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3 years ago

1. Insert a rational number and an irrational numbers between A)0.1 and 0.11

1 answer:

5 0

For this case we have the following definition:
A real number can be a rational number or an irrational number.
The rational numbers are those that can be expressed as the quotient of two integers, such as 3/4, -21/3, 5, 0, 1/2, while the irrational numbers are all the others.
Between 0.1 and 0.11 we have:
Rational number: 105/1000 = 0.105
Irrational number: 01057685429832687

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You find an interest rate of 10% compounded quarterly. Calculate how much more money you would have in your pocket if you had us

Elena-2011 [213]

Answer:

see the explanation

Step-by-step explanation:

we know that

step 1

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$r=10\%=10/100=0.10\\n=4$

substitute in the formula above

$A=P(1+\frac{0.10}{4})^{4t}$

$A=P(1.025)^{4t}$

Applying property of exponents

$A=P[(1.025)^{4}]^{t}$

$A=P(1.1038)^{t}$

step 2

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$r=10\%=10/100=0.10$

substitute in the formula above

$A=P(e)^{0.10t}$

Applying property of exponents

$A=P[(e)^{0.10}]^{t}$

$A=P(1.1052)^{t}$

step 3

Compare the final amount

$P(1.1052)^{t} > P(1.1038)^{t}$

therefore

Find the difference

$P(1.1052)^{t} - P(1.1038)^{t}$ ----> Additional amount of money you would have in your pocket if you had used a continuously compounded account with the same interest rate and the same principal.

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Andrei [34K]

Y=2 u=6

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Answer:

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Step-by-step explanation:

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kenny6666 [7]

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KonstantinChe [14]

Answer:

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