The temperature at 5 PM was -4 degrees. By 5 AM the temperature had decreased 8 degrees. What was the temperature at 5 AM?

Now suppose there are more goods and services and the economy is now open to trade with other countries. Based on this generated

OLEGan [10]

It should be noted that GDP deflator simply measures the changes in price for goods and services.

Your information is incomplete as the figures aren't given. Therefore, an overview will be given.

Gross domestic product simply means the total monetary value of the finished goods and services that are produced in a country.

The formula for calculating the GDP deflator will be:

= Nominal GDP/Real GDP × 100

Also, the formula fro calculating real GDP per capita will be:

= Real GDP/Total population

In conclusion, the real GDP is the GDP of a country that has been adjusted for inflation.

Learn more about GDP on:

brainly.com/question/1383956

3 0

2 years ago

Ruby's math teacher finds that there's roughly a linear relationship between the

Yuri [45]

It’s y= 61 because if 55 + 6 = 61 I need the points please

8 0

4 years ago

You walk 3 miles from your house to the store. At the store you meet up with a friend and walk with her 1 mile back towards your

jenyasd209 [6]

Answer:

2 miles from your house

Step-by-step explanation:

you walked 3 miles to the store so you are now 3 miles away.

you meet a friend and walk 1 mile in the direction of your house.

now there is 2 miles left.

3 - 1 = 2

5 0

3 years ago

There are 200 students in eleventh grade high school class. There are 40 students in the soccer team and 50 students in the bask

bekas [8.4K]

Answer:

P(A) = 0.2

P(B) = 0.25

P(A&B) = 0.05

P(A|B) = 0.2

P(A|B) = P(A) = 0.2

Step-by-step explanation:

P(A) is the probability that the selected student plays soccer.

Then:

$P(A)=\dfrac{40}{200}=0.2$

P(B) is the probability that the selected student plays basketball.

Then:

$P(B)=\dfrac{50}{200}=0.25$

P(A and B) is the probability that the selected student plays soccer and basketball:

$P(A\&B)=\dfrac{10}{200}=0.05$

P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:

$P(A|B)=\dfrac{P(A\&B)}{P(B)}=\dfrac{10}{50}=0.2$

P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.

4 0

4 years ago

Given that (y+z), (y+3) and (2y^2-1) are consecutive terms of an arithmetic progression. Find the possible values of y

Sphinxa [80]

Answer:

This problem actually says:

(y + 2), (y + 3) and (2*y^2 - 1) are consecutive terms of an arithmetic progression.

Now, the difference between two consecutive terms in an arithmetic progression is always the same, so we have:

(y + 3) - (y +2) = D

(2*y^2 - 1) - (y + 3) = D

From the first equation we have:

y + 3 - y - 2 = 1 = D

Now we can replace it in the other equation:

2*y^2 - 1 - y - 3 = 1

2*y^2 - y - 5 = 0

Now we need to solve that equation to find the possible values of y.

To solve a quadratic equation of the form:

a*x^2 + b*x + c = 0, we can use the Bhaskara's equation:

$y = \frac{-b +- \sqrt{b^2 -4*a*c} }{2*a}$

in this case, the solutions are:

$y = \frac{1 +-\sqrt[]{(-1)^2 - 4*2*(-5)} }{2*2} = \frac{1 +- \sqrt{61} }{4} = \frac{1+-6.4}{4}$

Then the possible values of y are:

y = (1 + 6.4)/4 = 1.85

y = (1 - 6.4)/5 = -1.35

5 0

4 years ago