Help me quick plz thanks
1 answer:
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Answer:
Choice number one:
.
Step-by-step explanation:
- Let
be the event that the number on the first card is even.
- Let
be the event that the number on the second card is even.
The question is asking for the possibility that event and
happen at the same time. However, whether
occurs or not will influence the probability of
. In other words,
and
are not independent. The probability that both
and
occur needs to be found as the product of
- the probability that event
- the probability that event
5 out of the ten numbers are even. The probability that event occurs is:
.
In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:
.
The probability that both and
occurs will be:
.
Answer:
0xFEE₁₆ = 7756₈
Step-by-step explanation:
Convert the hexadecimal number to decimal first
0xFEE
E = 14
F = 15
Then convert them to binary
Remainder = 0
Remainder = 1
Quotient = 1
Remainder = 1
Remainder = 1
Remainder = 1
Quotient = 1
Remainder = 1
0 F E E
0000 1111 1110 1110
Take groups of three
and multiply with where n is the place of the digit
111 111 101 110
4,2,1 4,2,1 4,2,1 4,2,1
4+2+1 4+2+1 4+0+1 4+2+0
7756
So, 0xFEE₁₆ = 7756₈