This situation has a constant rate of change because each sack weighs the same amount. After 5 sacks are removed, the total weight is 135 pounds. After 9 sacks are removed, the total weight is 87 pounds. From 5 to
9 sacks removed is 4 more sacks. Over the same interval, the total weight decreases by 135#87=48 pounds.
Since 4 sacks account for a 48-pound decrease, each sack must weigh 48÷4=12 pounds. This uses the linear idea that equal changes in the input produce equal changes in the output. Therefore, the weight of each sack is
12 pounds, which is answer A.

The model has:
r^2=0.42
This indicates a moderate fit, not a strong fit.
The prediction is being made at:
x=19.5
From the scatterplot, the observed data values appear to extend only to about:
x=16
So x=19.5is beyond the observed data range.
Because the model has only a moderate fit and the prediction is too far beyond the observed data, the prediction is not appropriate.
The correct interpretation is:
" No. The " r^2 " value indicates a moderate fit, but " x=19.5 " is more than " 25% " of the range beyond the maximum value. "

The function is:
f(x)=330× # 1.15 # ^x
Because xrepresents the number of years since 2005:
2006 # x=1
2011 # x=6
The average yearly rate of change from 2006 to 2011 is:
(f(6)-f(1))/(6-1)
First, find f(1):
f(1)=330× # 1.15 # ^1
f(1)=330×1.15
f(1)=379.50
Now find f(6):
f(6)=330× # 1.15 # ^6
f(6)#330×2.31306
f(6)#763.31
Now calculate the average rate of change:
(763.31-379.50)/(6-1)
=383.81/5
=76.762
Rounded to two decimal places:
76.76
So the average yearly rate of change of the artifact's value from 2006 to 2011 is approximately:
$76.76 " per year "
Therefore, the correct answer is:
# ( " A " )

The function is:
G(t)=1,600× # 1.31 # ^t
This is an exponential growth function because the base is greater than 1:
1.31 > 1
We need to find the population after 25 minutes, so substitute:
t=25
into the function:
G(25)=1,600× # 1.31 # ^25
Evaluate the exponent and multiply:
G(25)#1,367,421.41
Since the population is a number of bacteria cells, round to the nearest whole number:
G(25)#1,367,421

The maximum value of a graph is the highest output value the function reaches. Since P(t) represents yearly profit, the maximum value occurs where the store earns its greatest yearly profit. The graph is a downward- opening curve, so its highest point is the vertex. To answer the question, we need the t-value of the vertex, not the profit amount. From the graph, the highest point occurs between 4 and 5 years after opening, closest to 4.5 years. This means the virtual store reached its maximum yearly profit approximately 4.5 years after it opened.
The later values decrease after the vertex, so options involving 7.5 or 14 years are not correct. Therefore, the correct answer is B.