The daily linear demand curve in an internet store intercepts the price axis at 712 dollars. The slope of the linear demand curve is -4. In Uniform distribution: Mean = (b+a)/2 and Variance = [(b-a)²]/12.

In binomial distribution: Mean = np and Variance = np(1-p).

1. Compute the height, f(x), of the uniform distribution associated with this linear demand curve.

The potential answers are:

A: 0.001773

B: 0.001563

C: 0.001623

D: 0.001404

E: 0.002618

2. Compute the probability of sale if the price is set to 200 dollars.

The potential answers are:

A: 78.72%

B: 45.31%

C: 38.31%

D: 28.57%

E: 71.91%

3. Given the above price of 200 dollars, compute the expected (average) revenue from each individual visitor. The potential answers are:

A: 94.5 dollars.

B: 158.6 dollars.

C: 145.6 dollars.

D: 143.8 dollars.

E: 91.1 dollars.

4. Given the linear demand curve, compute the maximum daily demand (total number of visitors per day).

The potential answers are:

A: 141 units.

B: 160 units.

C: 154 units.

D: 126 units.

E: 178 units.

5. Given the above price of 200 dollars, and the maximum number of visitors per day, compute the average number of units sold per day.

The potential answers are:

A: 111 units.

B: 72.5 units.

C: 128 units.

D: 36 units.

E: 116 units.

6. Given the above price of 200 dollars, compute the average daily total revenue.

The potential answers are:

A: 13320 dollars.

B: 25375 dollars.

C: 22420 dollars.

D: 3240 dollars.

E: 25600 dollars.

7. Given the above price. compute the standard deviation of the number of units sold per day.

The potential answers are:

A: 6.7 units.

B: 5.8 units.

C: 6.4 units.

D: 6 units.

E: 6.9 units.