Well, this problem is best solved by setting up a system of two linear equations.
A linear equation can be defined
y=mx+b
where
b=initial value (when x=0), and
m=rate of increase or decrease.
In the given example, the x-axis represents hour, and the y-axis, number of cells.
Chemical #1
initial value = b = 12000 cells
rate = m = -4000 / hr
The equation is therefore
y1=-4000x+12000......................(1)
Similarly, for chemical #2
initial value = b = 6000 cells
rate = m = -3000 / hr
The equation is therefore
y2=-3000x+6000 .......................(2)
The time the two will have an equal sized colony would represent the solution of the system of equations (1) and (2), i.e. when y1=y2
which means
-4000x+12000 = -3000x+6000
transpose and solve for x
4000x-3000x = 12000-6000
1000x=6000
x=6 hours.
At 6 hours from the start,
y=-4000x+12000 = -4000*6+12000 = -24000+12000 = -12000 cells
So the solution is x=6, y=-12000, or (6,-12000)
Physical interpretation
Since cells cannot have a negative number, the two are actually equal before six hours, when they are both zero.
Case 1: y=0 when x=3
Case 2: y=0 when x=2
Therefore, after three hours, both trials will have zero cells.
You have to judge whether to give the mathematical solution (x=6,y=-12000) or the physical interpretation (x=3, y=0) as the answer.
Answer:
Range - 9
Mean - 16
Step-by-step explanation:
I thinkk
Answer:
119
Step-by-step explanation:
1/2x5x7x4+7x7
can I have the brainliest ?
75%.
3/4 is .75. Drag the decimal to the right two times and it is 75%.
Answer:
The population mean of at least one treatment effect are different.
Step-by-step explanation:
An analysis of variance (ANOVA) is conducted in order to determine if there are significant differences between the values of the population mean with respect to the response variable for the domains that under the effects of different treatments. A low p-value leads to reject the null hypothesis of the following hypothesis system:
Rejecting H0 means that this hypothesis is false and, in turn, allows us to conclude that the population mean of one of the domains is different from the others.
