A contractor is planning the acquisition of bulldozers needed for a new project in a remote area. From prior experience with this equipment he estimates that there is a 60% chance that each bulldozer will be operational for at least six months. If he purchases four bulldozers, it is desired to calculate the probabilities that exactly: 0, 1, 2, 3, and 4 bulldozers will be operational at the end of six months. Please document the following information: a. Count the number of possible outcomes of the status of each bulldozer at the end of six months:
b. List the five events listed above as a function of the status of each bulldozer at the end of six months documented in 1.); and
c. Calculate the probabilities of these events assuming that the status of each bulldozer is statistically independent of the status of the rest of bulldozers.

0.0256 ; 0.1536 ; 0.3456 ; 0.3456 ; 0.1296

Step-by-step explanation:

P, that each bulldozer will be operational for atleast 6 months = 0.6

p = 0.6 ; q = 1 - p = 1 - 0.6 = 0.4

Number of bulldozers purchased, n = 4

Probability that :

Exactly 0 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 0) = 4C0 * 0.6^0 * 0.4^4

P(x = 0) = 1 * 1 * 0.0256

P(x = 0) = 0.0256

Probability that :

Exactly 1 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 1) = 4C1 * 0.6^1 * 0.4^3

P(x = 1) = 4 * 0.6 * 0.064

P(x = 1) = 0.1536

Probability that :

Exactly 2 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 2) = 4C2 * 0.6^2 * 0.4^2

P(x = 2) = 6 * 0.36 * 0.16

P(x = 2) = 0.3456

Probability that :

Exactly 3 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 3) = 4C3 * 0.6^3 * 0.4^1

P(x = 3) = 4 * 0.216 * 0.4

P(x = 3) = 0.3456

Probability that :

Exactly 4 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 4) = 4C4 * 0.6^4 * 0.4^0

P(x = 1) = 1 * 0.1296 * 1

P(x = 1) = 0.1296

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1