The amount of time, in hours, that a computer functions before breaking downis continuous random variable with exponential density functionf(x) = .01e –x/100 , x ≥ 0; f(x) = 0 x < 0.

(a) what is the probability that a computer will function more than 100 hoursbefore breaking down ? (4points)

(b) if there are four such computers what is the probability that at most 3 computerswill function more than 100 hours before breaking down ? (6points)

0.6065

0.8647

Step-by-step explanation:

The first part of this question involves the exponential distribution with λ=1/100 while the next is binomial.

a. For the exponential distribution, P[at least t] = e-λt. In this case, it's P[at least 50] = e-50/100 = 0.6065.

b. This is really binomial. Each of the 4 computers has a 0.6065 chance of surviving at least 50 hours.

For the binomial distribution, P[x survive out of n total] = n! / (x!(n-x)!)px(1 - p)(n - x), where p is the chance that one survives. The easiest way to calculate P[x < 3] = 1 - P[x = 4].

The probability that 4 out of 4 computers survive is P[4 survive out of 4 total] = 4!/(4!(4-4)!)0.60654(1-0.6065)4-4=0.60654=0.1353. Therefore, the probability that 3 or fewer survive is 1-0.1353 = 0.8647.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

his chances of getting a coin of higher value is 4/19

step-by-step explanation: