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Mila [183]
3 years ago
15

Scientists released 5 foxes into a new habitat in year 0. Each year, there were

Mathematics
1 answer:
IRISSAK [1]3 years ago
8 0

Answer:

f(x)= 5*4^x

Step-by-step explanation:

Since every year there are four times as many foxes as the year before, every year we need to multiply the previous population by 4. This leads to having to multiply by 4^{n}after n years have gone by. So we are in the presence of a population growth that goes as an exponential function of base "4".

A function f(x) that represents an exponential growth of such type can be written as:

f(x)= N_0 * 4^x

where x represents the variable time in years, 4 is the multiplicative base, and N_0 is what is defined as the "initial value" of the population (in our case 5 - for the 5 foxes initially released).

Notice also that when we start counting - time zero (x=0) the exponential form 4^0 = 1 becomes a one (1) and the expression reduces to this initial value  N_0 (5 in our case).

Therefore, the function representing this population growth is: f(x)= 5*4^x

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Answer:

By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single

Step-by-step explanation:

let p1  be the proportion of elementary teachers who were single

let p2 be the proportion of high school teachers who were single

Then, the null and alternative hypotheses are:

H_{0}: p2=p1

H_{a}: p2>p1

We need to calculate the test statistic of the sample proportion for elementary teachers who were single.

It can be calculated as follows:

\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } } where

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  • p is the proportion of elementary teachers who were single (\frac{15}{100} =0.15)
  • N is the sample size (180)

Using the numbers, we get

\frac{0.2-0.15}{\sqrt{\frac{0.15*0.85}{180} } } ≈ 1.88

Using z-table, corresponding  P-Value is ≈0.03

Since 0.03>0.01 we fail to reject the null hypothesis. (The result is not significant at α = 0.01)

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