1. A certain manufacturing process to build phone batteries results in the production of defective batteries at a rate of 5% (0.05). Suppose that 10 batteries are produced. a. Find the expected number of defective batteries. b. Find the variance and standard deviation in the number of defective batteries. c. Find the probability that at least 1 battery is defective.

0.5 ; 0.475 ; 0.689 ; 0.4013

Step-by-step explanation:

Given that:

Rate of production of defective batteries p = 0.05

Number of batteries produced (n) = 10

The expected number of defective batteries = mean = n * p = 10 * 0.05 = 0.5 batteries

Variance of defective batteries :

Var(X) = n * p * q ; q = 1 - p

Hence,

Var(X) = 10 * 0.05 * 0.95 = 0.475

Standard deviation (X) = sqrt(variance) = sqrt(0.475) = 0.689

Probability that atleast 1 battery is defective :

Using the binomial probability function

P(x ≥ 1) = 1 - p(x = 0)

= 1 - q^n

= 1 - 0.95^10

= 1 - 0.59873693923837890625

= 0.40126306076162109375

= 0.4013

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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