Calculation of relative maxima and minima of a function f (x) in a range [a, b]:
We find the first derivative and calculate its roots.
We make the second derivative, and calculate the sign taken in it by the roots of the first derivative, and if:
f '' (a) <0 is a relative maximum
f '' (a)> 0 is a relative minimum
Identify intervals on which the function is increasing, decreasing, or constant. G (x) = 1- (x-7) ^ 2
First derivative
G '(x) = - 2 (x-7)
-2 (x-7) = 0
x = 7
Second derivative
G '' (x) = - 2
G '' (7) = - 2 <0 is a relative maximum
answer:
the function is increasing at (-inf, 7)
the function is decreasing at [7, inf)
Step-by-step explanation:
Tan30=4/adj
adj Tan 30= 4
adj=4/tan30
adj=6.93m
Area=L×b
4×6.93=27.72m
Answer:
7/12
Step-by-step explanation
We have both of the values of the dozen of doughnuts 5/6 of a dozen blueberry and 1/4 dozen pumpkin. In order to determine how many more blueberry muffins Min has, we have to set the denominators to be the same number. To do this we need a common denominator. The least common denominator between 6 and 4 is 12. We multiply the top and bottom of the fraction 5/6 by 2 to get 10/12. We then multiply 1/4 by 3 to get 3/12. 10-3 = 7, therefore Min has 7/12 more blueberry muffins
Answer:
7^4
Step-by-step explanation:
Answer:
a) Mean 0.11 and standard deviation 0.0044.
b) Mean 0.11 and standard deviation 0.0099.
c) Mean 0.11 and standard deviation 0.0198
Step-by-step explanation:
Central Limit Theorem:
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
In this question:
a. For a random sample of size n equals 5000.
Mean:
Standard deviation:
Mean 0.11 and standard deviation 0.0044.
b. For a random sample of size n equals 1000.
Mean:
Standard deviation:
Mean 0.11 and standard deviation 0.0099.
c. For a random sample of size n equals 250.
Mean:
Standard deviation:
Mean 0.11 and standard deviation 0.0198