Answer:
For the code we have 3 selections.
The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}
So we have 5 options.
The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)
So here we have 27 - 5 = 22 options
The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.
Here we have 22 - 1 = 21 options.
The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:
C = 5*22*21 = 2310.
There are 2310 different possible codes
I have a good record of guessing the right one so I would say #1.
You first need to change the percent into a decimal then multiply by 80 to get your answer
.35*80=28
Answer:
Step-by-step explanation:
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Lesson : Probability</u></h2><h2><u>
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- We have a bag with 4 white chips and 6 black chips
- Question : What is the probability of randomly choosing a white chip, not replacing it, and then randomly choosing another white chip?
4 white and 6 black
4 + 6 = 10
Choosing the first white chip : 
Since we took one out , now
3 white chips , 6 black chips
3 + 6 = 9
Choosing another white chip : 
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Step-by-step explanation:
3 groups with 6 flowers in each group
or
2 groups with 9 flowers in each group
or
6 groups with 3 flowers in each group
or
9 groups with 2 flowers in each one
