Title:
<h2>The revenue will be maximum for 253 passengers.</h2>
Step-by-step explanation:
Let, the number of passenger is x, which is more than 194.
In this case, the travel agency will charge [312 - (x - 194)] per passenger.
The total revenue will be ![x[312 - (x - 194)] = 506x - x^{2}](https://tex.z-dn.net/?f=x%5B312%20-%20%28x%20-%20194%29%5D%20%3D%20506x%20-%20x%5E%7B2%7D)
.
As x is the variable here, we can represent the revenue function by R(x). Hence, 
.
The revenue will be maximum when 
.
Answer:
y = 1.1x +4.46
y = 129.86 for x = 114
Step-by-step explanation:
The two-point form of the the equation for a line is useful for this.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (7.98 -2.7)/(3.2 -(-1.6))(x -(-1.6)) + 2.7
y = 5.28/4.8(x +1.6) + 2.7
y = 1.1x +1.76 +2.7
y = 1.1x +4.46
__
When x=114, the value of y is ...
y = 1.1(114) +4.46
y = 129.86
Answer:
60
Step-by-step explanation:
if 3 represents 5% and we know 5%*20 =100% then
3*20=60 represents 100%
It would be 34.34
Hope this helps! :)
Answer:
12 possibilities
Step-by-step explanation:
In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.
The same thing occurs in the second urn, as all balls have different labels.
The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).
For the first urn, we have a combination of 4 choose 2:
C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities
For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.
In total we have 6 + 6 = 12 possibilities.
