Whenever we divide a certain number by 10 or 100 or 1000... the decimal point of the number always moves to your left hand side.
for example: 25 or 25.00 is a number that has its decimal point after 5.
suppose we divide it by 10, the decimal point point moves as number of times as the number of zeros in the divisor. here, 10 has 1 zero. isn't it?
hence, decimal point moves 1 time or once to the left
2.500 or simply 2.5
on writing a percent as decimal..
we always have the divisor as 100.
number of zeros in 100 : 2
hence point moves 2 times or twice to the left.
25 / 100 = .25 or 0.25
on re-writing 0.25 as fraction, count the number of digits right to the point. it is 2 here.
hence there would be two zeros as well,i.e, 100
hence , 0.25 = 25/100
hope it helps you! :)
Complete Question
The complete question is shown on the first uploaded image
Answer:
First question
![Ab_1 = \left[\begin{array}{ccc}{-1}&{3}\\ 1 &4 \\5 &8\end{array}\right]\left[\begin{array}{ccc}5\\-2\\\end{array}\right]](https://tex.z-dn.net/?f=Ab_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-1%7D%26%7B3%7D%5C%5C%201%20%264%20%5C%5C5%20%268%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C-2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Second question
![Ab_1 = \left[\begin{array}{ccc}{-1}&{3}\\ 1 &4 \\5 &8\end{array}\right]\left[\begin{array}{ccc}5\\-2\\\end{array}\right] = \left[\begin{array}{ccc}{(-1 * 5 )+ (3* -2)}\\{(1 * 5)+ (4 * -2)}\\{(5 * 5) + (8*-2)}\end{array}\right] = \left[\begin{array}{ccc}{-11}\\{-3}\\{29}\end{array}\right]](https://tex.z-dn.net/?f=Ab_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-1%7D%26%7B3%7D%5C%5C%201%20%264%20%5C%5C5%20%268%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C-2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B%28-1%20%2A%205%20%29%2B%20%283%2A%20-2%29%7D%5C%5C%7B%281%20%2A%205%29%2B%20%284%20%2A%20-2%29%7D%5C%5C%7B%285%20%2A%205%29%20%2B%20%288%2A-2%29%7D%5Cend%7Barray%7D%5Cright%5D%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-11%7D%5C%5C%7B-3%7D%5C%5C%7B29%7D%5Cend%7Barray%7D%5Cright%5D)
Third question
![Ab_1 = \left[\begin{array}{ccc}{-1}&{3}\\ 1 &4 \\5 &8\end{array}\right]\left[\begin{array}{ccc}-3\\4\\\end{array}\right]](https://tex.z-dn.net/?f=Ab_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-1%7D%26%7B3%7D%5C%5C%201%20%264%20%5C%5C5%20%268%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Fourth question
![Ab_1 = \left[\begin{array}{ccc}{-1}&{3}\\ 1 &4 \\5 &8\end{array}\right]\left[\begin{array}{ccc}-3\\4\\\end{array}\right] = \left[\begin{array}{ccc}{(-1 * -3 )+ (3* 4)}\\{(1 * -3)+ (4 * 4)}\\{(5 * -3) + (8*4)}\end{array}\right] = \left[\begin{array}{ccc}{15}\\{13}\\{-23}\end{array}\right]](https://tex.z-dn.net/?f=Ab_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-1%7D%26%7B3%7D%5C%5C%201%20%264%20%5C%5C5%20%268%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C4%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B%28-1%20%2A%20-3%20%29%2B%20%283%2A%204%29%7D%5C%5C%7B%281%20%2A%20-3%29%2B%20%284%20%2A%204%29%7D%5C%5C%7B%285%20%2A%20-3%29%20%2B%20%288%2A4%29%7D%5Cend%7Barray%7D%5Cright%5D%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B15%7D%5C%5C%7B13%7D%5C%5C%7B-23%7D%5Cend%7Barray%7D%5Cright%5D)
Fifth question
The correct option is A
Step-by-step explanation:
From the question we are told that
The matrix A is ![A = \left[\begin{array}{ccc}{-1}&{3}\\ 1 &4 \\5 &8\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B-1%7D%26%7B3%7D%5C%5C%201%20%264%20%5C%5C5%20%268%5Cend%7Barray%7D%5Cright%5D)
The matrix B is ![B = \left[\begin{array}{ccc}5&{-3}\\{-2}&4\end{array}\right]](https://tex.z-dn.net/?f=B%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26%7B-3%7D%5C%5C%7B-2%7D%264%5Cend%7Barray%7D%5Cright%5D)
The first question is to set up the product 
, where 
is the first column of matrix B, this shown as
The second question is to calculate , this is evaluated as
The third question is to set up the product 
, where 
is the second column of matrix B, this shown as
The fourth question is to calculate , this is evaluated as
The fifth question is to determine the numerical expression for the first entry in the first column of AB using the row-column rule and from the calculation of we see that it is
Answer:
mean:-3
median:5
mode: all values appeared once
range:169
Step-by-step explanation:
Answer:
1300
Step-by-step explanation:
Given that :
Number of sections = 8
Number of seats per section, x ;
150 ≤ x ≤ 200
The possible Number of seats in the auditorium :
Number of seats per section * number of sections
150 * 8 ≤ x ≤ 200 * 8
1200 ≤ x ≤ 1600
The possible Number of seats will lie within 1200 and 1600
From the options, only 1300 lie within this range
Divide both sides by 5
e^x=3.152 take the natural log of both sides...
x=ln(3.152)
x≈1.148
