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Butoxors [25]
3 years ago
9

The nth term of a sequence is 3n + 2

Mathematics
2 answers:
kifflom [539]3 years ago
6 0
<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Simply substitute the first 3 values in (1, 2, 3)

A) 3(1) + 2 = 5

3(2) + 2 = 8

3(3) + 2 = 11

The first 3 terms would be 5, 8 and 11

B) To find the 10th term, subtitute 10 in

3(10) + 2 = 32

It would be 32

<h3><u>✽</u></h3>

➶ Hope This Helps You

➶ Good Luck

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

gavmur [86]3 years ago
3 0

Answer:

see explanation

Step-by-step explanation:

To calculate the first 3 terms substitute n = 1, 2, 3 into the n th term rule

a_{1} = (3 × 1) + 2 = 3 + 2 = 5

a_{2} = (3 × 2) + 2 = 6 + 2 = 8

a_{3} = (3 × 3) + 2 = 9 + 2 = 11

------------------------------------------------------------------

substitute n = 10 into the n th term rule

a_{10} = (3 × 10) + 2 = 30 + 2 = 32

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Answer:

3/14

Step-by-step explanation:

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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is
nadya68 [22]

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each with mean/expectation

E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i

and variance

\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2

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E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3

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\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]

\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2

We have

(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)

E[(W_1+W_2+W_3)^2]

=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])

because W_i and W_j are independent when i\neq j, and so

E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3

giving a variance of

\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3

and so the standard deviation is \sqrt{\dfrac{350}3}\approx\boxed{116.67}

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

\mathrm{Var}[W_1+W_2+W_3]

=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])

and since the W_i are independent, each covariance is 0. Then

\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]

\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3

and take the square root to get the standard deviation.

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