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Romashka [77]
2 years ago
11

A biologist is researching the population density of antelopes near a watering hole. The biologist counts 32 antelopes within a

radius of 34 km of the watering hole. What is the population density of antelopes? Enter your answer in the box. Use 3.14 for pi and round only your final answer to the nearest whole number.
Mathematics
1 answer:
Setler [38]2 years ago
7 0

Answer:

=0.3 per km²

Step-by-step explanation:

Population density is the population per uni area.

Area=πr² Where A is area and r is the radius of the circle.

A=3.14×34

=106.76km²

Population density = Population/Area

=32 antelopes/106.76 km²

=0.3 per km²

The population of antelopes within the given radius is 0.3 per km²

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attashe74 [19]

Answer:

the answer is E, right?

Step-by-step explanation:

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The figure shows the front side of a purse in the shape of a trapezoid.
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For this case, what you must do is take out the area of two triangles and add it to the area of a rectangle to find the total area.
 We have then:
 Triangle area:
 A = (1/2) * ((13-9) / (2)) * (10) = 10 in ^ 2
 Rectangle area:
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A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe
d1i1m1o1n [39]

Answer:

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

df=n-1=15-1=14  

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

\bar X=5.63 represent the sample mean  

s=1.61 represent the standard deviation for the sample  

n=15 sample size  

\mu_o =15/tex] represent the value that we want to test  
[tex]\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5.63  

Alternative hypothesis:\mu > 5.63  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=15-1=14  

P value

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

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Consider the graph of the function f(x) = 25
trasher [3.6K]

Considering it's horizontal asymptote, the statement describes a key feature of function g(x) = 2f(x) is given by:

Horizontal asymptote at y = 0.

<h3>What are the horizontal asymptotes of a function?</h3>

They are the limits of the function as x goes to negative and positive infinity, as long as these values are not infinity.

Researching this problem on the internet, the functions are given as follows:

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The limits are given as follows:

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\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 2(2)^x = 2(2)^{\infty} = \infty

Hence, the correct statement is:

Horizontal asymptote at y = 0.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

#SPJ1

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Answer:

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