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Anna71 [15]
4 years ago
10

A population of insects grows exponentially. Initially, there were 20 insects and the population of the insects grew by 50% ever

y week. Use a function P(t)P(t) that represents the insect population at the end of t weeks. What is the insect population at the end of week 12?
Mathematics
2 answers:
anzhelika [568]4 years ago
8 0
I just took this test, the answer is 2595
vladimir2022 [97]4 years ago
6 0
P(t) = Po(1 + r)^t; where Po is the initial population = 20. r is the rate = 50% = 0.5 and t is the time = 1 weeks

P(t) = 20(1 + 0.5)^12 = 20(1.5)^12 = 20(129.7) = 2,594.9 ≈ 2,595
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Jalyn plans to sell Algebra Nation t-shirts as a fundraiser for a new scholarship initiative. The wholesale t-shirt company char
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3 years ago
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0
3 years ago
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balu736 [363]

Answer:

=3.162

Step-by-step explanation:

x2  

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2×1

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5 0
3 years ago
33 1/3% of 360 please help I don’t understand
Mars2501 [29]

Answer:

33 1/3% of 360 is 120

Step-by-step explanation:

33 1/3 can be converted to 33.333% or simply just 1/3, divide 360 by 3 to get your answer which is 120.

3 0
3 years ago
Read 2 more answers
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