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3 years ago

Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0

1 answer:

5 0

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: $x^4+9x^2-4x^2-36$

Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses: $x^2(x^2+9)-4(x^2+9)$

Now you can rewrite this as $(x^2-4)(x^2+9)$ , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is $a^2-b^2=(a+b)(a-b)$ . Applying that here, we have $(x+2)(x-2)(x^2+9)$ . x^4 + 5x^2 - 36 is completely factored.

Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x: $2x^2+10x-x-5$

Next, factor 2x^2 + 10x and -x - 5 separately. Make sure that they have the same quantity on the inside: $2x(x+5)-1(x+5)$

Now you can rewrite the equation as $(2x-1)(x+5)$ . 2x^2 + 9x - 5 is completely factored.

<h3><u>Putting it all together, your factored expression is $(x+2)(x-2)(x^2+9)(2x-1)(x+5)=0$ </u></h3>

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Answer:

a) Figure and code attached

b) $E(X)=\mu = np = 8*0.2= 1.6$

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

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And we can calculate this with the complement rule.

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

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

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".

Solution to the problem

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

$X \sim Binom(n=8, p=0.2)$

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

We can use the following R code to generate the histogram for this case:

> x <- seq(0,8,by = 1)

> y <- dbinom(x,8,0.2)

> plot(x,y,type = "h",main="Histogram")

And as we can see we got the result on the figure attached. And the distribution seems to be skewed to the right.

Part b

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$E(X)=\mu = np = 8*0.2= 1.6$

The variance is given by:

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

And the standard deviation would be:

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

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For this case we have the following inequality:

$P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

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

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

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