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adelina 88 [10]
3 years ago
14

The world population is estimated to be 7.8 billion in 2020, and growing at an instantaneous rate of 1.05%. (a) Estimate the pop

ulation in 2100 if the growth rate remains constant at 1.05%. (b) Estimate the population in 2100 using the logistic function, assuming a carrying capacity of 13 billion. Also, what will be the growth rate in 2100
Mathematics
1 answer:
REY [17]3 years ago
3 0

Answer:

Population in 2100 is 17.99 billion.

Step-by-step explanation:

The population of the world in 2020  = 7.8 billion.

The growth rate = 1.05%

Now find the population after 2100. Use the below formula to find the population.

Population in 2100 = Population of 2020 (1 + growth rate)^n

Population in 2100 = 7.8 (1 + 0.0105)^80

Population in 2100 = 17.99 billions.

Now, find the growth rate in 2100.

dN/dt = [r N (K – N) ] / K

r = Malthusian parameter

K = carrying capacity.

Now divide both sides by K, now x = N/K then do the differential equation.

dx/dt = r x ( 1- x)

Now integrate, x(t) = 1/ [ 1 + (1/x – 1) c^-rt

From the first equation = dN/dt = (13 – 7.8) / 80 = (r × 7.8×(13 – 7.8) / 12

0.065 = (r × 7.8× 5.2) / 12

0.065 = r × 3.38

r = 1.92%

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P^2 - 4p - 32 / p + 4, simplify the rational expression. state any restrictions
dusya [7]

Answer:

The correct answer is: p - 8 ∧ restriction p ≠ -4

Step-by-step explanation:

(p² - 4 p - 32) / ( p + 4)

The existence of this rational algebraic expression is possible only if it is:

p + 4 ≠ 0  => Restriction is p ≠ -4

(p² - 4 p - 32) / ( p + 4) = (p² - 8 p + 4 p - 32) / (p + 4) =

= (p (p -8) + 4 (p -8)) / (p + 4)= (p - 8) (p + 4) / ( p +4) = p - 8

God with you!!!

7 0
3 years ago
Read 2 more answers
Last answer option is
wel

statements 1, 3, 4 are correct


6 0
3 years ago
A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n
mafiozo [28]

Answer:

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

p_v =P(z>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

\hat p=\frac{40}{120}=0.333 estimated proportion of strawberries damaged

p_o=0.25 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

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Elan Coil [88]
The store with the larger numbers
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Does anyone know how to change their username for Brainly????? If so HELP!!!!!!!!!! Will give 20 points to anyone I really don't
Len [333]

Answer:

no I don't is it not in the user settings?

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