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3 years ago

What are two ways the debt to GDP ratio can decrease?

Mathematics

2 answers:

bekas [8.4K]3 years ago

7 0

Answer:

Decrease government spending and stimulate economic growth.

Step-by-step explanation:

The debt to GDP ratio is essentially the relationship between a nation's government debt and it's productive value (goods and services produced). A lower ratio is generally considered good as that means the economy can cover its debts, but there are other important factors like the nature of the debt which influences the impact.

Decreasing government spending is the first way to lower the debt ratio. If spending is reduced, government will issue less bonds to finance that spending, thus lowering its debt.

Another way is to increase the output of the economy. The most common way to do that is to cut interest rates - it encourages commercial lending and spending, stimulating economic growth.

kobusy [5.1K]3 years ago

5 0

Think of debt to GDP as a fraction. Debt / GDP

So the two ways the ratio can decrease is if the nominator decreases (for example 2/3 goes to 1/3), or the denominator increases (1/3 goes to 1/4).
Dose this help?

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Bcc Medical Radiography program had 139 graduates. There were 25 more females than males. How many females and males graduated?

NISA [10]

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m+f=139, using f from above we have:

m+(m+25)=139

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4 years ago

2 1/3 divided by y= -1.27272727 what does y equal

Semmy [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Dennis_Churaev [7]

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4 years ago

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Lady_Fox [76]

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3 years ago

The number of fish in a small bay is modeled by the function F defined by F(t)=10 (t3 â 12t2 + 45t +100), where t is measured in

RSB [31]

The graph of the given function is attached showing characteristics of the function.

The correct responses are;

(a) F'(4) = -30 means that on day 4, the number of fish in the bay is decreasing at a rate of 30 fish per day.

(b) Over the interval 0 ≤ t ≤ 8 the absolute minimum number of fish in the bay is; 1,000.

(d) The rate of change of the number of pelicans flying near is; P' = 10·(3·c² - 24·c + 45)

Reasons:

The given function is; f(t) = 10·(t³ - 12·t² + 45·t + 100)

Where;

t = Number of days

0 ≤ t ≤ 8

(a) The derivative of the given function is presented as follows;

F'(t) = 10 × (3·t² - 2×12·t + 45) = 10·(3·t² - 24·t + 45)

Therefore;

F'(4) = 10 × (3 × 4² - 24 × 4 + 45) = -30

Therefore F'(4) = -30 means that on day 4, the number of fish in the bay is decreasing at 30 fish per day.

(b) The absolute minimum is given as follows;

At a minimum or maximum value, F'(t) = 10·(3·t² - 24·t + 45) = 0

Which gives;

3·(t - 5)·(t - 3) = 0

t = 5, or t = 3

At t = 5, we have;

f(5) = 10 × (5³ - 12 × 5² + 45 × 5 + 100) = 1,500

At t = 3, we have;

f(3) = 10 × (3³ - 12 × 3² + 45 × 3 + 100) = 1,540

Therefore, f(5) = 1,500 is a local maximum

However, at x = 0, we have;

f(0) = 10 × (0³ - 12 × 0² + 45 × 0 + 100) = 1000

At x = 8, we have;

f(8) = 10 × (8³ - 12 × 8² + 45 × 8 + 100) = 2,040

Therefore, the absolute minimum is given at t = 8, where f(t) = 1,000

bay is decreasing is between the local maximum at t = 3, and the local

minimum at t = 5, which gives;

The rate of change of the number of fish is decreasing for values of t in the range;

3 ≤ t ≤ 5

(d) P = 10·(t³ - 12·t² + 45·t + 100)

P' = 10·(3·t² - 24·t + 45)

At time t = c, we have;

P' = 10·(3·c² - 24·c + 45)

Based on a similar question online, we have;

(c) The interval over which the rate of change is decreasing

(d) The rate of change of the number of pelicans flying near the bay at t = c

Learn more about differentiation of functions here:

brainly.com/question/1422315

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2 years ago