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2 years ago

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What is p-7/12>3/10 please help

1 answer:

Karo-lina-s [1.5K]2 years ago

5 0

We have such inequality
$p- \frac{7}{12}> \frac{3}{10}$ /+ $\frac{7}{12}$
$p> \frac{3}{10}+ \frac{7}{12} \\ p> \frac{3*6}{10*6}+ \frac{7*5}{12*5} \\ p> \frac{18}{60}+ \frac{35}{60} \\ p> \frac{18+35}{60} \\ p> \frac{53}{60}$ - its the answer

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3 years ago

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3 years ago

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Anni [7]

Answer:

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Step-by-step explanation:

we know that

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

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

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$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

2 years ago