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
Answer:
%70
Step-by-step explanation:
<h2><u><em>
Brainliest Please!</em></u></h2>
Answer:
<h2>
11.4</h2>
Solution,
∆ ACB = ∆ EFD
finding the value of X,
Apply cross product property
Calculate the product
Divide both sides by 5
Calculate
Hope this helps...
Good luck on your assignment...
Answer:
A. 
(first option)
Step-by-step explanation:
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186.99999992 and that is the correct anwser i think
