Answer:
A. Initially, there were 12 deer.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. After 15 years, there will be 410 deer.
D. The deer population incresed by 30 specimens.
Step-by-step explanation:

The amount of deer that were initally in the reserve corresponds to the value of N when t=0


A. Initially, there were 12 deer.
B. 
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. 
C. After 15 years, there will be 410 deer.
D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:
ΔN=N(15)-N(10)
ΔN=410 deer - 380 deer
ΔN= 30 deer.
D. The deer population incresed by 30 specimens.
Answer:
x=3
Step-by-step explanation:
f(x) = 2^x
Let f(x) =8
8 = 2^x
Rewriting 8 as 2^3
2^3 = 2^x
The bases are the same so the exponents are the same
3=x
Answer:
3.33
Step-by-step explanation:
Solve for w:
p = (1.2 w)/h^2
(1.2 w)/h^2 = (6 w)/(5 h^2):
p = (6 w)/(5 h^2)
p = (6 w)/(5 h^2) is equivalent to (6 w)/(5 h^2) = p:
(6 w)/(5 h^2) = p
Multiply both sides by (5 h^2)/6:
Answer: w = (5 h^2 p)/6
Your collection's mean answers
median - 1930
range - 54
interquartile range - 50
mean absolute devitation - 23.75
Your friends collection mean answers
median - 1929.5
range - 15
interquartile range - 7
mean absolute devitation - 3.5
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