When a die is rolled, there are six posible results:
1, 2, 3, 4, 5, and 6
Total number of posible results: n=6
Divisible by 4 is only the result 4
Number of favorable results: f=1
P(divisible by 4)=f/n
P(divisible by 4)=1/6
P(divisible by 4)=0.1667
P(divisible by 4)=0.1667*100%
P(divisible by 4)=16.67%
P(divisible by 4)=1/6=0.1667=16.67%
Well let's see:
The first letter can be any one of 26 .
For each one . . .
The second letter can be any one of the remaining 25.
For each one . . .
The third letter can be any one of the remaining 24.
For each one . . .
The two digits can be any number from 01 to 98 ...
except 11, 22, 33, 44, 55, 66, 77, or 88. (No repetition.)
There are 90 of them.
So the total number of possibilities is (26 · 25 · 24 · 90) .
When I multiply that out, I get 1,404,000 .
I don't know how you got your number, so I can't comment on your
method, but I did find something interesting about your number:
If I assume that you did the three letters the same way I did, then
if I divide your number by (26·25·24), the quotient will show me
how you handled the two digits.
1,263,600 / (26·25·24) = 81 .
That's very intriguing, because it's so close to the 90 sets of digits
that I used. But I don't know what it means, or if it means anything
at all.
Answer:
($410 - $60) / $70 = 5
After subtracting the $60 base fee, you can just divide the hours paid for by the hourly wage and find out how many hours were worked. Hope this helps!
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
