Answer:
The answer is below
Step-by-step explanation:
1)
mean (μ) = 12, SD(σ) = 2.3, sample size (n) = 65
Given that the confidence level (c) = 90% = 0.9
α = 1 - c = 0.1
α/2 = 0.05
The z score of α/2 is the same as the z score of 0.45 (0.5 - 0.05) which is equal to 1.65
The margin of error (E) is given as:
The confidence interval = μ ± E = 12 ± 0.47 = (11.53, 12.47)
2)
mean (μ) = 23, SD(σ) = 12, sample size (n) = 45
Given that the confidence level (c) = 88% = 0.88
α = 1 - c = 0.12
α/2 = 0.06
The z score of α/2 is the same as the z score of 0.44 (0.5 - 0.06) which is equal to 1.56
The margin of error (E) is given as:
The confidence interval = μ ± E = 23 ± 2.8 = (22.2, 25.8)
Answer:100 off discount
Step-by-step explanation:
If you use the 20% discount you get only 95 dollars off the price
Answer:
a
Step-by-step explanation:
the equation for a circle centered at orgin is x^2+y^2=r where r is the radius. multiplying, adding, or subtracting any numbers to the x and y components such as the other choices here causes the circle to be translated about the graph.
10) 7(2m+3)
11) 3(3r-1)
12) 2p(5q+4)
Answer:
(A) Set A is linearly independent and spans 
. Set is a basis for .
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors ![A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D1%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%201%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%201%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, we are to decide which of the given statements is true:
In Matrix ![A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D%281%29%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%20%281%29%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%20%281%29%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans .
Therefore Set A is linearly independent and spans . Thus it is basis for .
