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tatuchka [14]
2 years ago
8

A1 = 10 and an =an -1 +2

Mathematics
1 answer:
Vesna [10]2 years ago
8 0

I assume you're asking to solve for the n-th term in the sequence, a_n.

From the given recursive rule,

a_n = a_{n-1} + 2 \implies a_{n-1} = a_{n-2} + 2

and by substitution,

\implies a_n = a_{n-2} + 2\times2

Similarly,

a_n = a_{n-1} + 2 \implies a_{n-2} = a_{n-3} + 2

\implies a_n = a_{n-3} + 3\times2

The pattern continues, so that we can write the n-th term in terms of the 1st one:

a_n = a_1 + (n-1)\times2 \implies a_n = 10 + 2(n-1) = \boxed{2n+8}

So the first few terms of the sequence are

{10, 12, 14, 16, 18, 20, …}

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You roll two fair dice, one green and one red. (a) Are the outcomes on the dice independent? Yes No (b) Find P(1 on green die an
ASHA 777 [7]

Answer:

a

  Yes

b

 P(1 \& 5) = \frac{1}{36}

c

 P(5 \& 1) = \frac{1}{36}

d

 P(5 \& 1 | 1\& 5  ) = \frac{1}{18}

Step-by-step explanation:

From the question we are told that

     Two fair dice, one green and one red were rolled

Generally the  outcomes on the dice independent because the outcome on the first dice is not affected by the second die

Generally the  probability of getting a 1  on a dice rolled is  P(1) =  \frac{1}{6}

the  probability of getting a 5 on a dice rolled is  P(5) =  \frac{1}{6}

Generally the probability of P(1 on green die and 5 on red die) is mathematically represented as

         P(1 \& 5) = \frac{1}{6}  *\frac{1}{6}

         P(1 \& 5) = \frac{1}{36}

Generally the probability of P(5 on green die and 1 on red die) is mathematically represented as

          P(5 \& 1) = \frac{1}{6}  *\frac{1}{6}

         P(5 \& 1) = \frac{1}{36}

Generally the probability of P((1 on green die and 5 on red die) or (5 on green die and 1 on red die)) is mathematically represented as

          P(5 \& 1 | 1\& 5  ) = \frac{1}{36} + \frac{1}{36}

         P(5 \& 1 | 1\& 5  ) = \frac{1}{18}

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3 years ago
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