A normal distribution has a mean of 10 and a standard
1 answer:
Answer:
15.87% probability of selecting a number that is at most 9
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
What is the probability of selecting a number that is at most 9?
This is the pvalue of Z when X = 9. So
has a pvalue of 0.1587
15.87% probability of selecting a number that is at most 9
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This is asking what can you raise 2 to in or order to get 4.
X=2
First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.
![\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] * \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-3%262%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D%20)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
![\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5Cx%26x%26x%5Cend%7Barray%7D%5Cright%5D%20)
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
![\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5C4%266%268%5Cend%7Barray%7D%5Cright%5D)
:)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
:)