The amount of time Mary spends on the phone each month can be thought of as a random variable that is approximately normally distributed. Mary spends an average of 40 hours on her phone with a standard deviation of 4 hours. Find the probability that Mary spends between 30 and 40 hours on the phone. ​

0.4938

Step-by-step explanation:

Given that:

Mean, m = 40

Standard deviation, s = 4

Probability of spending between 30and 40 hours on phone :

Zscore = (x - mean) / standard deviation

P(Z = 40) - P(Z= 30)

((40 - 40) / 4) - ((30-40) / 4)

(0/ 4) - (-10/ 4)

P(Z < 0) - P(Z <-2.5)

P(Z^ 0).=0.5 ; P(Z < - 2.5) = 0.0062097

0.5 - 0.0062097

P = 0.4938

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.

a

step-by-step explanation: