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Likurg_2 [28]
3 years ago
5

Learning outcome: Solve a linear system of equations in two and three variables using algebraic methods. Read and solve word pro

blems by modeling them with systems of linear equations. Activity: Let p represent the selling price for a bluetooth device. Suppose that the demand equation for this device can be modeled as the linear function D(p) = 200- 5p and that the supply equation for the same device as S(p) = 10p. Find the equilibrium price and quantity for this device keeping in mind that at market equilibrium demand equals supply.
Mathematics
1 answer:
Talja [164]3 years ago
8 0

Answer: Equilibrium price would be $13.33 and equilibrium quantity  would be 133.3 units.

Step-by-step explanation:

Since we have given that

Demand function is stated as

D(p)=200-5p

Supply function is stated as

S(p)=10p

We need to find the equilibrium price and quantity for this device.

For equilibrium, D(p) = S(p).

so, it becomes,

200-5p=10p\\\\200=10p+5p\\\\200=15p\\\\p=\dfrac{200}{15}\\\\p=\$13.33

and when we put the value of p in any function:

Q=10\times 13.33=133.3\ units

Hence, Equilibrium price would be $13.33 and equilibrium quantity  would be 133.3 units.

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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

Part 4) Is a exponential growth function

Part 5) A=6,000(1.07)^{t}  

Part 6) \$11,802.91  

Part 7) Is a exponential growth function

Part 8) A=5,000(e)^{0.10t}    or  A=5,000(1.1052)^{t}  

Part 9)  \$13,591.41

Part 10) Natalie has the most money after 10 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

y=100x+300

see part 1)

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

Remember that

1 year is equal to 12 months

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10 years=10(12)=120 months

For x=120 months

substitute in the linear equation

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

Part 4) 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 5) Write the model equation for Sally’s situation

A=6,000(1.07)^{t}  

see the Part 4)

Part 6) 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 7) 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 8) Write the model equation for Natalie’s situation

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

see Part 7)

Part 9) 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 10) 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

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