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Gre4nikov [31]
2 years ago
12

Osama starts with a population of 1,000 amoebas that increases 30% in size every hour for a number of hours, h. The expression 1

,000(1+0.3)h finds the number of amoebas after h hours. Which statement about this expression is true?
A. It is the product of the initial population and the growth factor after h hours.


B. It is the sum of the initial population and the percent increase.


C. It is the initial population raised to the growth factor after h hours.


D. It is the sum of the initial population and the growth factor after h hours.
Mathematics
2 answers:
Blizzard [7]2 years ago
3 0

Answer: The correct option is A, itis the product of the initial population and the growth factor after h hours.

Explanation:

From the given information,

Initial population = 1000

Increasing rate or growth rate = 30% every hour.

No of population increase in every hour is,

1000\times \frac{30}{100} =1000\times 0.3

Total population after h hours is,

1000(1+0.3)^h

It is in the form of,

P(t)=P_0(t)(1+r)^t

Where P_0(t) is the initial population, r is increasing rate, t is time and [tex(1+r)^t[/tex]  is the growth factor after time t.

In the above equation 1000 is the initial population and (1+0.3)^h is the growth factor after h hours. So the equation is product of of the initial population and the growth factor after h hours.

Therefore, the correct option is A, itis the product of the initial population and the growth factor after h hours.

Bogdan [553]2 years ago
3 0

Answer:

the answer should be A

Step-by-step explanation:

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The  the probability that a camera has a lens defect given that it has a charging defect is calculated as:

P(L|C)=\frac{P(C)P(L)}{P(C)}\\\\=\frac{0.025\times 0.03125}{0.03125}\\\\=0.025

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A university is trying to determine what price to charge for tickets to football games. At a price of ​$ per​ ticket, attendance
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This question is incomplete, the complete question is;

A university is trying to determine what price to charge for tickets to football games. At a price of ​$24 per​ ticket, attendance averages 40,000 people per game. Every decrease of ​$4 adds 10,000 people to the average number. Every person at the game spends an average of ​$4 on concessions. What price per ticket should be charged in order to maximize​ revenue? How many people will attend at that​ price?

Answer:

a) price per ticket = $18

b) 55,000 will attend at that price

Step-by-step explanation:

Given that;

price of ticket = $24

so cost of ticket = 24 - x

every decrease of $4 adds 10,000 people

for $1 adds 2,500 people

Total people = 40000 + 2500x

Revenue function = people × cost + people × 4

so

R(x) = (40000 + 2500x)(24 - x) + (40000 + 2500x)4

R(x) = 960,000 - 40000x + 60000x - 2500x² + 160,000 + 10000x

R(x) = 1120000 + 30000x - 2500x²

R(x) = -2500x² + 30000x + 1120000

now for maximum revenue; R'(x) is to zero

so

R'(x) = d/dx(  -2500x² + 30000x + 1120000 ) = 0

⇒ -5000x + 30000 + 0 = 0

⇒ -5000x + 30000 = 0

⇒ 5000x = 30000

x = 30000 / 5000

x = 6

R'(X) = -5000 < 0 ⇒ MAXIMUM

∴ Revenue is maximum at x = 6

so cost of ticket ;

price per ticket = (24 - x) = 24 - 6 = $18

so price per ticket = $18

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= 40000 + (2500 × 6)

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Answer:

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Intervals of increase and decrease: increasing on (-∞, 0) and (1.25, 3.25), decreasing on (0, 1.25) and (3.25, ∞)

Positive and negative intervals: positive on (2, 4), negative on (-∞, 0), (0, 2), and (4, ∞)

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