Answer:
The largest possible number of x intercept is 9 while the largest possible number of relative max/min is 8
Step-by-step explanation:
For any polynomial of degree n with distinct and real solutions, it can have at most n different x intercepts. This would imply it can have at most 9 distinct real solutions.
It can also have at most n-1 relative max/min in alternating order. This is best illustrated when such polynomial is sketched on a graph.
For example a quadratic expression is a polynomial of degree 2 and has at most 2 distinct solutions and 1 relative max/min.
In this question, for the polynomial, its degree (n) = 9
So it can have at most 9 x intercepts and at most 8 relative max/min.
Answer:
a. The expected or average costs for all weekly rat purchases is $20.00
Step-by-step explanation:
a. A mean value of $20.00 means that over a period of 52 weeks, the company can expect to spend $20.00 per week on rat purchases.
b. This is incorrect since individual values don't interfere in the mean. For instance, if half the weeks had a cost of $19.00 and the other half had a cost of $21.00, the mean cost would still be $20.00 even though no particular week had a $20.00 cost
c. Incorrect. The median is the central value in a distribution; the median and the mean are not necessarily the same.
d. Incorrect, same as item b.
Answer:
The average velocity the entire trip was 43.25 mi/h
Step-by-step explanation:
In this case we have to calculate an average based on the miles traveled. First we have to calculate the total of miles traveled, then calculate the portion of the total travel of each and with this calculate the average speed during the trip. First the total miles traveled:
Now the percentages:
At 57 mi/h were 
At 44 mi/h were 
At 37.8 mi/h were 
Now multiplying the speed by the portion and summing them we can have the average velocity:
The average velocity the entire trip was 43.25 mi/h
Hello,
I note (a,b,c) the result of a quarters, b dimes and c pennies:
2 solutions:
106=( 3, 3, 1)=( 1, 8, 1)
106=( 0, 0, 106) but : 100= 0*25+ 0*10+ 100
106=( 0, 1, 96) but : 100= 0*25+ 1*10+ 90
106=( 0, 2, 86) but : 100= 0*25+ 2*10+ 80
106=( 0, 3, 76) but : 100= 0*25+ 3*10+ 70
106=( 0, 4, 66) but : 100= 0*25+ 4*10+ 60
106=( 0, 5, 56) but : 100= 0*25+ 5*10+ 50
106=( 0, 6, 46) but : 100= 0*25+ 6*10+ 40
106=( 0, 7, 36) but : 100= 0*25+ 7*10+ 30
106=( 0, 8, 26) but : 100= 0*25+ 8*10+ 20
106=( 0, 9, 16) but : 100= 0*25+ 9*10+ 10
106=( 0, 10, 6) but : 100= 0*25+ 10*10+ 0
106=( 1, 0, 81) but : 100= 1*25+ 0*10+ 75
106=( 1, 1, 71) but : 100= 1*25+ 1*10+ 65
106=( 1, 2, 61) but : 100= 1*25+ 2*10+ 55
106=( 1, 3, 51) but : 100= 1*25+ 3*10+ 45
106=( 1, 4, 41) but : 100= 1*25+ 4*10+ 35
106=( 1, 5, 31) but : 100= 1*25+ 5*10+ 25
106=( 1, 6, 21) but : 100= 1*25+ 6*10+ 15
106=( 1, 7, 11) but : 100= 1*25+ 7*10+ 5
106=( 1, 8, 1) is good
106=( 2, 0, 56) but : 100= 2*25+ 0*10+ 50
106=( 2, 1, 46) but : 100= 2*25+ 1*10+ 40
106=( 2, 2, 36) but : 100= 2*25+ 2*10+ 30
106=( 2, 3, 26) but : 100= 2*25+ 3*10+ 20
106=( 2, 4, 16) but : 100= 2*25+ 4*10+ 10
106=( 2, 5, 6) but : 100= 2*25+ 5*10+ 0
106=( 3, 0, 31) but : 100= 3*25+ 0*10+ 25
106=( 3, 1, 21) but : 100= 3*25+ 1*10+ 15
106=( 3, 2, 11) but : 100= 3*25+ 2*10+ 5
106=( 3, 3, 1) is good
106=( 4, 0, 6) but : 100= 4*25+ 0*10+ 0
