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Llana [10]
3 years ago
15

Find the sum of the first 18 odd positive integers.

Mathematics
1 answer:
faltersainse [42]3 years ago
5 0

Answer:

The sum of the first 18 odd positive integers is 324.

Step-by-step explanation:

Given : The first 18 odd positive integers.

To find : The sum of the first 18 odd positive integers?

Solution :

The odd positive integers are 1,3,5,.... forming an Arithmetic sequence .

Where, First term is a=1

Common difference is d=2

Number of terms n=18

The sum of A.P formula is

S_n=\frac{n}{2}[2a+(n-1)d]

Substitute the values in the formula,

S_{18}=\frac{18}{2}[2(1)+(18-1)2]

S_{18}=9[2+(17)2]

S_{18}=9[2+34]

S_{18}=9[36]

S_{18}=324

Therefore, The sum of the first 18 odd positive integers is 324.

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In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p
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Answer:

\mathbf{P(X=5) =0.0888}    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

\mathtt{P(X=x) =(^{n}_{x} )   \  \pi^x \  (1-\pi)^{n-x}}

\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} )   \  \pi^x \  (1-\pi)^{n-x}}

where;

n = 8 and π = 0.36

For x = 5

The probability \mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} )   \  0.36^5 \  (1-0.36)^{8-5}}

\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} )   \  0.36^5 \  (0.64)^{3}}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =({8 \times 7 } )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =0.0887645}

\mathbf{P(X=5) =0.0888}     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})

{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times  (0.36)^0  \times  (1-0.36)^8  \ )  +  \dfrac{8!}{1!(7!)} \times  (0.36)^1  \times  (1-0.36)^7  \ +\dfrac{8!}{2!(6!)} \times  (0.36)^2  \times  (1-0.36)^6  \ +  \dfrac{8!}{3!(5!)} \times  (0.36)^3  \times  (1-0.36)^5 +  \dfrac{8!}{4!(4!)} \times  (0.36)^4  \times  (1-0.36)^4  \  +  \dfrac{8!}{5!(3!)} \times  (0.36)^5  \times  (1-0.36)^3  \ )

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

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