Answer: Option 'c' is correct.
Step-by-step explanation:
Since we have given that
the optimized solution of a linear program to an integer as it does not affect the value of the objective function.
As if we round the optimized solution to the nearest integer, it does not change the objective function .
while it is not true that it always produces the most optimal integer solution or feasible solution.
Hence, Option 'c' is correct.
Answer:
The tap drains approximately half the water from the tank in 15 minutes
The tank initially has 50 gallons of water
Step-by-step explanation:
<u>The tap drains approximately half the water from the tank in 15 minutes. </u><u>True.</u>
This is true because see that the point (15,25) lies on the graph of the line.
Which means after 15 minutes the amount of water left in the tank is 25 gallons and 25 is half of 50.
The tap drains exactly one gallon from the tank every minute. False
This is false because the slope of the graph is 
<u>The tank initially has 30 gallons of water. </u><u>False.</u>
The y-intercept is the initial gallons of water which is 50.
<u>The tank initially has 50 gallons of water. </u><u>True.</u>
The y-intercept is the initial gallons of water which is 50.
<u>The tank takes 50 minutes to drain completely. </u><u>False</u><u>.</u>
It is false because the x-intercept tells us the tank is drained completely and it is not 50 minutes but rather 30 minutes.
The Gradient would be negative therefore making it start by going down then up, making a sad mouth. the x intercepts will be the same. Frankly it just flipped upside down. I remember it like if the gradient is positive then happy smile, if its negative then sad smile. Hope I helped.
Answer: don't know sorry
Step-by-step explanation:
Answer:
<em>250 cameras / day</em>
<em>The daily profit would be $300</em>
Step-by-step explanation:
<u>Modeling With Equations</u>
Some situations in real life are adequate for being modeled as functions of the variables they depend on. Equations can be of great help for economy calculations since we could determine optimum levels of production, revenue, costs, and other useful information.
The fixed cost our camera manufacturer is $1500 each day and a variable cost of $9 per camera sold. It can be written as
C(x)=1500+9x
Being C(x) the total cost of manufacturing, and x the number of cameras sold each day
If the company sells the cameras for $15 apiece, then the revenue for x cameras will be
R(x)=15x
a)
We want to find out how many cameras must be sold to equal its daily cost, so
1500+9x=15x
6x=1500
x=250 cameras / day
b) Given the manufacturer can increase production by 50 cameras per day (x=300), then the revenue will be
R(300)=15(300)=$4500
And the cost
C(300)=1500+9(300)=4200
The daily profit would be $4500-$4200=$300
