Determine the amount of time it would take for an investment to double if interest is earned at a rate of 3.75% compounded continuously. Use a graph to find your answer.
A. 1.85 years
B. 1.93 years
C. 18.48 years
D. 18.83 years
E. 19.80 years*
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Cobalt-60 is used for radiotherapy. It has a half-life of 5.26 years. If 4 g of cobalt-60 is administered, how much remains in 3 years?
A. 1.2 g*
B. 2.7 g
C. 3.3 g
D. 2.1 g
E. 0.2 g
Could someone kindly help me double check these answers to these questions? Thank you all so much!!!!!! :) :D πβ»βΊπ γ :-)π
For those wondering, the correct answers to these questions were actually 18.48 and 2.7 respectively
They both turned out to be right when I submitted my assessment or assignment! Thank you so much, oobleck, for your help!!!!!!!! :) :D π
#1. You want t where
e^(.0375t) = 2
Not E. What calculation did you use?
#2. Not A
since less than one half-life has passed, more than half of the original amount must remain. What calculation did you use?
So #1 is C or C., and #2 is B or B.?
Because I am very confident in these two new answers of mine! Hehe.
Sure! Let's take a look at the questions:
Question 1: To determine the amount of time it would take for an investment to double, we can use the formula for continuous compounding: A = P * e^(rt), where A is the final amount, P is the initial principal, r is the interest rate, and t is the time in years.
Let's set up the equation: 2 = e^(0.0375t), where 0.0375 is the interest rate of 3.75% in decimal form.
To solve for t, we can take the natural logarithm (ln) of both sides of the equation: ln(2) = ln(e^(0.0375t)).
Using the property of logarithms, we can bring down the exponent: ln(2) = 0.0375t * ln(e).
Since ln(e) is equal to 1, we can simplify the equation: 0.0375t = ln(2).
Now, we can solve for t: t = ln(2) / 0.0375 β 18.48 years.
Therefore, the correct answer is C: 18.48 years.
Question 2: The half-life of cobalt-60 is 5.26 years, which means that after each half-life, the amount remaining is reduced by half.
To find out how much remains in 3 years, we need to determine the number of half-lives that have passed. Dividing the time elapsed (3 years) by the half-life (5.26 years), we get 3 / 5.26 β 0.57.
Since half of the original amount remains after each half-life, we can calculate the remaining amount by multiplying the original amount (4 g) by 0.5 raised to the power of the number of half-lives: 4 * (0.5)^0.57 β 1.2 g.
Therefore, the correct answer is A: 1.2 g.
I hope that helps! Let me know if you have any more questions. :)
Sure! I can definitely help you with that.
Question 1: Determine the amount of time it would take for an investment to double if interest is earned at a rate of 3.75% compounded continuously.
To find the amount of time it takes for an investment to double, we can use the formula for continuous compound interest:
A = P * e^(rt)
Where:
A = the final amount or future value
P = the initial amount or present value
e = the base of the natural logarithm, approximately 2.71828
r = the interest rate
t = time in years
In this case, since we want to find the time it takes for the investment to double, we can set A = 2P.
2P = P * e^(rt)
Next, we can cancel out the common factor of P:
2 = e^(rt)
To solve for t, we take the natural logarithm of both sides:
ln(2) = rt
Finally, we can solve for t by dividing both sides by r:
t = ln(2) / r
Substituting the given interest rate of 3.75% (or 0.0375 in decimal form) into the equation, we get:
t = ln(2) / 0.0375
We can use a calculator or graphing software to find the value of ln(2) / 0.0375. In this case, it is approximately 18.48 years.
Therefore, the correct answer is option C: 18.48 years.
Question 2: Cobalt-60 has a half-life of 5.26 years. If 4 g of cobalt-60 is administered, how much remains in 3 years?
To solve this question, we can use the formula for exponential decay:
N(t) = Nβ * (1/2)^(t / T)
Where:
N(t) = the amount remaining after time t
Nβ = the initial amount
t = time in years
T = the half-life of the substance
In this case, Nβ is given as 4 g, t is given as 3 years, and T is given as 5.26 years. Plugging these values into the formula, we get:
N(3) = 4 * (1/2)^(3 / 5.26)
Evaluating this expression, we find that N(3) is approximately 1.19 g.
Therefore, the amount of cobalt-60 remaining after 3 years is approximately 1.19 g.
The closest answer option to this value is option A: 1.2 g, which is correct.
I hope this helps! If you have any further questions, feel free to ask.