Acar’s stopping distance in feet is modeled by the equation d(v)= 2.15v^2/58.4f where v is the initial velocity of the car in miles per hour and f is a constant related to friction. if the initial velocity of the car is 47 mph and f = 0.34, what is the approximate stopping distance of the car?
a. 21 feet
b. 21 miles
c. 239 feet
d. 239 miles

C is the correct answer.The approximate stopping distance of the car =239 feet.

Step-by-step explanation:

Given: A car’s stopping distance in feet is modeled by the equation

where v is the initial velocity of the car in miles per hour and f is a constant related to friction.

If the initial velocity of the car v= 47 mph and f = 0.34 then

The distance of the car

≈239 feet           [ round to nearest ones]

Therefore ,the approximate stopping distance of the car =239 feet.

correct answer:

a + bi

required rate of return = 0.166 = 16.6%

step-by-step explanation:

under the capital asset pricing model (capm), the required or expected rate of return for a particular security or asset is given using the formula:

required rate of return = risk-free rate + beta*(market risk premium)

we are given that;

risk-free rate = 4% = 0.04

beta = 1.4

market risk premium = 9% = 0.09

we simply plug these values into the above formula;

required rate of return = 0.04 + 1.4*(0.09)

required rate of return = 0.04 + 0.126

required rate of return = 0.166 = 16.6%