Answer:
34 + 8n
Step-by-step explanation:
a₁ = 42
a₂ = a₁ + 8 = 42 + 8 = 50
a₃ = a₂ + 8 = 50 + 8 = 58
a₄ = a₃ + 8 = 58 + 8 = 66
Arithmetic sequence: 42, 50 , 58 , 66, ......
a₁ = 42 ; d = 8
= a₁ + d(n-1)
= 42 + 8(n - 1)
= 42 + 8n - 8
= 34 + 8n
What is 2y + x (3y) = - 15
1 answer:
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Answer:
135 and 135
Step-by-step explanation:
The computation is shown below:
The number of examiners who passed in only one subject is as follows
= n(E) - n(E ∩M) + n(M) - n(E ∩M)
= (80 - 60 + 70 - 60)%
= 30%
Now the number of students who passed in minimum one subject is
n(E∪M) = n(E) + n(M) - n(E ∩M)
= 80 - + 70 - 60
= 90%
Now the number of students who failed in both subjects is
= 100 - 90%
= 10% of total students
= 45
So total number of students appeared for this 450
So, those who passed only one subject is
= 450 × 30%
= 135
Now the Number of students who failed in mathematics is
= 100% - Passed in Mathematics
= 100% - 70%
= 30% of 450
= 135
