If there was 20% taken off you add 20% back on and that is how you will get your answer. So take 760 plus the 20% which gives you 912 dollars
Answer:
Vertical compression
Step-by-step explanation:
To find the transformation, compare the function to the parent function and check to see if there is a horizontal or vertical shift, reflection about the x-axis or y-axis, and if there is a vertical stretch.
Answer:
- 64%: 1024
- 96%: 1536
- 132%: 2112
- 232%: 3712
Step-by-step explanation:
<u>64% of 1600</u>
64% of 1600 = 2 × (32% of 1600) = 2×512
64% of 1600 = 1024
<u>96% of 1600</u>
96% of 1600 = 3 × (32% of 1600) = 3×512
96% of 1600 = 1536
<u>132% of 1600</u>
132% of 1600 = (100% of 1600) + (32% of 1600) = 1600 + 512
132% of 1600 = 2112
<u>232% of 1600</u>
232% of 1600 = (100% of 1600) + (132% of 1600) = 1600 +2112
232% of 1600 = 3712
Answer:
False
Step-by-step explanation:
The totals of the rows and columns of a two way table are not called marginal distributions. They are called conditional distributions.
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 .
