Step-by-step explanation:
<em><u>I </u></em><em><u>thin</u></em><em><u>k</u></em><em><u> </u></em><em><u>mode</u></em><em><u>,</u></em><em><u>range</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>median</u></em><em><u> </u></em><em><u>dpo</u></em><em><u> </u></em><em><u>ako</u></em><em><u> </u></em><em><u>sure</u></em>
Answer:
8/27
Step-by-step explanation:
Answer:
C. No. The sum of the dimensions of the eigenspaces equals nothing and the matrix has 3 columns. The sum of the dimensions of the eigenspace and the number of columns must be equal.
Step-by-step explanation:
Here the sum of dimensions of eigenspace is not equal to the number of columns, so therefore A is not diagonalizable.
Answer:
6,400,000,000 (6.4 billions)
Step-by-step explanation:
We have a total of 10 digits, and each digit was 10 possibilities of being filled, except the first digit of the area code and first digit of the telephone number, that have only 8 possibilities.
So if 8 digits have 10 possible values and 2 digit has 8 possible values, the number of possible telephone numbers is:
10^8 * 8^2 = 6,400,000,000 (6.4 billions)
The answer is 6. this is because when you find the range, it's the difference between the highest and the lowest value in the set
