Answer:
E = 0 x (4/6) + 60 x (1/6) + 120 x (1/6) = 30
Step-by-step explanation:
You're looking for the extreme values of
subject to the constraint
.
The target function has partial derivatives (set equal to 0)


so there is only one critical point at
. But this point does not fall in the region
. There are no extreme values in the region of interest, so we check the boundary.
Parameterize the boundary of
by


with
. Then
can be considered a function of
alone:



has critical points where
:



but
for all
, so this case yields nothing important.
At these critical points, we have temperatures of


so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.
Answer: The distance between the girls is 362.8 meters.
Step-by-step explanation:
So we have two triangle rectangles that have a cathetus in common, with a length of 160 meters.
The adjacent angle to this cathetus is 40° for Anna, then the opposite cathetus (the distance between Anna and the tower) can be obtained with the relationship:
Tan(A) = opposite cath/adjacent cath.
Tan(40°) = X/160m
Tan(40°)*160m = 134.3 m
Now, we can do the same thing for Veronica, but in this case the angle adjacent to the tower is 55°
So we have:
Tan(55°) = X/160m
Tan(55°)*160m = X = 228.5 m
And we know that the girls are in opposite sides of the tower, so the distance between the girls is equal to the sum of the distance between each girl and the tower, then the distance between the girls is:
Dist = 228.5m + 134.3m = 362.8m
Answer:
6 hours as a babysitter and 12 hours as a hostess
Answer:
Option C - Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in
Step-by-step explanation:
We are given;
n = 15
t-value = 1.66
Significance level;α = 0.05
So, DF = n - 1 = 15 - 1 = 14
From the one-sample t - table attached, we can see that the p - value of 0.06 at a t-value of 1.66 and a DF of 14
Now, since the P-value is 0.06,it is greater than the significance level of 0.05. Thus we do not reject the null hypothesis. We conclude that there is not sufficient evidence that the true diameter differs from 0.5 in.