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3 years ago

What is the factored form of x^3 -1

Mathematics

1 answer:

lianna [129]3 years ago

8 0

Answer:

The factored form of x^3 -1 will be:

$x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)$

Step-by-step explanation:

Given the expression

$x^3-1$

Rewrite 1 as 1³

$=x^3-1^3$

$\mathrm{Apply\:Difference\:of\:Cubes\:Formula:\:}x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)$

$x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)$

$=\left(x-1\right)\left(x^2+x+1\right)$

Thus, the factored form of x^3 -1 will be:

$x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)$

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3 years ago

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Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

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Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

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(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

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Ivan

Answer:

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Step-by-step explanation:

This question makes no sense so my answer makes no sense.

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