The given sequence is
a₁ = 29
a₂ = 39
a₃ = 49
a₄ = 59
This sequence is an arithmetic sequence. Th first term is a₁ = 29, and the common difference is d= 10.
The n-th term is
The 33-rd termis
a₃₃ = 29 + (33 - 1)*10
= 29 + 320
= 349
Answer: a₃₃ = 349
Answer:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30
Answer:
(x-7)/(x+2)
Step-by-step explanation:
(x-7)(x+7)/(x+7)(x+2)
(x-7)/(x+2)
(20 + 12) - 8, since the question states n is EIGHT less than the sum of 20 and 12.
Answer:
2:9
Step-by-step explanation
4:18 divided by 2 is the lowest it can go so 4 divided by 2 is 2
18 divided by 2 is 9 so,
2:9
