#1.) The population of bacteria after t hours is given by P(t) = 500/(1+83.33e^-0.162t)
a.) What is the initial population?
b.) When will the amount of bacteria in the population be 12?
#2.) Which investment yields the greater amount after 4 years?
a.) $900 at 9% interest compounded monthly
b.) $700 at 12.5% interest compounded quarterly
c.) $1050 at 5% compounded continuously
#3.) The half-life of a radioactive isotope if 4 days.
a.) Write the formula to represent the half-life after t days.
b.) if 12.6 g are present now, how much is present after 9 days?
Thank you for any help!
1a. Po = 500 / (1 + 83.33 * e^0= =
500 / (1 + 83.33) ,
500 / 84.33 = 5.935.93.
1b. 12 = 500 / (1 + 83.33 * e^-0.162t
Cross multiply:
12(1 + 83.33 * e^-0.162t) = 500,
Divide both sides by 12:
1 + 83.33 * e^-0.162t = 500 / 12 = 41.67,
83.33 * e^-0.162t = 40.67,
Divide both sides by 83.33:
e^-0.162t = 40.67 / 83.33 = 0.500,
-0.162t = ln(0.50),
-0.162t = -0.6931,
t = 4.28 hrs.
2a. Pt = Po * (1 + r)^n,
Pt = 900 *(1.0075)^48 = 1288.26,
2b. Pt = 700 * (1.03125)^16 = 1145.31,
2c. Pt = 1050 * e^(0.05*4) = 1282.47.
a. = Highest yield.
28km
84.33
1a.) To find the initial population, we can look at the equation P(t) = 500/(1+83.33e^-0.162t). The initial population refers to the population at time t=0. So, we substitute t=0 into the equation:
P(0) = 500/(1+83.33e^-0.162*0)
P(0) = 500/(1+83.33)
P(0) = 500/84.33 ≈ 5.93 (rounded to two decimal places)
Therefore, the initial population is approximately 5.93.
1b.) To find the time when the amount of bacteria in the population is 12, we can solve the equation P(t) = 12. Rearrange the equation:
12 = 500/(1+83.33e^-0.162t)
Multiply both sides by (1+83.33e^-0.162t):
12(1+83.33e^-0.162t) = 500
1+83.33e^-0.162t = 500/12
1+83.33e^-0.162t = 41.67
Now isolate the exponential term:
83.33e^-0.162t = 41.67-1
83.33e^-0.162t = 40.67
Divide both sides by 83.33:
e^-0.162t = 40.67/83.33
e^-0.162t ≈ 0.48792
To solve for t, we take the natural logarithm ln of both sides:
ln(e^-0.162t) ≈ ln(0.48792)
-0.162t ≈ ln(0.48792)
Divide both sides by -0.162:
t ≈ ln(0.48792)/-0.162
t ≈ -2.016
Therefore, the amount of bacteria in the population will be approximately 12 after 2.016 hours.
2.) To compare the investments, we need to calculate the final amount for each investment option using the compound interest formula:
a.) $900 at 9% interest compounded monthly:
The formula for compound interest is A = P(1 + r/n)^(n*t), where A is the final amount, P is the principal amount, r is the interest rate (as a decimal), n is the number of times compounded per year, and t is the number of years.
For investment a:
P = $900
r = 9% = 0.09
n = 12 (compounded monthly)
t = 4 years
A = 900(1 + 0.09/12)^(12*4) ≈ $1185.04 (rounded to two decimal places)
b.) $700 at 12.5% interest compounded quarterly:
For investment b:
P = $700
r = 12.5% = 0.125
n = 4 (compounded quarterly)
t = 4 years
A = 700(1 + 0.125/4)^(4*4) ≈ $1115.51 (rounded to two decimal places)
c.) $1050 at 5% compounded continuously:
For investment c, we can use the formula A = P * e^(r*t), where e is Euler's number.
P = $1050
r = 5% = 0.05
t = 4 years
A = 1050 * e^(0.05*4) ≈ $1241.00 (rounded to two decimal places)
Therefore, the investment yielding the greater amount after 4 years is investment c) $1050 at 5% compounded continuously with approximately $1241.00.
3a.) The formula to represent the half-life after t days is given by A(t) = A₀ * (1/2)^(t/h), where A(t) is the amount after t days, A₀ is the initial amount, t is the time elapsed, and h is the half-life.
Therefore, the formula for the half-life after t days is A(t) = A₀ * (1/2)^(t/4).
3b.) To find the amount that is present after 9 days, given that 12g is present now, we substitute the values into the half-life formula:
A(t) = A₀ * (1/2)^(t/4)
A(9) = 12 * (1/2)^(9/4)
A(9) ≈ 12 * 0.31623 (rounded to five decimal places)
A(9) ≈ 3.795
Therefore, approximately 3.795g will be present after 9 days.