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2 years ago

Factor: 3x^2 - 288y^2 - 36

Mathematics

2 answers:

erica [24]2 years ago

7 0

Answer:

3x^2-288y^2-36

Rewrite -36 as 12×3

Rewrite -288 as 96×3

3x^2+96.3y^2+12.3

factor out common term 3

=3(x^2-96y^2-12)

ZanzabumX [31]2 years ago

6 0

Answer:

3(x^2-96y^2-12)

Step-by-step explanation:

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Answer:

$(a)\ P(x = 0) = 0.2725$

$(b)\ P(x \ge 1) =0.7275$

$(c)\ P(x \le 2) = 0.8948$

Step-by-step explanation:

Given

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$p = 15\%$ --- proportion that one-time fling

This question is an illustration of binomial probability, and it is represented as:

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

Solving (a): P(x = 0) --- None has done one time fling

$P(x = 0) = ^8C_0* (15\%)^0 * (1 - 15\%)^{8-0}$

$P(x = 0) = 1* 1 * (1 - 0.15)^{8}$

$P(x = 0) = 0.85^8$

$P(x = 0) = 0.2725$

Solving (b): $P(x \ge 1)$

To do this, we make use of compliment rule:

$P(x = 0) + P(x \ge 1) =1$

Rewrite as:

$P(x \ge 1) =1 - P(x = 0)$

$P(x \ge 1) =1 - 0.2725$

$P(x \ge 1) =0.7275$

Solving (c): $P(x\le 2)$ --- Not more than 2 has one time fling

This is calculated as:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

We have:

$P(x = 0) = 0.2725$

$P(x = 1) = ^8C_1* (15\%)^1 * (1 - 15\%)^{8-1}$

$P(x = 1) = 8* (0.15) * (1 - 0.15)^7$

$P(x = 1) = 0.3847$

$P(x = 2) = ^8C_2* (15\%)^2 * (1 - 15\%)^{8-2}$

$P(x = 2) = 28* (0.15)^2 * (1-0.15)^6$

$P(x = 2) = 0.2376$

So:

$P(x\le 2) = P(x = 0) + P(x =1) + P(x = 2)$

$P(x \le 2) = 0.2725 + 0.3847 + 0.2376$

$P(x \le 2) = 0.8948$

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2 years ago

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2 years ago

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2 years ago

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Answer:

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Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

$P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92$

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of $P(A\cap B\cap C\cap D)$ .

If two events X and Y are independent, then

$P(X\cap Y)=P(X)\times P(Y)$

Assume the probability of one part working does not depend on the functionality of any of the other parts.

$P(A\cap B\cap C\cap D)=P(A)\times P(B)\times P(C)\times P(D)$

Substitute the given values.

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2 years ago