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Mumz [18]
3 years ago
12

n many population growth problems, there is an upper limit beyond which the population cannot grow. Many scientists agree that t

he earth will not support a population of more than 16 billion. There were 2 billion people on earth at the start of 1925 and 4 billion at the beginning of 1975. If yy is the population, measured in billions, tt years after 1925, an appropriate model is the differential equation
Mathematics
1 answer:
givi [52]3 years ago
3 0

Answer:

\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})

Step-by-step explanation:

The logistic function of population growth, that is, the solution of the differential equation is as follows:

P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}

We use this equation to find the value of r.

In this problem, we have that:

K = 16, P_{0} = 2, P(50) = 4

So we find the value of r.

P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}

4 = \frac{16*2e^{50r}}{16 + 2*(e^{50r} - 1)}

4 = \frac{32e^{50r}}{14 + 2e^{50r}}

56 + 8e^{50r} = 32e^{50r}}

24e^{50r} = 56

e^{50r} = 2.33

Applying ln to both sides of the equality

50r = 0.8459

r = 0.017

So

The differential equation is

\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})

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The hypotenuse of a right triangle is four times the length of one of the legs. The length of the other leg is 1500⎯⎯⎯⎯⎯⎯⎯⎯√ fee
Mrac [35]

Answer:

Step-by-step explanation:

So this is just playing with the pythagorean theorem

a^2 + b^2 = c^2 with a and b as legs and c as the hypotenuse.

Let's say a is equal to 1500 so then c is 4 times b

1500^2 + b^2 = (4b)^2  Now we just treat this like a normal algebraic expression

2,250,000 = 16b^2 - b^2

2,250,000 = 15b^2

150,000 = b^2

sqrt(150,000) = b

100sqrt(15) = b

You can check this now.

1500^2 + sqrt(150,000)^2 = c^2

sqrt(2,250,000 + 150,000) = c

sqrt(2,400,000) = c

400sqrt(15) = c Which fits the requirement of c = 4b

7 0
3 years ago
Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. Thi
Dovator [93]

Complete question:

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.

A) Find the mean and standard deviation for the number of girls in groups of 36 births.

B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.

C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?

Answer:

a) mean = 18

Standard deviation =3

b) low range = 12

High range = 24

c) The result of 33 girls is significantly high. Yes, the method is effective.

Step-by-step explanation:

Given:

p = 0.5

n = 36

a) The mean is the product of n and p

Mean u = np

u = 36 * 0.5 = 18

The standard deviation is the square root of the product of n and p&q.

S.d ó = \sqrt{npq}

= \sqrt{np(1-p)}

= \sqrt{36(0.5)(1-0.5)} = \sqrt{9} = 3

b) To find the range rule of thumb:

• For low range

Low range = u - 2ó

= 18 - (2 * 3)

= 12

• High range

= u + 2ó

= 18 + (2*3)

= 24

c) The result is significantly high, because 33 is greater than 24 girls.

A result of 33 girls would prove the method as effective.

8 0
3 years ago
-3.3=x/6 solve for x
Leni [432]

-3.3=\frac{x}{6}

Multiply both sides of the equation by 6 to isolate x.

-19.8=x

Your answer is -19.8.

I hope this helps :)

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satela [25.4K]

Answer:

x = 2/3, -2/3

Step-by-step explanation:

Solve the equation for  x  by finding  a ,  b , and  c  of the quadratic then apply the quadratic formula.

7 0
3 years ago
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timurjin [86]

Answer:

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Step-by-step explanation:

When the -4 is plugged in, it needs to be multiplied and then added to 39. That equals -45.

8 0
3 years ago
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