Direct variation means their ratio is a constant
x/y is 9/45=1/5
so when x=2, y=5*2=10
Answer:
$118.22
Step-by-step explanation:
Subtract $130.00-$8.00= $122.00
Then subtract $122.00- $3.78 = $118.22
9514 1404 393
Answer:
340
Step-by-step explanation:
The ratio of natural to synthetic is give as ...
5 : 8
Then the ratio of natural to total would be ...
5 : (5+8) = 5 : 13
The amount of natural oil in 884 total liters of oil will be ...
(5/13)(884 L) = 340 L
884 liters of Petrolyn oil will contain 340 liters of natural oil.
Answer:
City @ 2017 = 8,920,800
Suburbs @ 2017 = 1, 897, 200
Step-by-step explanation:
Solution:
- Let p_c be the population in the city ( in a given year ) and p_s is the population in the suburbs ( in a given year ) . The first sentence tell us that populations p_c' and p_s' for next year would be:
0.94*p_c + 0.04*p_s = p_c'
0.06*p_c + 0.96*p_s = p_s'
- Assuming 6% moved while remaining 94% remained settled at the time of migrations.
- The matrix representation is as follows:
- In the sequence for where x_k denotes population of kth year and x_k+1 denotes population of x_k+1 year. We have:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_k = x_k_+_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_k%20%3D%20x_k_%2B_1)
- Let x_o be the populations defined given as 10,000,000 and 800,000 respectively for city and suburbs. We will have a population x_1 as a vector for year 2016 as follows:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o = x_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_o%20%3D%20x_1)
- To get the population in year 2017 we will multiply the migration matrix to the population vector x_1 in 2016 to obtain x_2.
![x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_o)
- Where,
![x_o = \left[\begin{array}{c}10,000,000\\800,000\end{array}\right]](https://tex.z-dn.net/?f=x_o%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%2C000%2C000%5C%5C800%2C000%5Cend%7Barray%7D%5Cright%5D)
- The population in 2017 x_2 would be:
![x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}10,000,000\\800,000\end{array}\right] \\\\\\x_2 = \left[\begin{array}{c}8,920,800\\1,879,200\end{array}\right]](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%2C000%2C000%5C%5C800%2C000%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cx_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2C920%2C800%5C%5C1%2C879%2C200%5Cend%7Barray%7D%5Cright%5D)
Answer:
The minimum number of assignment statements needed is 5
Step-by-step explanation:
To write the algorithm, we apply the strategy of interchanging the values of variables in the assignment statements.
Assume "tmp" is the new variable, let assign tmp to w
The algorithm is:
Procedure exchange (w,x,y,z: integers)
tmp := w
w := x
x := y
y := z
z := tmp
return (w,x,y,z)
end
From the algorithm, it is obvious that there will be a minimum of 5 assignment statements needed.