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eduard
3 years ago
13

If interest rates are at a level of 1% and expected inflation is 2%, would you prefer saving or spending your money?

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
6 0
<span>If interest rates are at a level of 1% and expected inflation is 2%, it would be preferable to spend your money instead of saving it.

</span>Suppose<span> you have $100 and you save it in a savings account that pays a 1% interest rate. After a year, you will have $101 in your account.

During this period, if inflation runs 2%, you would have to have $102 to make up for the impact of higher prices.

Since you will only have $101 in your account, you have actually lost some purchasing power.

If your savings don’t grow to reflect this rise in prices over time, the effect will be as though you are actually losing money.

This means that if you have $100 which you can use to buy a TV set, and you saved the money instead is a savings account that pays 1% interest.
After 1 year, because of inflation of 2%, the TV set now costs $102 whereas the money in your bang account wil be $101.

Thus, you actually need to get an extra $1 from somewhere to fund the TV set you could have been able to buy a year ago.
</span>
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Add, -11.5x and move it over to 8.5x

Then, subtract them together and you get 20x + 51.86 = 51.76

And you move (subtract) 51. 86 over to 51.76, subtract that and you get             20x = -0.1

Divide, and your answer is -200
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3 years ago
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Answer:

?

Step-by-step explanation:

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) The density of oil in a circular oil slick on the surface of the ocean at a distance of r meters from the center of the slick
Feliz [49]

Answer:

Therefore the mass of the of the oil is 409.59 kg.

Step-by-step explanation:

Let us consider a circular disk. The inner radius of the disk be r and the outer diameter of the disk be (r+Δr).

The area of the disk

=The area of the outer circle - The area of the inner circle

= \pi (r+\triangle r)^2- \pi r^2

=\pi [r^2+2r\triangle r+(\triangle r)^2-r^2]

=\pi [2r\triangle r+(\triangle r)^2]

Since (Δr)² is very small, So it is ignorable.

∴A=2\pi r\triangle r

The density \delta (r)= \frac{40}{1+r^2}

We know,

Mass= Area× density

        =(2r \pi \triangle r)(\frac{40}{1+r^2}})

Total mass M=\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i

Therefore

\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i=\int_0^5 \frac{80r\pi }{1+r^2}dr

                      =40\pi[ln(1+r^2)]_0^5

                      =40\pi [ln(1+5^2)-ln(1+0^2)]

                     =40\pi ln(26)

                     = 409.59 kg (approx)

Therefore the mass of the of the oil is 409.59 kg.

3 0
3 years ago
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