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gtnhenbr [62]
3 years ago
15

Using the same data set, four models are estimated using the same response variable, however, the number of explanatory variable

s differs. Which of the following models provides the best fit? Model 1 Model 2 Model 3 Model 4 Multiple R 0.993 0.991 0.936 0.746 R Square 0.987 0.982 0.877 0.557 Adjusted R Square 0.982 0.978 0.849 0.513 Standard Error 4,043 4,463 11,615 20,878 Observations 12 12 12 12 Multiple Choice Model 1 Model 2 Model 3 Model 4
Mathematics
1 answer:
kirill [66]3 years ago
8 0

Answer:

Model A

Step-by-step explanation:

Given the table :

___________M 1 ____ M 2 ____ M 3 ____M 4

Multiple R _ 0.993 ___ 0.991 ___0.936__ 0.746

R Square __0.987___ 0.982 ___0.877 __0.557

Adj R² ____ 0.982___ 0.978 __ 0.849 ___0.513

S E_______ 4,043 __ 4,463 ___11,615 __20,878 Observations_ 12 _____ 12 _____ 12 ____12

Based on the detains of the model given, we could use the R value, R² and standard error values to evaluate the performance of the different models.

The best model will be one with Correlation Coefficient (R value) closet to 1. The model with the highest R value will also have the highest Coefficient of determination, R² value. The a best model is one which has a low a standard error value.

From the table, Model A has the highest R and R² values. It also has the lowest standard error value. Hence, we can conclude that model A provides the best fit.

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Forty-seven percent of fish in a river are catfish. Imagine scooping out a simple random sample of 25 fish from the river and ob
WARRIOR [948]

Answer:

Therefore the correct option is a.) The standard deviation is 0.0998. The 10% condition is met because it is very likely there are more than 250 catfish in the river.

Step-by-step explanation:

i) Let p = 0.47

ii) therefore q = 1 - 0.47 = 0.53

iii) sample size, n =25

iii) standard deviation = \sqrt{\frac{p \times q}{n} }  = \sqrt{\frac{0.47 \times 0.53}{25} } = 0.0998

Therefore the correct option is

a.) The standard deviation is 0.0998. The 10% condition is met because it

    is very likely there are more than 250 catfish in the river.

8 0
3 years ago
At the dealership where she works, Sade fulfilled 3/10 of her quarterly sales goal in January and another 1/10 of her sales goal
skelet666 [1.2K]

Hi There!

----------------------------------------

Problem #1:

At the dealership where she works, Sade fulfilled 3/10 of her quarterly sales goal in January and another 1/10 of her sales goal in February. What fraction of her quarterly sales goal had Sade reached by the end of February?

3/10 + 1/10 = 4/10

4/10 of her goal.

----------------------------------------

Problem #2:

Shane has been monitoring his mileage. According to last week's driving log, he drove 1/10 of a mile in his car and 9/10 of a mile in his truck. How far did Shane drive last week in all?

1/10 + 9/10 = 10/10 = 1

Shane drove 1 mile.

----------------------------------------

Problem #3:

For a class experiment, Vina's class weighed a log before and after subjecting it to termites. Before subjecting it to termites, the log weighed 7/10 of a pound. After the termites, the log weighed 3/10 of a pound. How much weight did the termites take from the log?

7/10 - 3/10 = 4/10

The termites took 4/10 pound away from the log.

----------------------------------------

Hope This Helps :)

6 0
3 years ago
Distribute: 9(3x^3-2)
allochka39001 [22]
The answer is 27x^3 - 18
7 0
3 years ago
Please help with step by step
gregori [183]
f(x)=10e^{\frac{x-1}{2}}

substitute f(x) for y
y=10e^{\frac{x-1}{2}}

solve for x

\frac{y}{10}=e^{\frac{x-1}{2}}

ln( \frac{y}{10}) = \frac{x-1}{2}

2ln( \frac{y}{10})=x-1

2ln( \frac{y}{10})+1 =x

2ln( \frac{y}{10})   is esquivent to 2ln(y) - 2ln(10)

2ln(y)-2ln(10)+1=x

Change x to y and y to x

2ln(x)-2ln(10)+1=y

Flip formula and substitute y for g(x)

g(x)=2ln(x)-2ln(10)+1
7 0
3 years ago
Which conclusion can best be made from this scatter plot?
Lana71 [14]

Answer:

ang sagot ay nasa pic

sana po ay makatulong

6 0
3 years ago
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