4
One in 1200 are not particularly good odds. On the other hand, winning the lotto is 1 chance in 13,000,000 which if you've ever played the lotto you know that those odds are good enough to insure that if you played for the rest of your life and you are 18 not expect to live to 80 and you have 104 [given 2 draw a week] chances of winning per year, it likely won't happen. One in 1200 is better but still not good, especially with only 1 draw.
3
As a fraction her probability of winning is 1/2000 which is 0.000833333 as a decimal. You can put that in as
1
÷
1200
=
if you are not sure how your calculator works.
2
Sample Space = {1,2,3,4 .... 1198,1199,1200}
The outcome depends on sophies number. Either 1 number can be chosen or all of them can.
1
The sample space is the integers from 1 to 1200 inclusive.
The question is an illustration of combination and there are 729 potential pass codes available
<h3>How to determine the number of potential pass codes?</h3>
The given parameters are
Symbols available = 9
Length of pass code = 3
From the question, we understand that a symbol may be entered any number of times.
This means that each of the 9 available symbols can be used three times
So, the number of potential pass codes is
Passcodes = 9 * 9 * 9
Evaluate the product
Passcodes = 729
Hence, there are 729 potential pass codes available
Read more about combination at:
brainly.com/question/11732255
#SPJ1
Answer:
b. 40,960
Step-by-step explanation:
There is probably some equation you could do for this but I find it simpler just to take 10 and multiply it by 2 and then keep multiplying the answers by 2 twelve times so:
10*2=20
20*2=40
40*2=80
80*2=160
160*2=320
320*2=640
640*2=1280
1280*2=2560
2560*2=5120
5120*2=10240
10240*2=20480
20480*2=40960
(1/3)/(5/6)=(1/3)/(6/5)=6/15
