Answer:
![\left[\begin{array}{ccc}16&10&39\\-32&-30&-18\\6&6&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D16%2610%2639%5C%5C-32%26-30%26-18%5C%5C6%266%26-19%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given that A = ![\left[\begin{array}{ccc}-2&-4&2\\5&1&5\\-1&-3&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-4%262%5C%5C5%261%265%5C%5C-1%26-3%26-4%5Cend%7Barray%7D%5Cright%5D)
and B = ![\left[\begin{array}{ccc}2&3&-5\\5&-4&2\\-1&-1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%26-5%5C%5C5%26-4%262%5C%5C-1%26-1%263%5Cend%7Barray%7D%5Cright%5D)
.
Therefore, BA = × = (Answer)
For, the 1st row-1st column element is obtained by multiplying the 1st row of B matrix and 1st column of A matrix.
Similarly, the 1st row-2nd column element is obtained by multiplying the 1st row of B matrix and 2nd column of A matrix.
And so on. (Answer)
Answer: k = -1/3
Step-by-Step Explanation:
=> 3x - 2ky + 4 = 0
Value of ‘x’ = -2
Value of ‘y’ = 3
Substitute values of ‘x’ and ‘y’ :-
=> 3x - 2ky + 4 = 0
= 3(-2) - 2k(3) + 4 = 0
= -6 - 6k + 4 = 0
= -6k = 6 - 4
= -6k = 2
= k = 2/-6
=> k = -2/6 = -1/3
Therefore, k = -1/3
If your decimal is 1/2 or more u round up if it is less than half u round down
Answer: The probability in (b) has higher probability than the probability in (a).
Explanation:
Since we're computing for the probability of the sample mean, we consider the z-score and the standard deviation of the sampling distribution. Recall that the standard deviation of the sampling distribution approximately the quotient of the population standard deviation and the square root of the sample size.
So, if the sample size higher, the standard deviation of the sampling distribution is lower. Since the sample size in (b) is higher, the standard deviation of the sampling distribution in (b) is lower.
Moreover, since the mean of the sampling distribution is the same as the population mean, the lower the standard deviation, the wider the range of z-scores. Because the standard deviation in (b) is lower, it has a wider range of z-scores.
Note that in a normal distribution, if the probability has wider range of z-scores, it has a higher probability. Therefore, the probability in (b) has higher probability than the probability in (a) because it has wider range of z-scores than the probability in (a).
