To get a number to decrease by 21% you multiply it by 0.79 (1-0.21)
r = 0.79
Answer:
![\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The <em>transpose of a matrix </em>
is one where you swap the column and row index for every entry of some original matrix 
. Let's go through our first matrix row by row and swap the indices to construct this new matrix. Note that entries with the same index for row and column will stay fixed. Here I'll use the notation 
and 
to refer to the entry in the i-th row and the j-th column of the matrices 
and 
respectively:
Constructing the matrix from those entries gives us
![P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=P%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
which is option a. from the list.
Another interesting quality of the transpose is that we can geometrically represent it as a reflection over the line traced out by all of the entries where the row and column index are equal. In this example, reflecting over the line traced from 2 to 1 gives us our transpose. For another example of this, see the attached image!
To evaluate the probability that the lifespan will be between 1440 and 1465 hours will be given by:
P(1440<x<1465)
using the z-score formula we obtain:
z=(x-μ)/σ
where:
μ=1450
σ=8.5
hence
when x=1440
z=(1440-1450)/8.5
z=-1.18
P(z<-1.18)=0.1190
when x=1465
z=(1465-1450)/8.5
z=1.77
P(z<1.77)=0.9625
hence:
P(1440<x<1465)
=0.9625-0.1180
=0.8445
Answer:
£2,121.8
Step-by-step explanation:
Given the following;
Principal P = £2000
Rate r = 3%
Time t = 2 years
n = 1 (time of compounding)
Using the compound interest formula;
A = P(1+r)^t
A = 2000(1+0.03)^2
A = 2000(1.03)^2
A = 2000(1.0609)
A = 2,121.8
hence the amount that will be in his account after 2 years is £2,121.8
Answer:
B
Step-by-step explanation:
