The number of cars sold by a car salesperson during each of the last 25 weeks is the following: Number Sold Frequency 0 10 1 10 2 5 What is the probability that the salesperson will sell one car during a week

μ=0.8

Step-by-step explanation:

The expected value of X is calculated as E(X) = μ = ∑xi P(X = xi)

Given the table below

Sold(x)0 1.2

Freq(X)10___ 105

E(X) = μ = ∑xi P(X = xi)

Total outcome is 10+10+5=25

Then, probability of each car sold is

P(0) =10/25

P(1) =10/25

P(2) =5/25

The corresponding probabilities are 10/25, 10/25 and 5/25. Thus, the expected value of X equals

μ = x1•P(X=0) + x2•P(X=1) + x3•P(X=2)

μ = 0(10/25) + 1(10/25) + 2(5/25)

μ = 0 + 0.4 + 0.4

μ = 0.8

Thus, the expected value the person will sell one car is 0.8

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

hope this is what you were asking for!

step-by-step explanation: