a) Write the regression equation with parameters from the R output.b) Suppose that the number of manufacturing enterprises employing 20 or more workers in Irvine is 250, could you predict that the annual mean concentration of sulfur dioxide in Irvine?c) What is the residual if in Irvine the annual mean concentration of sulfur dioxide is 15 micrograms per cubic meter.d) What is the value of the correlation coefficient?e) Calculate a 95% confidence interval for the slope of the model.f) Based on the confidence interval, is there a linear relationship between X and Y?
Answer:
Y = 0.0315x + 9.4764
Residual = 2.35
Correlation Coefficient = 0.847
Step-by-step explanation:
From the R output given :
Intercept = 9.4764
Slope = 0.0315
x = number of manufacturing enterprise employing 20 or more workers
y = annual mean concentration of Sulphur dioxide
The regression equation :
y = bx + c
b = slope ; c = intercept
y = 0.0315x + 9.4764
Prediction using the regression equation :
The predicted y value, when x = 250
y = 0.0315(250) + 9.4764
y = 17.3514
The residual, if actual annual concentration = 15
Y residual = 17.35 - 15 = 2.35
The correlation Coefficient value, R
R = √R²
R = √0.717
R = 0.847
B) 5000(1.011117)t
C) 5000(0.987783)t
D) 5000(1.048883)t
Can someone explain to me how to solve this please
Answer:
C) 5000(0.987783)^t
Step-by-step explanation:
The monthly interest rate is the APR divided by 12, so is 22%/12 ≈ 0.018333.
Each month, the previous balance (B) has interest charges added to it, so the new balance is ...
balance with interest charges = B + (22%)/12×B = 1.018333×B
The minimum payment is 3% of this amount, so the new balance for the next month is ...
balance after payment = (1.018333B)(1 - 0.03) = 0.987783B
Since the balance is multiplied by 0.987783 each month, after t payments, the balance starting with 5000 will be ...
5000×0.987783^t . . . . . . . . . matches choice C
Given
Jessie has 5 dogs
2 are male
3 are female
Procedure
What percent of the dogs are male?
Male dogs
40 % of dogs are male
Vegan 85.00 5.20 1.08
Omnivore 91.00 5.65 1.10
Calculate a 99% CI for the difference between the population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores. (Use μvegan−μomnivore). Round to three decimal places.)
Interpret the interval.
a. We are 99% confident that the true average cholesterol level for vegans is less than that of omnivores by an amount within the confidence interval.
b. We are 99% confident that the true average cholesterol level for vegans is greater than that of omnivores by an amount within the confidence interval.
c. We are 99% confident that the true average cholesterol level for vegans is greater than that of omnivores by an amount outside the confidence interval.
d. We cannot draw a conclusion from the given information.
Answer: hey
Step-by-step explanation:
plz hlep
...and put as an exact decimal or simplified fraction
Answer:
-2.85
Step-by-step explanation:
I looked it up tbh
Answer:
-2.85
Step-by-step explanation:
Google it or Edge it or Firefox it