Answer:
0.7233
Step-by-step explanation:
We want to find the area between the z-scores z=-0.95 and z=1.25.
We first find the area to the left of each z-score, and subtract the smaller area from the bigger one.
For the area to the left of z=-0.95, we read -0.9 under 5 from the standard normal distribution table.
This gives P(z<-0.95)=0.1711
Similarly the area to the left of z=1.25 is
P(z<1.25)=0.8944
Now the area between the two z-scores is
P(-0.25<z<1.25)=0.8944-0.1711=0.7233
Answer:
Step-by-step explanation:
Given
![A = \left[\begin{array}{cc}-2&6\\3&5\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%266%5C%5C3%265%5Cend%7Barray%7D%5Cright%5D)
Required
Determine the determinant
For a two by two matrix, A such that:
![A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant |A| is:
So, in
The determinant is:
Answer:
11,8,5
Step-by-step explanation:
an=11-3(n-1)
Let n=1
a1 = 11 - 3(1-1)
= 11 -0
=11
Let n=2
a2 = 11-3(2-1)
= 11 -3(1)
= 8
Let n=3
a3 = 11-3(3-1)
= 11 -3(2)
= 11 -6
=5
