Of the members of the Spring Lake Bowling Lanes, 57% have a lifetime membership
and bowl regularly (three or more times a week). If 70% of the club members bowl
regularly, find the probability that a randomly selected member is a lifetime member,
given that he or she bowls regularly.
a) 0.056
b) 0.814
c) 0.533
d) 0.538

The probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly is 0.814, option b

Step-by-step explanation:

Let "M" denotes the event that a club member has a lifetime membership and "B" be denotes the event that the club member bowls.

57% of the members have lifetime membership and bowl regularly. This means, the probability that a randomly selected member has a lifetime membership and bowls regularly =

70% of the club members bowl. This means, the probability that a randomly chosen member bowls = P(B) = 70% = 0.70

We have to find the probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly. This is conditional probability which can be expressed mathematically as P(M | B)

The formula to calculate the conditional probability is:

So, the formula for  our case will be:

Using the values in this formula we get:

This means, the probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly is 0.814

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.

all you have to do is multiply them both to see how many out comes you can put:

3 x 2 = 6

you can make 6 diferent combinations of crusts and toppings.

~batmans wife dun dun