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Alisiya [41]
3 years ago
15

Jan has 3 1/3 feet of rope to use on a camping trip. Assume 1 foot = 30.48 centimeters. What is the best estimate of the centime

ters of rope Jan has?
A. 9 cm

B. 12 cm

C. 90 cm

D. 160 cm
Mathematics
2 answers:
Otrada [13]3 years ago
7 0
D is the answer because if you calculate the answer correctly. that will be so.
jenyasd209 [6]3 years ago
6 0
<span>The answer would be 90, D I worked it out by taking the 3 feet and multiplying it by 30, in my head. I then added in the rest, after multiplying the .48 times 30. It actually is above the possible answer, but there was not another choice and that was the closest to the actual answer.</span>
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Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti
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Answer:

(a)D(x)=-2,500x+60,000

(b)R(x)=60,000x-2500x^2

(c) x=12

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500

Therefore, we have:

y=-2500x+b

At point (12,30000)

30000=-2500(12)+b\\b=30000+30000\\b=60000

Therefore:

D(x)=-2,500x+60,000

(b)Revenue

R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0

Therefore:

  • Optimal ticket price:$12
  • Maximum Revenue:$360,000

3 0
3 years ago
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