Answer:
C. 85 degrees
Step-by-step explanation:
(set the equations equal to each other):
6x-25 = 4x+15
(solve for x):
2x = 40
x = 20
(plug in x back into the equation):
6(20)-25 = 120-25 = 95
(subtract your value from 180 degrees (180 degrees is a line)):
180-95 = 85 degrees
Answer:
Step-by-step explanation:
The formula for determining the confidence interval for the difference of two population means is expressed as
Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)
Where
x1 = mean sale amount for mail order sales = 82.70
x2 = mean sale amount for internet sales = 66.9
s1 = sample standard deviation for mail order sales = 16.25
s2 = sample standard deviation for internet sales = 20.25
n1 = number of mail order sales = 17
n2 = number of internet sales = 10
For a 99% confidence interval, we would determine the z score from the t distribution table because the number of samples are small
Degree of freedom =
(n1 - 1) + (n2 - 1) = (17 - 1) + (10 - 1) = 25
z = 2.787
Margin of error =
z√(s²/n1 + s2²/n2) = 2.787√(16.25²/17 + 20.25²/10) = 20.956190
Answer:
Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:
Where
and
Since the distribution of X is normal then we know that the distribution for the sample mean
is given by:
And we have;


Step-by-step explanation:
Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of
Round your answers to two decimal places.
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:
Where
and
Since the distribution of X is normal then we know that the distribution for the sample mean
is given by:
And we have;


Answer:
The matrix is not invertible.
Step-by-step explanation:
We are given the following matrix in the question:
![A =\left[\begin{array}{ccc}-5&0&1\\-1&3&2\\0&10&6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%260%261%5C%5C-1%263%262%5C%5C0%2610%266%5Cend%7Barray%7D%5Cright%5D)
Condition for invertible matrix:
A matrix is invertible if and only if the the determinant is non-zero.
We can find the determinant of the matrix as:
![|A| = -5[(3)(6)-(2)(10)]-0[(-1)(6)-(2)(0)] + 1[(-1)(10)-(3)(0)]\\|A| = -5(18-20)+(-10)\\|A| = 10-10\\|A| = 0](https://tex.z-dn.net/?f=%7CA%7C%20%3D%20-5%5B%283%29%286%29-%282%29%2810%29%5D-0%5B%28-1%29%286%29-%282%29%280%29%5D%20%2B%201%5B%28-1%29%2810%29-%283%29%280%29%5D%5C%5C%7CA%7C%20%3D%20-5%2818-20%29%2B%28-10%29%5C%5C%7CA%7C%20%3D%2010-10%5C%5C%7CA%7C%20%3D%200)
Since the determinant of the given matrix is zero, the given matrix is not invertible.