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Murljashka [212]
2 years ago
13

Given log4(3) =0.7925 and log4(5) =1.1610, use log properties to find log4(300).

Mathematics
1 answer:
Ede4ka [16]2 years ago
5 0

Answer:

4.1145

Step-by-step explanation:

You can use the following formula to solve this kind of problems,

log_{a}(b)  +  log_{a}(c)  =  log_{a}(bc)

So, let's start of with log4_(300). Since the question have already gave log4_(3) =0.7925 and log4_(5) =1.1610. It means that you have to split the 300 into the simplest form where almost all of the numbers are 3 and 5.

log_{4}(300)  =  log_{4}(4 \times 5 \times 3 \times 5)  \\  =  log_{4}(4)  + log_{4}(5)  + log_{4}(3)  + log_{4}(5)  \\   = 1 + 1.1610 + 0.7925 + 1.1610 \\  = 4.1145

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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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