9514 1404 393
Answer:
- red division: 6 teams
- blue division: 5 teams
Step-by-step explanation:
We can let r and b represent the numbers of teams in the red and blue divisions, respectively. The total number of goals scored in each division will be the average for that division times the number of teams in that division.
r - b = 1 . . . . . . there is 1 more red team than blue
4.5r +4.2b = 48 . . . . . . total goals scored per week
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Solving by substitution, we have ...
r = b +1
4.5(b +1) +4.2b = 48 . . . . substitute for r
8.7b +4.5 = 48 . . . . . . . . simplify
8.7b = 43.5 . . . . . . . . . . subtract 4.5
b = 43.5/8.7 = 5 . . . . . divide by 8.7
r = b +1 = 6 . . . . . . . . . find r
There are 6 red teams and 5 blue teams.
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<em>Additional comment</em>
The basic idea is that you make an equation for each relation given in the problem statement. For a problem like this, you do need to have an understanding of how the average number of goals would be calculated and how that relates to the total goals.
The van is traveling about 55.9 miles per hour.
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>
Answer:
f(- 2) = - 1
Step-by-step explanation:
To evaluate f(- 2) substitute x = - 2 into f(x) , that is
f(- 2) = 3 - (- 2)² = 3 - 4 = - 1
Answer:
84.3%
Step-by-step explanation:
You first find the frequency. You do this by multiplying the frequency density by the class length.