Question

(a) which curve fits the data best?
figure 1
figure 2
figure 3
(b) use the equation of the best fitting curve from part (a) to predict the amount
of the radioactive substance after 70 days. round your answer to the nearest
hundredth.
milligrams

Explanation:

Step1: Analyze the nature of radioactive decay

Radioactive decay is an exponential process. The general form of an exponential function is \(y = a(b)^x\) where \(a\) is the initial amount and \(b\) (\(0 < b<1\)) is the decay factor.

Step2: Check the functions for the form of radioactive decay

• For \(y = 516(0.98)^x\), it is in the form of an exponential function \(y = a(b)^x\) with \(a = 516\) and \(b=0.98\) (\(0 < 0.98<1\)), which is consistent with the form of radioactive - decay models.
• For \(y = 0.03x^{2}-8x + 575\), it is a quadratic function. Quadratic functions \(y=ax^{2}+bx + c\) (\(a

eq0\)) do not model exponential decay (radioactive decay).

• For \(y=-0.26x + 380\), it is a linear function of the form \(y = mx + b\) (\(m=-0.26\), \(b = 380\)). Linear functions do not model exponential decay (radioactive decay).

Step3: Predict the amount after \(x = 70\) days using the exponential function

Substitute \(x = 70\) into \(y = 516(0.98)^x\).

$$y=516\times(0.98)^{70}$$

First, calculate \((0.98)^{70}\). Using the formula \(a^{n}=e^{n\ln(a)}\), \(\ln(0.98)\approx-0.0202\), and \(n = 70\). Then \(n\ln(a)=70\times(- 0.0202)=-1.414\), and \(e^{-1.414}\approx0.243\).

$$y = 516\times0.243$$

$$y=516\times0.243=125.388\approx125.39$$

Answer:

(a) Figure 1; (b) \(125.39\) milligrams