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3 years ago

Plz I need some help this is a quick one and I'm having trouble focusing!

Mathematics

2 answers:

Ivenika [448]3 years ago

3 0

Answer:

Juan:34

Kim:30

Larry:36

Step-by-step explanation:

Alexus [3.1K]3 years ago

3 0

Answer:

Kim = 30

Juan = 34

Larry = 36

Step-by-step explanation:

30+34+36=100

if Kim has 30 cards and Juan has 4 more than her while Larry has 6 more then Juan would have 34 and Larry would have 36 making this equation true.

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2 years ago

A set of equations is given below:

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3 years ago

A frame is marked down by 22 percent from an original price of $14.50. What is the new price?

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3 years ago

Read 2 more answers

A plot of land has been surveyed for a new housing development with borders AB, BC, DC, and DA. The plot of land is a right trap

jolli1 [7]

Answer:

90 approximately because if AB is 80 then DC should be 90 because AB come before DC in letters

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2 years ago

Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti

spayn [35]

Answer:

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

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

(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