Scientists are studying the population of a particular type of fish. The table below shows the data gathered over a five-month time period. Use the data to answer questions 5-9.

number of months number of fish
0 8
1 39
2 195
3 960
4 4,738
5 23,375

5. what does the scatterplot of the data show?
1. strong positive linear relationship
2. a strong negative linear relationship
3. a curve that represents exponential growth
4. a curve that reprents ecponential decay
I select number 3

6. Complete an exponential transformation on the y-values. What is the new value of y when x+5?

1. 4.3688
2. 3.6756
3. 0.6990
4. 3.3757
I select #3

7. Find the linear transformation Model
1. log y hat=0.6935•logx + 0.9013
2. log y hat=0.9013x+0.6935
3. log y hat=0.6935x+0.9013
4.
log y hat=0.9013•logx+0.6935

8. Use the linear transformation model to predict the number of fish in 12 months.
this is where I need help.

9. A power model is shown below. Determine the residual for the observed data x=7 and y=70

log y hat=1.6+0.31log x
1.71.37
2.1.37
3.1.85
4.-1.37
I select 4

To answer question #5, we need to analyze the scatterplot of the data. From the given data, we can plot the number of months on the x-axis and the number of fish on the y-axis. Plotting the points, we can determine the shape of the graph.

In this case, as the number of months increases, the number of fish also increases exponentially. Therefore, the scatterplot of the data shows a curve that represents exponential growth. Hence, the correct answer is option 3.

To answer question #6, we are asked to complete an exponential transformation on the y-values. If we observe the data, we can see that the number of fish approximately doubles every month. This suggests that we can use a base of 2 for the exponential transformation.

Using the formula y = ab^x, where a is the initial value at x=0 (8 in this case) and b is the base (2 in this case), we can substitute the values to find the new y-value when x+5.

Calculating y when x+5, we have:
y = 8 * (2)^((x+5))
y = 8 * (2)^(x+5)
y = 8 * (2)^x * (2)^5
y = 8 * 32 * (2)^x
y = 256 * (2)^x

Now, let's substitute x+5 into the equation and calculate the new value of y:

y_new = 256 * (2)^(x+5)
y_new = 256 * (2)^(2+5)
y_new = 256 * (2)^7
y_new = 256 * 128
y_new = 32,768

Therefore, the new value of y when x + 5 is 32,768. Hence, the correct answer is option 3.

To answer question #7, we need to determine the linear transformation model for the data. In this case, we can use logarithmic transformations. Taking the logarithm (base 10 or ln) of both sides of the equation:

log(y_hat) = bx + a

Based on the given options, we can see that log y_hat = 0.6935x + 0.9013. Therefore, the correct answer is option 3.

To answer question #8, we can use the linear transformation model to predict the number of fish in 12 months. Substituting x = 12 into the equation:

log(y_hat) = 0.6935(12) + 0.9013
log(y_hat) = 8.322 + 0.9013
log(y_hat) = 9.2233

Now, we need to solve for y_hat by taking the antilog of both sides:

y_hat = 10^(9.2233)
y_hat = 8,050.8

Therefore, the predicted number of fish in 12 months is approximately 8,050.8.

To answer question #9, we are asked to find the residual for the observed data x = 7 and y = 70. The power model is given as log y_hat = 1.6 + 0.31 log x.

Substituting the values into the model, we have:
log y_hat = 1.6 + 0.31 log 7

Calculating the logarithm, we get:
log y_hat = 1.6 + 0.31(0.8451)
log y_hat = 1.6 + 0.2618
log y_hat = 1.8618

Now, we need to solve for y_hat by taking the antilog of both sides:
y_hat = 10^(1.8618)
y_hat = 70.88

The observed value of y is 70. Therefore, the residual is calculated as (observed value - predicted value):
Residual = 70 - 70.88
Residual = -0.88

Therefore, the residual for the observed data x = 7 and y = 70 is approximately -0.88. Hence, the correct answer is option 4.