An amusement park has 24,000 visitors per day when it charges $30 per person. For each $1 increase in price, the park loses about 500 visitors. How much should the park charge to maximize daily revenue? What is the
maximum daily revenue?

R(x)=-100x^2+1200x+16000

(-8,0) and (20,0)

$14, cost that gives maximum revenue.

1400 visitors that give max rev.

x: price increase

When price = 8, visitors = 2000

For each increase(x) of 1 in price, visitors decrease by 100

price = 8 + x

visitors = 2000 - 100x

1. Revenue = price * visitors

R(x) = (8+x) (2000-100x)

R(x) = 16,000 - 800x + 2,000x - 100x²

R(x) = -100x² + 1,200x + 16,000

2. Coordinates of maximum point

R(x) is parabola curving down. Maximum point is at vertex

Vertex has x-coordinate = -b/2a

x = -1200/-200 = 6

R(6) = -100*36 + 1,200*6 + 16,000 = 19,600

Maximum point (6, 19600)

3. Admission cost / maximum revenue

Maximum revenue occurs when price increase = 6

Cost = 8 + 6 = 14

Admission cost of $14 gives maximum revenue

4. Number of visitors:

Visitors = 2000 - 100x = 2000 - 600 = 1400

hdjd

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;