Question

learning goal from lesson 14.1
i can represent and interpret categorical and quantitative data based on two variables (independent and dependent).
represent means:

• i can show two - variable data on a scatter plot.
• i can describe the relationship between the variables.
• i can identify a function of best fit for the data set.

lesson reflection (circle one)
starting... getting there... got it!
lesson 14.1 checkpoint

• once you have completed the above problems and checked your solutions, complete the lesson checkpoint below.
• complete the lesson reflection above by circling your current understanding of the learning goal.

1. a new medication is being studied to see how quickly it is metabolized in the body. starting with an initial dose of 15 milligrams in the bloodstream, it declined to 10.7 milligrams after 5 hours. complete the table to determine which of these equations could approximate the amount of milligrams.

equation	yes	no
$m = 10.7(0.93)^h$	$square$	$square$
$m = 15(0.93)^h$	$square$	$square$
$m = 15(0.71)^h$	$square$	$square$

1. the data gives the estimated value in dollars of a model of a classic car over several years.

15300

16100

17300

18400

19600

20700

22000

find an approximate exponential model for the car’s value by averaging the successive ratios of the value. you can solve by using a scientific calculator or use the exponential regression tool.

Explanation:

Step1: Analyze problem 1

We know the initial amount $m_0 = 15$ milligrams and at $h = 5$, $m = 10.7$ milligrams. We substitute $h = 5$ into each equation to check.

Step2: Check $m = 10.7(0.71)^h$

When $h = 5$, $m=10.7\times(0.71)^5=10.7\times0.19847171\approx2.13$ which is not $10.7$, so No.

Step3: Check $m = 10.7(0.93)^h$

When $h = 5$, $m = 10.7\times(0.93)^5=10.7\times0.69568801\approx7.44$ which is not $10.7$, so No.

Step4: Check $m = 15(0.93)^h$

When $h = 5$, $m = 15\times(0.93)^5=15\times0.69568801\approx10.44\approx10.7$ (approximate match), so Yes.

Step5: Check $m = 15(0.71)^h$

When $h = 5$, $m = 15\times(0.71)^5=15\times0.19847171\approx2.98$ which is not $10.7$, so No.

Step6: Analyze problem 2

To find an exponential model $y = ab^x$, we first find the successive - ratios of the values to estimate the growth factor $b$. Then we use the initial value as $a$.

Answer:

1.

• For $m = 10.7(0.71)^h$: No
• For $m = 10.7(0.93)^h$: No
• For $m = 15(0.93)^h$: Yes
• For $m = 15(0.71)^h$: No

1. First, find the successive - ratios:

• $\frac{16100}{15300}\approx1.052$
• $\frac{17300}{16100}\approx1.075$
• $\frac{18400}{17300}\approx1.063$
• $\frac{19600}{18400}\approx1.065$
• $\frac{20700}{19600}\approx1.056$
• $\frac{22000}{20700}\approx1.063$
• The average of these ratios is $\frac{1.052 + 1.075+1.063 + 1.065+1.056+1.063}{6}\approx1.061$
• Let the initial value $a = 15300$ and the growth factor $b\approx1.061$. The exponential model is $y = 15300(1.061)^x$ where $x$ is the number of years.