A car insurance company has determined that 8% of all drivers were involved in a car accident last year. If 14 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?

0.0958

Step-by-step explanation:

Percentage involved in accident = 8% = 0.08

Number of drives selected = 14

Probability that 3 or more who were involved in a car accident last year?

P(x>=3) = P(x=3) or P(x>3)

Similarly;

P(x>=3) = 1-P(0<= x <=2)

P(x>=3) = 1-P(0<= x <=2)

Using the binomial probability calculator :

Cummulative probability :

P(X >= x) = 0.09583481948

= 0.0958

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)