6a³-18a²-1620a
Collect 6a:
6a(a²-3a-270)
6a(a-18)(a+15)
Answer:
x=2.125
y=0
C=19.125
Step-by-step explanation:
To solve this problem we can use a graphical method, we start first noticing the restrictions
and
, which restricts the solution to be in the positive quadrant. Then we plot the first restriction
shown in purple, then we can plot the second one
shown in the second plot in green.
The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.
So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.
Answer:
Null hypothesis; H0: µ ≥ 52 mpg
Alternative hypothesis; Ha: µ < 52 mpg
Step-by-step explanation:
We are told that the newest model gets an average of 52 miles per gallon and that the buyers believe that the average mpg was overstated.
This means the alternative hypothesis will have an average less than 52. Thus;
Null hypothesis; H0: µ ≥ 52 mpg
Alternative hypothesis; Ha: µ < 52 mpg
The answer is 2
Because 2-3=-1
Answer:
Relative frequency of male nonparticipation is O.42
Step-by-step explanation:
Figure 1 is your partially filled frequency table.
1. Complete the table
(a) Total No
Total Yes + Total No = Total
102 + Total No = 187
Total No = 187 - 102
= 85
(b) Female No
Male No + Female No = Total No
40 + Female No = 85
Female No = 85 - 40
= 45
(c) Female Yes
Female Yes + Female No = Female Total
Female Yes + 45 = 95
Female Yes = 95 - 45
= 50
(d) Male Yes
Male Yes + Female Yes = Total Yes
Male Yes + 50 = 102
Male Yes = 102 - 50
= 52
(e) Male Total
Male Yes + Male No = Male Total
52 + 40 = Male Total
Male Total = 92
Figure 2 shows the completed table.
2. Frequency of Male No
There are 92 males, of whom 40 do not participate in an after-school activity.
The relative frequency of male nonparticipation is
40/92 = 0.43