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3 years ago

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When a class took a math test, 15% of the class failed, 25% made some mistakes (but didn’t fail), and 24 students got perfect sc

ores. How many students were in the class?

2 answers:

gavmur [86]3 years ago

7 0

In the class there would be a total of 40 students

Degger [83]3 years ago

4 0

15%+25%=40%, meaning that 40% of the class didn’t get a perfect score and 60% did. If 24 students got a perfect score and 24 students is equal to 60% of the class, you can solve by setting up a proportion.

24 students X students
——————- = —————-
60% of class 100% of class

Next, cross multiply to get 2400 = 60X. Divide both sides by 60 and you’ll get 40. There are 40 students in the class.

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<span>Karen got an average of 0.444 hits out of 45 bats.
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Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

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Part 9) Is a exponential growth function

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Part 12) $\$6,107.01$

Part 13) Natalie has the most money after 10 years

Part 14) Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

$y=mx+b$

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$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

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$y=100x+300$

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John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

$y=100x+300$

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

$10\ years=10(12)=120 months$

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

so

$2\ years=2(12)=24\ months$

For x=24 months

substitute in the linear equation

$y=100(24)+300=\$2,700$

Part 5) What type of exponential model is Sally’s situation?

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

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}\\ A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

$A=6,000(1.07)^{t}$

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

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$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 9)

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For t=10 years

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Part 12) How much money will Natalie have after 2 years?

For t=2 years

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Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

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