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3 years ago

What issues may arise if nominal gross domestic product was used instead of real gross domestic product?

Advanced Placement (AP)

1 answer:

AVprozaik [17]3 years ago

7 0

Answer: See explanation

Explanation:

Real gross domestic product is simply refered to the economic output of a particular country which has been adjusted for price changes as inflation was taken into consideration.

Nominal gross domestic product is the measurement of the gross domestic product of a particular country which makes use of current prices, and isn't inflation adjusted.

The issue that may arise when nominal gross domestic product was used instead of real gross domestic product is that the nominal GDP leads to the inflation of the growth figure in the economy. This is because the nominal GDP doesn't take inflation into effect.

This leads to the misleading of the GDP since there'll be an overstatement of the GDP even though it was actually a rise in the inflation rate for the particular economy.

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Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

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To answer the question

Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have

For Chloe's velocity

$C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2$

Finding the boundaries of the function gives;

$0\times e^{4-0^2} = 0$ and $2\times e^{4-2^2} = 2$

At t = 1, we have $1\times e^{4-1^2} = e^{3} = 20.086$

We find the maximum point as follows;

$\frac{\mathrm{d} \left (t\times e^{4-t^2} \right )}{\mathrm{d} x}=0$

From which we have;

$\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0$

$e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0$

$e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\$

$e^{4-t^2}=0$ or $(1 - 2\times t^2 })=0$

∴ 1 = 2·t² and from which t = (√2)/2

Hence the function C(x) is decreasing from t = (√2)/2 to t = 2

For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;

For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2

and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5

Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.

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3 years ago