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spin [16.1K]
3 years ago
6

Jim orders prints from a website. The site charges him $6.95 a month and $0.04 for each print he orders.

Mathematics
2 answers:
Alexxandr [17]3 years ago
8 0
P=$17.79-$6.95
P=$10.84
P=$10.84 divide by 0.04
P= 271
Jim ordered 271 prints last month
9966 [12]3 years ago
6 0

Answer:

271 prints.

Step-by-step explanation:

So to answer the question, we need to first write a formula to find the answer.

We know that the starting price per month is 6.95 dollars.

The full price is 17.79 dollars.

So we can start this off by writing:

17.79=6.95

Now, every time Jim prints, he charges 0.04 dollars. This will add to the orginal cost, which is 6.95. And remember this isnt a single payment, but a payment that varies depending on the amount of paper, p. So we can write this as:

17.79=6.95+0.04p

6.95 wont ever change, hense why it doesnt have a variable. 0.04 however, varies depending on the amount of paper, and thats why we included the variable p.

These two values combined make 17.79, hense why it is on the opposite side of the equals sign.

So, now we need to find how many papers, p, were printed.

To find this, lets just take our equation, and solve for P like an ordinary problem:

17.79=6.95+0.04p

So first off, lets subtract the 6.95 from both sides, because that iscolates the 0.04p, which contains the variable.

Now we have:

10.84=0.04p

Now, we could divide both sides by 0.04.

But remember, the whole point of doing this is to make 0.04p into 1p.

So another way of doing this is multiplying the 0.04 by 25, since 0.04*25=1.

So lets multiply both sides by 25:

10.84*25 is 271.

Since 10*25 is 250

And

.84*25 is 21.

So now we know:

271=1p

So 271 is your answer!

Hope this helps!

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2 years ago
John, Sally, and Natalie would all like to save some money. John decides that it
brilliants [131]

Answer:

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

Part 2)  y=100x+300

Part 3) \$12,300

Part 4) \$2,700

Part 5) Is a exponential growth function

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

Part 7) \$11,802.91

Part 8)  \$6,869.40

Part 9) Is a exponential growth function

Part 10) A=5,000(e)^{0.10t}    or  A=5,000(1.1052)^{t}

Part 11)  \$13,591.41

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

The slope is equal to

m=\$100\ per\ month

The y-intercept or initial value is

b=\$300

so

y=100x+300

therefore

John’s situation is modeled by a linear equation

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

see part 1)

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

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 

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute  the value of t in the exponential growth function

A=6,000(1.07)^{2}=\$6,869.40

Part 9) What type of exponential model is Natalie’s situation?

we know that

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  

P=\$5,000\\r=10\%=0.10

substitute in the formula above

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

Applying property of exponents

A=5,000(1.1052)^{t}

 therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

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

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

A=5,000(e)^{0.10*10}=\$13,591.41

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

For t=2 years

substitute

A=5,000(e)^{0.10*2}=\$6,107.01

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

Sally has the most money after 2 years

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