Probability of distinct 3-digits is 1/6.
Probability of both the first digit and the last digit are even numbers is 1/5.
<u>Step-by-step explanation</u>:
Given set of numbers {1,2,3,5,8,9} = 6
Total number of 3-digits can be formed = 654 = 120.
<u>Probability of the three-digit numbers with distinct digits</u> :
The number of 3-digit numbers with distinct digits = (654) / (321) = 20.
P(distinct 3-digits) = no. of distinct 3-digits / Total no.of 3-digits
⇒ 20/120
⇒ 1/6
<u>Probability that both the first digit and the last digit of the three-digit number are even numbers</u> :
The even numbers in the set are 2 and 8 = 2 possibilities.
The number of 3-digits with 1st and 3rd digit are even = (262) = 24.
P(1st and 3rd digits even)= no. of 1st and 3rd digit even/Total no. of 3-digits.
⇒ 24/120
⇒ 1/5
Answer:
Step-by-step explanation:
Each term is 4 more than the preceding term.
Common difference d = 4
nth term = a₁ + (n-1)d = -6 + (n-1)4
a₁₀ = -6 + 9⋅4 = 30
I would say A scale which measures by pounds
Y-70
X-30
Z-80
You’re welcome