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3 years ago

Rico thinks of a number between 30 and 40 that is represented by

1 answer:

7 0

Hello,

x,y,z∈ IN

$x=\frac{3}{4}*y$
$y=\frac{1}{3}*z$

$140$
$46\leq\ y\ \leq\ 50 ==>\ \frac{46*3}{4}\leq\frac{3y}{4}\leq\frac{50*3}{4}$
$35\ \leq\ z\ \leq\ 37$ ==>
x∈{35,36,37}
y∈{47,48,49}
z∈{141,144,147}

We must go futher:

if x=35 then y=4*35/3=46.666 not good , not whole
if x=36 then y=4*36/3=48 ok
if x=37 then y=37*4/3=49.333.. not good

So there is only one solution (36,48,144) for (x,y,z)

No problem with z for z=3/1*4/3*x=4x ==>whole.

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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

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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.