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amm1812
1 year ago
9

Which function has two x-intercepts, one at (0, 0) and one at (4, 0)? f(x) = x(x − 4) f(x) = x(x + 4) f(x) = (x − 4)(x − 4) f(x)

= (x + 4)(x + 4)
Mathematics
2 answers:
Firdavs [7]1 year ago
6 0

Answer:

The answer is D on edge

Step-by-step explanation:

Fofino [41]1 year ago
5 0

If you want a polynomial to have the x-intercepts of your choice, you simply have to build a polynomial by multiplying pieces like (x-x_k) where x_k is the x-intercept you want.

So, if you want x-intercepts 0 and 4, you have to build the polynomial

(x-0)(x-4) = x(x-4)

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1 year ago
Let Xn be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is t
Firlakuza [10]

Answer:

the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)

Step-by-step explanation:

If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1

therefore

Xn =X1-X2 = X1- (n-X1) = 2X1-n

thus

E(Xn) =∑ (2*X1-n) p(X1) =  2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0

the variance will be

σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) -  n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²

since

σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4

and

σ²(X1) = E(X1²) - [E(X1)]²

n/4 = E(X1²) - (n/2)²

E(X1²) = n(n+1)/4

therefore

σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²

= n - 2n² = n(1-2n)

σ²(Xn) = n(1-2n)

4 0
1 year ago
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