Understanding Decay Rate and Investment Growth

The video tutorial covers the concepts of decay rate and half-life, explaining how to quickly find one if the other is known using the formula K x T = ln(2). It provides a practical example by calculating the half-life of Plutonium-239, emphasizing the importance of converting percentage decay rates to decimal form. The tutorial also includes a financial example, demonstrating how to calculate the initial investment needed to reach a future value of $10,000 with continuous compounding interest. The video concludes with a summary of the key points discussed.

What is the relationship between decay rate and half-life in the formula discussed?

• K x T = ln(2)

• K / T = ln(2)

• K + T = ln(2)

• K - T = ln(2)

How can you find the decay rate if you know the half-life?

• Multiply natural log by half-life

• Divide natural log by half-life

• Add natural log to half-life

• Subtract natural log from half-life

What is the first step in calculating the half-life of Plutonium 239?

• Multiply decay rate by 100

• Convert decay rate to decimal

• Divide decay rate by 100

• Convert decay rate to percentage

What is the half-life of Plutonium 239 when the decay rate is 0.0028% per year?

• 48,510.52 years

• 36,132.89 years

• 24,755.26 years

• 12,377.63 years

Why is the slow decay rate of Plutonium 239 concerning?

• It makes it less effective as a fuel

• It remains hazardous for a long time

• It decays too quickly to be useful

• It is not a concern at all

What is the formula used for calculating present value with continuous compounding interest?

• P = P0 / e^(KT)

• P = P0 - e^(KT)

• P = P0 + e^(KT)

• P = P0 x e^(KT)

What is the interest rate used in the example of continuous compounding?

• 7%

• 6%

• 5%

• 4%

How many years into the future do the parents want to have $10,000 for their child?

• 18 years

• 20 years

• 10 years

• 15 years

What is the calculated present value needed to achieve $10,000 in 18 years?

• $4,395.93

• $3,395.93

• $2,395.93

• $5,395.93

What is the value of e raised to the power of 0.06 x 18 in the investment example?

• 3.9447

• 1.9447

• 2.9447

• 4.9447