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finlep [7]
3 years ago
15

An = 3n2 + 4n +5 find the first 5 terms of the following sequence

Mathematics
1 answer:
creativ13 [48]3 years ago
5 0

Answer:

              a_1=12\,,\ \ a_2=25\,\ \ a_3=44\,,\ \ a_4= 69\,,\ \ a_5= 100

Step-by-step explanation:

a_1 = 3\cdot1^2+4\cdot1+5= 3+4+5=12\\\\a_2= 3\cdot2^2+4\cdot2+5= 3\cdot4+8+5=25\\\\a_3= 3\cdot3^2+4\cdot3+5= 3\cdot9+12+5=44\\\\a_4= 3\cdot4^2+4\cdot4+5= 3\cdot16+16+5=69\\\\a_5= 3\cdot5^2+4\cdot5+5= 3\cdot25+20+5=100

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Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

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So, probability of event 'H' is given as:

P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, n(S)=6

Now, probability of rolling a 3 is given as:

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Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083

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