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love history [14]

3 years ago

12

Which off the following represents the general term for the sequence 1, 3, 5, 7, 9, ...?

2 answers:

tino4ka555 [31]3 years ago

8 0

The general term for the sequence is n+2.

Basile [38]3 years ago

3 0

Answer:

2n-1

Step-by-step explanation:

We are given a sequence:

1,3,5,7,9,...

We are asked its general term

Clearly this is a series of odd numbers

So, the general term is 2n-1

n=1 i.e. first term = 2-1=1

n=2 second term=4-1=3

n=3 third term= 6-1=5

and so on

Hence, general term of the sequence 1,3,5,7,9,... is:

2n-1

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A geologist has collected 5 specimens of basaltic rock and 7 specimens of granite. The geologist instructs a laboratory assistan

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The rocks are chosen without replacement, which means that the hypergeometric distribution is used to solve this question. First we get the parameters, and then we answer the questions. From this, we get that:

$E(X) = 5.25, Var(X) = 0.5966$
P(X < 6) = 0.9545
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Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The mean and the variance are:

$\mu = \frac{nk}{N}$

$\sigma^2 = \frac{nk(N-k)(N-n)}{N^2(N-1)}$

We have that:

5 + 7 = 12 rocks, which means that $N = 12$

9 are chosen, which means that $n = 9$

7 are granite, which means that $k = 7$

Question a:

$E(X) = \mu = \frac{9\times7}{12} = 5.25$

$Var(X) = \sigma^2 = \frac{9\times7(12-7)(12-9)}{12^2(12-1)} = 0.5966$

Thus:

$E(X) = 5.25, Var(X) = 0.5966$

Question b:

Since there are only 5 specimens of basaltic rock, at least 9 - 5 = 4 specimens of granite are needed, which means that:

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,12,9,7) = \frac{C_{7,4}*C_{5,5}}{C_{12,9}} = 0.1591$

$P(X = 5) = h(5,12,9,7) = \frac{C_{7,5}*C_{5,4}}{C_{12,9}} = 0.4773$

$P(X = 6) = h(6,12,9,7) = \frac{C_{7,6}*C_{5,3}}{C_{12,9}} = 0.3181$

Thus

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1591 + 0.4773 + 0.3181 = 0.9545$

So P(X < 6) = 0.9545.

Question c:

5 of basaltic and 4 of granite: 0.1591 probability.

7 of granite is P(X = 7), in which

$P(X = 7) = h(7,12,9,7) = \frac{C_{7,7}*C_{5,2}}{C_{12,9}} = 0.0455$

0.1591 + 0.0455 = 0.2046, thus:

P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

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