Answer:
![a=\frac{m}{(q+r^2)}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bm%7D%7B%28q%2Br%5E2%29%7D)
Step-by-step explanation:
We begin with the equation
for which we need to solve for a.
The first thing that we can do is to factor out the a from both terms on the left side of the equation
![qa+r^2a=m\\\\a(q+r^2)=m](https://tex.z-dn.net/?f=qa%2Br%5E2a%3Dm%5C%5C%5C%5Ca%28q%2Br%5E2%29%3Dm)
Then we can divide each side by
to isolate a
![a(q+r^2)=m\\\\a=\frac{m}{(q+r^2)}](https://tex.z-dn.net/?f=a%28q%2Br%5E2%29%3Dm%5C%5C%5C%5Ca%3D%5Cfrac%7Bm%7D%7B%28q%2Br%5E2%29%7D)
Answer:
Step-by-step explanation:
<u>Part A:</u>
Lets take the value of a=10
Thus a+1=b
10+1=11 , it means the value of b is 11
11>10 or b>a
Now take the value of a as -3
Therefore a+1=b
-3+1= -2
-2>-3 or b>a
The pair of values for a and b are: a=10 then the value of b would be 11;
And a=-3; then the value of b would be: -2.
<u>Part B:</u>
It is not possible to create a pair of values for a and be, in which the numerical relationship shown in the given conditional statement is false, therefore b>a if a+1=b....
Answer:
A = 100 B = 130 C = 150
Step-by-step explanation:
Translated into a system of equations:
A+B+C = 380
B = A+30
C = B + 20
So we find B from the last one:
B = C-20
Plug it in the second one:
C-20 = A+30
A = C-50
Plug it in the first:
C-50 + C-20 + C = 380
3C - 70 = 380
3C = 450
C=150
A = C - 50 = 150 - 50 = 100
B = C - 20 = 150 - 20 = 130
3:8 is the ratio
hope this helped
Answer:
![\bar X= \frac{1930055}{110}=17545.95](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%20%5Cfrac%7B1930055%7D%7B110%7D%3D17545.95)
Step-by-step explanation:
For this case we can construct the following table in order to find the mean for the grouped data:
Interval Frequency (fi) Midpoint (Xi) Xi *fi
5001-10000 27 7500.5 202513.5
10001-15000 23 12500.5 287511.5
15001-20000 12 17500.5 210006
20001-25000 18 22500.5 405009
25001-30000 30 27500.5 825015
Total 110 1930055
And the mean is calculated with the following formula:
![\bar X= \frac{\sum_{i=1}^n x_i f_i}{n}](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20x_i%20f_i%7D%7Bn%7D)
Where ![n = \sum_{i=1}^n f_i = 110](https://tex.z-dn.net/?f=%20n%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20f_i%20%3D%20110)
So then if we replace into the formula we got:
![\bar X= \frac{1930055}{110}=17545.95](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%20%5Cfrac%7B1930055%7D%7B110%7D%3D17545.95)