What arithmetic sequence with a_{4}=18 and a_{10}=48
1 answer:
Answer:
= 5n - 2
Step-by-step explanation:
the nth term of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
given a₄ = 18 and a₁₀ = 48 , then
a₁ + 3d = 18 → (1)
a₁ + 9d = 48 → (2)
subtract (1) from (2) term by term to eliminate a₁
6d = 30 ( divide both sides by 6 )
d = 5
substitute d = 5 into (1) and solve for a₁
a₁ + 3(5) = 18
a₁ + 15 = 18 ( subtract 15 from both sides )
a₁ = 3
Then
= 3 + 5(n - 1) = 3 + 5n - 5 = 5n - 2
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Answer:
S = {0,2,3,4}
P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001
Mean = 0.879
Standard Deviation = 1.033
Step-by-step explanation:
Let the number of people having same birth month be = x
The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)
The number of ways of distributing their birthdays = 12⁴
The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)
P(X = 0) =
P(X=0) = 0.573
P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)
P(X=2) =
P(X=2) = 0.401
P(X=3) =
P(x=3) = 0.025
P(X=4) =
P(X=4) = 0.001
Mean,
Standard deviation,
SD = 1.033
Answer:
s> 144 2/3
Step-by-step explanation:
I took the test in k12 and i got it right so yea !
