Answer:
cycle=33 students, driven=132 students, walk=55 students
Step-by-step explanation:
<h2>solution</h2><h2>60+15+x=100 to get the remaining percentage</h2><h2>x=25</h2><h2>15÷100×220=33</h2><h2>60÷100×220=132</h2><h2>25÷100×220=55</h2>
First break= 1/2
second break = 1/2 of half= 1/4
third break = half of 1/4 = 1/8
forth break = 1/2 of 1/8 = 1/16
fifth break = 1/2 of 1/16 =1/32
fraction painted = 1 -1/32 =31/32
portion painted = 150*31/32 = 145.3125 ft^2
$462
3/5 of 35 gallons = 21 gallons
21 gallons x $22 = $462
Given that:
mean,μ=35.6 min
std deviation,σ=10.3 min
we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.
using z-table, the z-score that will give us 0.2296 is:-1.99
therefore:
z-score is given by:
(x-μ)/σ
hence:
-1.99=(x-35.6)/10.3
-20.497=x-35.6
x=35.6-20.497
x=15.103
Answer:
The image of ![\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)
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
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 :
![T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%265%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
![T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%261%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-2%267%5Cend%7Barray%7D%5Cright%5D)
(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 ![\left[\begin{array}{c}4&-4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
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
