Suppose that this person drives at r mph going to the mountains, and gets there in 12 hours. Returning, this person drives at (r+20) mph and gets home in 8 hours. We don't know the distance yet, but can solve for the initial speed, r, by setting
d = 12r = (r+20)(8). Solving for r, r=40 mph (going) and (40+20)mph = 60 mph (returning. Since d=12 r, d = (12 hrs)(40 mph) = 480 miles (answer).
Here is all you need for b)
Answer: 
Step-by-step explanation:
The first step is to find the ratio of the lengths.
According to the information given in the exercise, one the solids has edges of 12 feet and the other solid has edges of 24 feet.
Therefore, the ratio of the length of the smaller solid to the length of the is the following:
Now, the ratio to the volumes of the smaller solid to the other one is the following:
Then, knowing that the volume of the smaller solid is:
You get that the volime of the larger solid is:
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332
This can happen if you add another independent variable to your regression model that is strongly correlated to some other variable already in the model.
This is called multicollinearity.
If there is a high correlation between your independent variables can lead to problems.
<span>It can lead to increased variance of the coefficient estimates and make the estimates very sensitive to minor changes in the model.</span>
