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VARVARA [1.3K]
2 years ago
15

Select the correct answer.

Mathematics
2 answers:
Alexeev081 [22]2 years ago
6 0

Answer: The answer is C. Divide the GDP by the size of the population.

Step-by-step explanation:

faust18 [17]2 years ago
4 0
The answer is C - Divide the GDP by the size of the population.
You might be interested in
18n squared + 57n -10
aniked [119]
Most people use the decomposition method but i dont know how to do that so i use the Joes method a method similar to decomp. but easier.
canadian way

Find gcf: None
complet trinomial: 18n^2+57-10
find the product=10(18)
                       =-180
find the sum     =57
-3 and 60 goes into both meaning if you multiply 3 and60 you get -180 and if you add them you get 57.

so (n-3)(n+60)
divide by a in this case it 18
so (n<u>-3</u>)(n+<u>60</u>)
       18     18
do not divide. treat it like a fraction so you reduce it to lowest terms
(n<u>-3</u>)(n<u>-3)
</u>   18   10
<u /> at this point its reduced to lowest terms so now you take the deniminator and move it beside the "n"

=(18n-3)(10n-3)

therefore your answer is (18n-3)(10n-3)
I hoped this helped :)
3 0
3 years ago
Mrs. Smith wants to purchase new markers for
olga_2 [115]

Answer:

off course D

Step-by-step explanation:

follow me say a thanks mark me branliest

7 0
2 years ago
Am I correct? If so why?
babunello [35]

Answer: its correct

Step-by-step explanation: and its correct cuase its the answer you choose

7 0
2 years ago
Gils aunt cut his birthday cake into 33 equal pieces eighteen pieces were eaten at his birthday part what fraction of the cake w
kozerog [31]

Answer:

15/33

Step-by-step explanation:

total pieces=33

Pieces eaten=18

Left=33-18

=15

Fraction=15/33

7 0
3 years ago
Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin
Artemon [7]
"<span>Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find \bar{d}, s_{d}, the t-test statistic, and the critical values to test the claim that \mu_{d} = 0"

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find </span><span>\bar{d}
You are asked to find the mean difference between the two variables, which is given by the formula:
\bar{d} =  \frac{\sum (x - y)}{n}

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example: 
</span><span>\bar{d} =  \frac{6}{8} = 0.75</span>

B) Find <span>s_{d}
</span><span>You are asked to find the standard deviation, which is given by the formula:
</span>s_{d} =  \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }

These are the steps to follow:
1) Subtract the mean difference from each pair's difference 
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
s_{d} = \sqrt{ \frac{101.5}{8-1} }
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
t =  \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}

where SE = standard error is given by the formula:
SE =  \frac{ s_{d} }{ \sqrt{n} }

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract  \mu_{d}
6) divide by the standard error

In our example:
SE = 3.81 / √8
      = 1.346

The problem gives us <span>\mu_{d} = 0, therefore:
t = [(9.75 - 9) - 0] / 1.346</span>
  = 0.56

D) Find t_{\alpha / 2}
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find <span>t_{\alpha / 2} = 1.895</span>

Since our t-value is less than <span>t_{\alpha / 2}</span> we can reject our null hypothesis!!

7 0
3 years ago
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