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Katyanochek1 [597]

3 years ago

10

How do you calculate a nation's per capita income? A. Divide the total income by the size of the population. B. Divide the size

of the population by the GDP. C. Divide the GDP by the size of the population.

2 answers:

nikdorinn [45]3 years ago

5 0

Answer:

Divide the total income by the size of the population.

Step-by-step explanation:

The formula is GDP/Population is : Divide the total income by the size of the population.

Per capita income is defined as a ratio of the amount of all of a region's income divided by its population.

Hence, a nation's per capita income, is calculated by dividing the area's total income by its total population.

lara [203]3 years ago

3 0

Answer:

A

Step-by-step explanation:

The per capita income is the income of a person in a nation. The per capita is calculated by dividing the gross income of the nation by total population.

Let

Pci = Per Capita Income

I = Total income

and

P = Total population

Then,

Pci = I/P

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3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

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Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

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Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

<u>1) Finding $(\overline{x},\overline{y})$ </u>

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

<u>2) Finding the line through $(\overline{x},\overline{y})$ with slope m.</u>

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

<u>3) Find the squared vertical distances between this line and the three points.</u>

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

<u>4) Find the value of m that makes R minimum.</u>

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$

$\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)$

now to find the minimum value we'll just use a condition that $\dfrac{dR}{dm}=0$

$0=2(m-1)+2(2-m)(-1)$

now solve for m:

$0=2m-2-4+2m$

$m=\dfrac{3}{2}$

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

5 0

3 years ago

The half life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am

Veronika [31]

Using an exponential function, it is found that:

a) $N(t) = 75(0.5)^{\frac{t}{3.8}}$

b) 37.5 grams of the gas remains after 3.8 days.

c) The amount remaining will be of 10 grams after approximately 11 days.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

$A(t) = A(0)(1 - r)^t$

In which:

A(0) is the initial value.

r is the decay rate, as a decimal.

Item a:

We start with 75 grams, and then work with a half-life of 3.8 days, hence the amount after t daus is given by:

$N(t) = 75(0.5)^{\frac{t}{3.8}}$

Item b:

This is N when t = 3.8, hence:

$N(t) = 75(0.5)^{\frac{3.8}{3.8}} = 37.5$

37.5 grams of the gas remains after 3.8 days.

Item c:

This is t for which N(t) = 10, hence:

$N(t) = 75(0.5)^{\frac{t}{3.8}}$

$10 = 75(0.5)^{\frac{t}{3.8}}$

$(0.5)^{\frac{t}{3.8}} = \frac{10}{75}$

$\log{(0.5)^{\frac{t}{3.8}}} = \log{\frac{10}{75}}$

$\frac{t}{3.8}\log{0.5} = \log{\frac{10}{75}}$

$t = 3.8\frac{\log{\frac{10}{75}}}{\log{0.5}}$

$t \approx 11$

The amount remaining will be of 10 grams after approximately 11 days.

More can be learned about exponential functions at brainly.com/question/25537936

4 0

2 years ago