Answer:
Since it's increasing add 3.5% to 100%
= 103.5% of £16470.45
= 103.5/100 × 16470.45
= 17046.92
= £17046.92
Hope this helps.
any of the 34 cars can finish first
any of the remaining 33 cars can finish second
any of the remaining 32 cars can finish third
Thus, the top 3 finishes can be done in (34)(33)(32) = 35904 ways
Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) can serve.
An example of such system of equation is
x > 0
y > 0
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.
Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.
Part C:
Given that </span><span>Natalie
can only attend a school in her designated zone and that Natalie's zone is
defined by y < −2x + 2.
To identify the schools that
Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.
For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true
For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true
For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false
For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true
For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false
For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.
</span>
9×8-29+30/15-15 equals 30
Answer:
Para trasladar 300 cajones, las dos opciones cuestan los mismo. Son indiferentes entre si.
Step-by-step explanation:
<u>Dada la siguiente información:</u>
<u></u>
Cantidad de cajones= 300
Opción A:
Camiones con capacidad para 50 cajones que cobran 150 por viaje.
Cantidad de camiones= 300/50= 6
Opción B:
Camiones con capacidad de 60 cajones por 180 por viaje.
Cantidad de camiones= 300/60= 5
<u>Debemos calcular el costo total de cada opción y elegir la de menor costo.</u>
Costo total A= 150*6= $900
Costo total B=180*5= $900
Para trasladar 300 cajones, las dos opciones cuestan los mismo. Son indiferentes entre si.
