Answer:
Actually, you have the correct answer. I think you are confused as to what the exact answer might be.
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% !
Answer:
6,000 dollars
Step-by-step explanation:
We first need to find out how much it would cost Danielle for a month if she chose the second option
200 + 40 = 240
240 x 30 = 7,200
Now that we know how much Danielle would pay if she chose the second option, now we need to find out how much she <em>saved</em>.
7,200 - 1,200 = 6,000
So Danielle saved 6,000 dollars by choosing the first option instead of the second one.
It is related to econometrics, however using simple algebraic logic we can solve this problem. General formula with residual is
, where r is a residual. Plugging the pair into the formula, we obtain that
and r=0.5
