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grandymaker [24]
3 years ago
7

What is the growth factor of the following example? Assume time is measured in the units given.

Mathematics
2 answers:
qwelly [4]3 years ago
7 0
The answer is B because 1% means 0.01
Bad White [126]3 years ago
7 0
B. 0.01 per day !

Because since it’s a 1 digit there has to be a “0” in front of it and then your number!
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A manufacturer has been selling 1400 television sets a week at $450 each. A market survey indicates that for each $21 rebate off
GuDViN [60]

Correct question is;

A manufacturer has been selling 1400 television sets a week at $450 each. A market survey indicates that for each $21 rebate offered to a buyer, the number of sets sold will increase by 210 per week

A) Find the demand function (price p as a function of units sold x).

p(x)=______

(B) How large a rebate should the company offer the buyer in order to maximize its revenue?

$=_____

Answer:

A) p(x) = (-1/10)x + 590

B) Rebate = $170

Step-by-step explanation:

A) We are told that p(x) is the demand function and x is the number of TV sets sold per week.

Now, since for each $21 rebate offered, the number of sets sold increases by 210 per week, it means that the slope here of this demand function is; m = -21/210 = -1/10

Now, he has been selling 1400 TV sets a week at $450 each. This means; p(1400) = $450

Thus, the demand function will be;

p(x) - 450 = (-1/10)(x - 1400)

Expanding the RHS;

p(x) - 450 = (-1/10)x + 140

Add 450 to both sides to get;

p(x) = (-1/10)x + 140 + 450

p(x) = (-1/10)x + 590

B) Formula for revenue is;

R = price × quantity sold

Our demand function is p = (-1/10)x + 590

Making x the subject, we have;

x = 5900 - 10p

x is quantity sold.

Thus,

R = p(5900 - 10p)

R = 5900p - 10p²

Maximum price will occur at dR/dP = 0

Thus;

dR/dP = 5900 - 20p

At dR/dP = 0,we have;

20p = 5900

p = 5900/20

p = $280

Thus, rebate = 450 - 280 = $170

6 0
2 years ago
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