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aleksandrvk [35]
3 years ago
15

Zoey purchased a new car in 1992 for \$28,000$28,000. The value of the car has been depreciating exponentially at a constant rat

e. If the value of the car was \$9,600$9,600 in the year 1996, then what would be the predicted value of the car in the year 1998, to the nearest dollar?
Mathematics
2 answers:
Andreas93 [3]3 years ago
7 0

Answer:

5621

Step-by-step explanation:

elena-s [515]3 years ago
3 0
400$
Explanation:
In 1992, the price is 28000$ and 9600$ in 1996 so in 4 years the value the car lost is 18400$
Dividing 18400 by 4 years we get 4600 depreciation value per year.
Between 1996 and 1998 there is 2 years
So in 1998 the value of the car will be:
9600 (value in 1996) - 2x4600 = 400$
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[1 -2]

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Sidana [21]

Answer:

When the ticket price is $3 or $4 the production will be in break even

Step-by-step explanation:

<u><em>The correct question is</em></u>

The revenue function for a production by a theatre group is R(t) = -50t^2 + 300t where t is the ticket price in dollars. The cost function for the production is C(t) = 600-50t. Determine the ticket price that will allow the production to break even

we know that

Break even is when the profit is equal to zero

That means

The cost is equal to the revenue

we have

R(t)=-50t^2+300t

C(x)=600-50t

Equate the cost and the revenue

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solve for t

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using a graphing tool

the solution is t=3 and t=4

see the attached figure

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When the ticket price is $3 or $4 the production will be in break even

5 0
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choli [55]

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Step-by-step explanation:

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