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3 years ago

Innovative School System

Mathematics

1 answer:

svp [43]3 years ago

7 0

Answer:

is not , something other than red, 45%, 55%

Step-by-step explanation:

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The Students at Winwood elementary school collected 574 cans of food in 20 days for a food drive. What was average number of can

Nonamiya [84]

Answer:

About 29

Step-by-step explanation:

575 cans of food / 20 days = About 28.75 cans a day, rounded to 29.

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3 years ago

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12. Sean lives 2.25 miles<br> from school. Write this<br> distance as a fraction in simplest form.

mars1129 [50]

It would be 2 1/4 because .25 in fraction form is 1/4 because it is a quarter and then 2

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3 years ago

If I roll a standard number cube (labeled 1 to 6), and I roll it twice, what is the probability that I roll an odd number each t

suter [353]

Answer:

It is a 50% chance that you will roll and odd number each time.

Step-by-step explanation:

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3 years ago

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Plz help me well mark brainliest if correct!!...

valina [46]

Answer:

You should stick with option D

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3 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

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

Part A) Annual

in this problem we have

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

substitute in the formula above

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

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

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

3 years ago