1.the number of bacteria present in a colony is 180 at 12 noon and the bacteria grows at a rate of 22% per hour. how many will be present at 8pm?
2.A 1970 comic book has appreciated 10% per year and originally sold for $0.35. What will it be worth on 2010?
3.The cost of a high definition television now averages $1200 but the cost is decreasing about 15% per year. What will it be worth in 6 years?
4.a savings bond of $1000 pays 6.5% annual interest compounded yearly. What is the balance after the certificate matures in 5 years?
Please help me with my homework...use this nethod for each problem thanks it means alot!!!😌
Growth/decay:
Growth/decay factor:
Initial value:
Function:
Answer:
#1
growth
factor: 1.22
initial: 180
at 8pm: 180 * 1.22^8
Consider how I derived the math from the words, and work on the others. Let us know which items seem difficult.
To solve these problems using the same method, we can follow these steps:
1. Growth/Decay: Determine if the problem involves growth or decay. If the quantity is increasing over time, it is a growth problem. If the quantity is decreasing over time, it is a decay problem.
2. Growth/Decay Factor: Find the growth or decay factor, which is the percentage given in the problem plus 100% (or 1). For growth problems, the growth factor will be greater than 1, and for decay problems, the factor will be between 0 and 1.
3. Initial Value: Identify the initial value given in the problem. This is the starting quantity before any growth or decay occurs.
4. Function: Use the growth/decay factor and initial value to form an exponential function. The general form is: y = initial value * (growth/decay factor)^t, where y represents the final value, t represents the number of time periods, and the caret (^) denotes exponentiation.
5. Answer: Use the exponential function to calculate the final value by plugging in the given number of time periods into the function.
Let's solve each problem step-by-step:
1. Growth Problem:
Growth/Decay: Growth
Growth/Decay Factor: 1 + 0.22 = 1.22 (since the growth rate is 22%)
Initial Value: 180 bacteria
Function: f(t) = 180 * (1.22)^t
Answer: f(8) = 180 * (1.22)^8 = 180 * 3.50998 ≈ 631 bacteria
2. Growth Problem:
Growth/Decay: Growth
Growth/Decay Factor: 1 + 0.10 = 1.10 (since the appreciation rate is 10%)
Initial Value: $0.35
Function: f(t) = 0.35 * (1.10)^t
Answer: f(2010-1970) = 0.35 * (1.10)^40 ≈ $6.87
3. Decay Problem:
Growth/Decay: Decay
Growth/Decay Factor: 1 - 0.15 = 0.85 (since the cost is decreasing by 15%)
Initial Value: $1200
Function: f(t) = 1200 * (0.85)^t
Answer: f(6) = 1200 * (0.85)^6 ≈ $577.25
4. Growth Problem:
Growth/Decay: Growth
Growth/Decay Factor: 1 + 0.065 = 1.065 (since the interest rate is 6.5%)
Initial Value: $1000
Function: f(t) = 1000 * (1.065)^t
Answer: f(5) = 1000 * (1.065)^5 ≈ $1338.23
I hope this helps! Let me know if you have any more questions.
To solve each of these problems, we can use the concept of exponential growth or decay. I will explain the general method and then show you how to apply it to each problem.
Growth/Decay Method:
1. Determine if it is a growth or decay problem.
2. Find the growth or decay factor.
3. Determine the initial value.
4. Write the function that represents the situation.
5. Use the function to find the answer.
Let's apply this method to each problem:
1. Growth Rate of Bacteria:
a) It is a growth problem because the bacteria is increasing.
b) The growth factor is 1 + (growth rate) = 1 + 0.22 = 1.22. (Since the bacteria grows at a rate of 22% per hour, we add 1 to the percentage to get 1.22)
c) The initial value is 180 bacteria at 12 noon.
d) The function is: f(t) = (initial value) * (growth factor)^t, where t is the number of hours since 12 noon.
e) To find the answer at 8 pm (8 hours after 12 noon), substitute t = 8 into the function:
f(8) = 180 * 1.22^8 ≈ 180 * 3.345 ≈ 601.06
Therefore, there will be approximately 601 bacteria at 8 pm.
2. Comic Book Appreciation:
a) It is a growth problem because the value of the comic book is appreciating.
b) The growth factor is 1 + (growth rate) = 1 + 0.10 = 1.10.
c) The initial value is $0.35.
d) The function is: f(t) = (initial value) * (growth factor)^t, where t is the number of years since the original price.
e) To find the value in 2010 (40 years after 1970), substitute t = 40 into the function:
f(40) = 0.35 * 1.10^40 ≈ 0.35 * 44.168 ≈ $15.41
Therefore, the comic book will be worth approximately $15.41 in 2010.
3. High Definition Television Cost:
a) It is a decay problem because the cost of the television is decreasing.
b) The decay factor is 1 - (decay rate) = 1 - 0.15 = 0.85.
c) The initial value is $1200.
d) The function is: f(t) = (initial value) * (decay factor)^t, where t is the number of years since now.
e) To find the cost in 6 years, substitute t = 6 into the function:
f(6) = 1200 * 0.85^6 ≈ 1200 * 0.374 ≈ $449.28
Therefore, the television will be worth approximately $449.28 in 6 years.
4. Savings Bond Interest:
a) It is a growth problem because the balance is increasing due to interest.
b) The growth factor is 1 + (growth rate) = 1 + 0.065 = 1.065.
c) The initial value is $1000.
d) The function is: f(t) = (initial value) * (growth factor)^t, where t is the number of years until maturity.
e) To find the balance after 5 years, substitute t = 5 into the function:
f(5) = 1000 * 1.065^5 ≈ 1000 * 1.348 ≈ $1348
Therefore, the balance after 5 years will be approximately $1348.
I hope this explanation helps you solve your homework problems! Let me know if you have any further questions.