3-(1/3) to the 4th power
1 answer:
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The value of a₃₁ of the arithmetic sequence exists 77.4.
<h3>How to find the value of a₃₁ of the arithmetic sequence?</h3>Given: a₅ = 12.4 and a₉ = : 22.4
For the arithmetic sequence a₁, a₂, a₃, ..., the n-th term exists
where d = common difference
a₅ = 12.4,
a₁ + 4d = 12.4 .........(1)
Because a₉ = 22.4,
a₁ + 8d = 22.4 .........(2)
Subtract (1) from (2), we get
a₁ + 8d - (a₁ + 4d) = 22.4 - 12.4
4d = 10
Dividing throughout by 4, we get
d = 2.5
From (1), we get
a₁ = 12.4 - 4 2.5 = 2.4
a₃₁ = 2.4 + 30 2.5 = 77.4
Therefore, the correct answer is a₃₁ = 77.4
To learn more about the arithmetic sequence refer to; brainly.com/question/6561461
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Answer:
51%
Step-by-step explanation:
Given P(passing math) = 59%, P(passing physics) = 26%, and P(passing both) = 17%, you want to find the probability of passing only one of the courses.
<h3>Probability relations</h3>We can record the given probabilities in a 2-way table (values shown in blue). The table is completed by making sure the totals add up (values shown in black).
The probability of passing one course and failing the other is the sum of the probabilities with a yellow background:
42% +9% = 51%
The probability of passing one or the other is 51%.
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<em>Additional comment</em>
We can also get there using the relation ...
P(A+B) = P(A) +P(B) -P(AB)
The union of A and B also includes their overlap:
P(A+B) = P(AB') +P(A'B) +P(AB)
In other words, the probability of interest is ...
P(AB') +P(A'B) = P(A) +P(B) -2×P(AB) = 59% +26% -2(17%)
P(AB') +P(A'B) = 51%
