To determine the probability that one or the other circumstances will occur, you will count the number of possible outcomes and divide it by all the possible outcomes.
#of female teaching assistants + # of male teaching assistants + # of female professors
6 + 16 + 11 = 33
Total = 41
33/41 = 80
There is a approximate 80% probability that one or the other will occur.
You can take the log of the left and right hand side, and then apply the <span>logarithm rules:
log(a</span>ˣ) = x·log(a)
log(ab) = log(a) + log(b)
log(9^(x-1) * 2^(2x+2)) = log(6^(3x))
log(9^(x-1)) + log(2^(2x+2)) = 3x log(6)
(x-1) log(9) + (2x+2) log(2) - 3x log(6) = 0
x(log9 + 2log2 - 3log6) = log9 - 2log2
x = (log9 - 2log2) / (log9 + 2log2 - 3log6)
simplifying by writing log9 = 2log3 and log6 = log2+log3
x= 2(log3 - log2) / (2log3 + 2log2 - 3log2 - 3log3) =
x= -2(log3 - log2) / (log3 + log2) = -2 log(3/2) / log(6)
So 6^x = 4/9
Answer:
The answer is 36
Step-by-step explanation:
- First you have to divide 288 by 8 and
- then you have to see how many times 8 will go into 2 it can't so see how many times it will go into 28 which is 3 and you will get 24
- then you subtract 24 from 28 and you get 4
- then bring down the 8 an it 48 now see how many times 8 will go into 48 which is 6 times and you get 48
- then you subtract 48 by 48 and you get 0 so the answer is 36
- HOPE THIS HELPED
A(-7,-4) B(-2,0)
√[(x'-x)^2+(y'-y)^2]
√(-2-(-7)^2+(0-(-4)^2
√(5^2)+(4^2)
√25+16
√41
the distance is approximately 6.4 units
The answer is fifteen (15)
Hope this helps!
