A group of friends spent 1, 0, 2, 3, 4, 3, 6, 1, 0, 1, 2, and 2 hours online last night. What are the 1st and 3rd quartiles for
Ostrovityanka [42]
Arrange the numbers from smallest to largest:
0,0,1,1,1,2,2,3,3,4,6
Split the data in the middle, since there is an odd amount of numbers take an even amount on each side:
0,0,1,1,1,2,2,3,3,4,6
Lower half: 0,0,1,1,1
Higher half: 2,3,3,4,6
1st quartile is the middle value of the lower half: 1
3rd quartile is the middle value of the higher half: 3
Answer:
db / dt = kb
this becomes b(t) = Ce^(kt)
C = 100, the initial population
P(1) = 420 = 100 e^(1k)
4.2 = e^k
ln 4.2 = k
a) thus, b(t) = 100 e^(t ln 4.2)
b) b(3) = 100 e^(3 ln 4.2)
c) growth constant will still be ln 4.2 (constant percentage of populatioin)
d) 10000 = 100 e^(t ln 4.2)
100 = e^(t ln 4.2)
ln 100 = t ln 4.2
t = ln 100 / ln 4.2
Step-by-step explanation:
Answer:
8 days to sell all the games
Step-by-step explanation:
;)
Answer:
366
Step-by-step explanation:
2×3+4×100-50+10
PEMDAS says multiply and divide from left to right
6 + 400 - 50 +10
Then add and subtract
406-50+10
356+10
366
