In 1993, Moose Population: 3280
In 1999, population became: 4960
P (Population) , t (years)
t = 6 —> 4960 - 3280 = 1680
Average Change —> 1680/6 = 280 moose/year
• In terms of 1990:
t = 3 —> 3280-3 (280)
P(1990) = 2440
P(t) = 2440 + 280t
• In 2003; t = 13
P(13) = 2440 + 280 (13)
P(13) = 2440 + 3640
P(13) = 6080
• Moose population in 2003
= 6080
Answer:
y=-2x-3
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-5-(-3))/(1-0)
m=(-5+3)/1
m=-2/1
m=-2
y-y1=m(x-x1)
y-(-3)=-2(x-0)
y+3=-2(x)
y+3=-2x
y=-2x-3
Answer:
C Both the mean and median are appropriate measures of center.
Step-by-step explanation:
2/6 + 2/6 + 2/6
1/6 + 5/6
4/6 + 2/6
Answer:
The lines will intersect 595 times.
Step-by-step explanation:
We can do it in 2 methods.
<u>1st method:-</u> Let us count the number of intersections by drawing lines one by one.
1st line - 0 points=0
2nd line- new 1 point=1
3rd line - old 1 point + new 2 points with two old lines = 1+2
4th line - old (1+2) points+ new 3 points= 1+2+3
5th line - old (1+2+3) points + new 4 points=1+2+3+4
Like that from 35th line-1+2+3+4......34 = 
<u>2nd method:-</u>
We have to choose 2 lines from 35 lines.
We can do it in 
ways = 595 points
