Given:
PV = 13,440
i = 5.86% , compounded monthly
t = 4 years
13,440(0.0586/12))/(1-(1+0.0586/12)^-48= 15,109.44
15,109.44 + 156.60 = 15,266.04
15,266.04 - 13,440.00 = 1,826.04
<span>1,826.04/15,266.04 = 11.96 % Percentage total of Finance Charge of the total loa</span>
Answer:
- <u><em>Option b. just below 30%</em></u>
<u><em></em></u>
Explanation:
Please, see attached the <em>histogram that represents the distribution of acceptance rates (percent accepted) among 25 business schools in 2004. </em>
<em />
The<em> median</em> is the value that separates the lower 50% from the upper 50% of the data.
Since there are 25 business schools, the middle value is the number 13.
The height of each bar is the<em> frequency</em> or number of business school for that acceptace rate:
- The first bar has frequency of 1 school
- The second bar has frequency of 3 schools: cummulative frequency: 1+3=4.
- The third bar has frequency 5 schools: cummulative frequency 4 + 5 = 9.
- The fourth bar has frequency 3 schools: cummulative frequency: 9+3=12.
Then, the 13th value is on the next bar, the fifth bar.
The fifth bar has acceptance rates 25 ≤ rate < 30.
That means that the median acceptance rate is greater than or equal to 25 and less than 30.
Thus, the choice is the option <em>b. just below 30%.</em>
AFB = CFD = 31
BFC = 90 - (31+31) = 28
DFE = BFC = 28
CFE = CFD + DFE = 31 + 28 = 59
Answer:
The image of
through T is ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We know that
→
is a linear transformation that maps
into
⇒

And also maps
into
⇒

We need to find the image of the vector ![\left[\begin{array}{c}4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D)
We know that exists a matrix A from
(because of how T was defined) such that :
for all x ∈ 
We can find the matrix A by applying T to a base of the domain (
).
Notice that we have that data :
{
}
Being
the cannonic base of 
The following step is to put the images from the vectors of the base into the columns of the new matrix A :
(Data of the problem)
(Data of the problem)
Writing the matrix A :
![A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now with the matrix A we can find the image of
such as :
⇒
![T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
We found out that the image of
through T is the vector ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
<span>angle measures in order from greatest to least.
<BAC, <ABC, <ACB
hope it help.</span>