Answer:
The answers are
EBF and FBC
ABD and DBE
Step-by-step explanation:
Umm what do you mean can you explain more of it 2 me plz<span />
Answer:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
Dividing fractions is equivalent to taking the reciprocal of the second and multiplying them. In this case, we have 1/2*8/5=8/10=4/5.
Answer:
1090*.08=87.2. 87.2*5years=$436.
Step-by-step explanation:
hope this helps
