Question

The table shows the estimated number of bees, y, in a hive x days after a pesticide is released near the hive.

A 2-column table with 6 rows. The first column is labeled number of days with entries 0, 10, 20, 30, 40, 50. The second column is labeled estimated number of bees with entries 10,000; 7,500; 5,600; 4,200; 3,200; 2,400.
Which function best models the data?

y = 9,958(0.972)x
y = 0.972(9,958)x
y = 9,219x– 150
y = –150x + 9,219

Answer 1

Answer:

A.

Step-by-step explanation:

The given table is

Days     Bees

  0        10,000

 10        7,500

 20       5,600

 30       4,200

 40       3,200

 50       2,400

Where $x$ represents days and $y$ represents bees.

The exponential function that models this problem must be like

$y=a(1-r)^{x}$, which represenst an exponential decary, because in this case, the number of bees decays.

We nned to use one points, to find the rate of decay. We know that $a=10,000$, because it starts with 10,000 bees.

Let's use the points (10, 7500)

$y=a(1-r)^{x}\\7500=10000(1-r)^{10}$

Solving for $r$, we have

$\frac{7500}{10000}=(1-r)^{10} \\(1-r)^{10} =0.75$

Using logarithms, we have

$ln((1-r)^{10}) =ln(0.75)\\10 \times ln(1-r)=ln(0.75)\\ln(1-r)=\frac{ln(0.75)}{10} \approx -0.03\\e^{ln(1-r)}=e^{-0.03}\\1-r =e^{-0.03}\\r=-e^{-0.03}+1 \approx 1.97$

Replacing all values in the model, we have

$y=10000(1-1.97)^{x}\\y=10000(0.97)^{x}$

Therefore, the right answer is the first choice, that's the best approximation to this situation.

Answer 2

Answer:

A. y = 9,958(0.972)x

Step-by-step explanation: