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Alexus [3.1K]
3 years ago
7

HELP PLEASE! DUE TMR! Answer whichever ones you want and i can try to deal with the rest. I just have alot of work and need extr

a help
1) what are the roots of p(x)=64 - x^3


2)What are the solutions of x^3 (x^2 + 8x - 20)


3) find the zeros to the polynomial x^3 - 3x^2 - 10x


4) create an equation with the following roots: -4 , 4 , -3 , 2/5 (fraction)


5) create an equation with the following roots: ± 2i , -7 , 1


6) find all the zeros on the following polynomial: 2x^3 - 11x^2 - 16x + 105

Given zero: 5
Mathematics
1 answer:
Yuri [45]3 years ago
4 0

Answers

1) 4

2) 0, 2, -10

3) 0, 5, -2

Step-by-step explanation:

1) x³ - 64 = 0

x³ = 64

x = 4

2) x³(x² + 8x - 20) = 0

x³(x² + 10x - 2x - 20) = 0

x³(x(x + 10) - 2(x + 10)) = 0

x³(x + 10)(x - 2) = 0

x = 0, 2, -10

3) x³ - 3x² - 10x = 0

x(x² - 3x - 10) = 0

x(x² - 5x + 2x - 10) = 0

x(x(x - 5) + 2(x - 5)) = 0

x(x - 5)(x + 2) = 0

x = 0, 5, -2

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Find the mean, variance &amp;a standard deviation of the binomial distribution with the given values of n and p.
MrMuchimi
A random variable following a binomial distribution over n trials with success probability p has PMF

f_X(x)=\dbinom nxp^x(1-p)^{n-x}

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1

The mean is given by the expected value of the distribution,

\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}
\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}
\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}

The remaining sum has a summand which is the PMF of yet another binomial distribution with n-1 trials and the same success probability, so the sum is 1 and you're left with

\mathbb E(x)=np=126\times0.27=34.02

You can similarly derive the variance by computing \mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2, but I'll leave that as an exercise for you. You would find that \mathbb V(X)=np(1-p), so the variance here would be

\mathbb V(X)=125\times0.27\times0.73=24.8346

The standard deviation is just the square root of the variance, which is

\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834
7 0
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