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.
#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!
3 answers
28km
84.33
2 answers