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Alexxx [7]
3 years ago
10

The sequence {an } is defined by a of o = 1 and a of n + 1 = 2 a of n + 2 for n = 0, 1, 2.... What is the value of a of 3?

Mathematics
1 answer:
maria [59]3 years ago
3 0
\begin{cases}a(0)=1\\a(n+1)=2a(n)+2&\text{for }n\ge0\end{cases}

Use the second part of the definition (the recursive one):


a(1)=2a(0)+2=4
a(2)=2a(1)+2=10
a(3)=2a(2)+2=22
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(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

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3 years ago
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postnew [5]

Answer:

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Step-by-step explanation:

This answer is correct or not please tell me

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Read 2 more answers
AN EMPTY SET IS EQUIVALENT TO ZERO?
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Answer:

In mathematics, the empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero.

Hope this helps!

Step-by-step explanation:

7 0
3 years ago
The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 oun
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Answer: A) .1587

Step-by-step explanation:

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i.e. \mu=12.30 and \sigma=0.20

Let x denotes the amount of soda in any can.

Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.

Then, the probability that a randomly selected can will need to go through the mentioned process =  probability that a randomly selected can has more than 12.50 ounces of soda poured into it =

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Hence, the required probability= A) 0.1587

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