 = -7

1 + 6 = 7

2 + 3 = 5

3 + 2 = 5

6 + 1 = 7

Hope this helped, plz mark brainest and have a good day :)

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Find the value of ab when <br> a = 8/11 and b= 4/5<br> (Simplify your answer.)

Harrizon [31]

Step-by-step explanation:

ab = a×b

= 8/11 × 4/5

= 32/55 or 1 whole 23/32

Hope this helped and please mark brainliest if you get it right!!

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3 years ago

Read 2 more answers

A particular concentration of a chemical found in polluted water has been found to be lethal to 20% of the fish that are exposed

torisob [31]

Answer:

a) $P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

b) $P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

c) $P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

d) $E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=1-0.2=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Part a

For this case we want this probability:

$P(X=14)$

And using the mass function we have this:

$P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

Part b

For this case we want this probability:

$P(X \geq 10)$

And we can find this using the complement rule:

$P(X\geq 10) = 1-P(X$

$P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}$

$P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}$

$P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}$

$P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}$

$P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}$

$P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}$

$P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}$

$P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}$

$P(X=8)=(20C8)(0.8)^{8} (1-0.8)^{20-8}=8.65x10^{-5}$

$P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046$

And if we replace we got:

$P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

Part c

For this case we want this probability:

$P(X \leq 16)$

And we can use the complement rule like this:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115$

And if we replace we got:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

Part d

The expected value is given by:

$E(X)= np = 20 *0.8 = 16$

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

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(attached is a graph of the data)

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St John's | 174000 | 173000
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Trois-Riv. | 140000 | 138000

(attached is a graph of the data)

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