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3 years ago

In the data set below, what is the variance? 9 8 9 5 9 If the answer is a decimal, round it to the nearest tenth. variance (σ2):

Mathematics

1 answer:

victus00 [196]3 years ago

7 0

Answer:

2.4

Step-by-step explanation:

We would first calculate the mean, because we would need it to calculate our varance.

Mean= sum of the number of given data set/ total number

Mean = (9 +8 +9 +5 +9 )/5

= 40/5= 8

variance (σ2):= [(9-8)^2 + (8-8)^2 + (9-8)^2 +(5-8)^2 +(9-8)^2]/5

=( 1+ 0 + 1 +9 +1)/5

= 12/5

variance (σ2)= 12/5

=2.4

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Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]

iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

$x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6$

First, we can expand this power using the binomial theorem:

$(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}$

After that, we can apply De Moivre's theorem to expand each summand: $(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)$

The final step is to find the common factor of i in the last expansion. Now:

$x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6$

$=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6$

$=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))$

The last part is to multiply these factors and extract the imaginary part. This computation gives:

$Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288$

$Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288$

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

A calculator simplifies the imaginary part Im(x⁶) to 0

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3 years ago

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Elena-2011 [213]

Answer:

(2, - 3)

Step-by-step explanation:

given the 2 equations

5x + 3y = 1 → (1)

7x + 3y = 5 → (2)

multiply all terms in (1) by - 1

- 5x - 3y = - 1 → (3)

add (2) and (3) term by term to eliminate the term in y

(7x - 5x) + (3y - 3y) = (5 - 1)

2x = 4 ( divide both sides by 2 )

x = 2

Substitute x = 2 in either (1) or (2) for corresponding value of y

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3y = - 9 ( divide both sides by 3 )

y = - 3

solution is (2, - 3)

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3 years ago

10x + 9y=0 8x + 18y=0

Luba_88 [7]

Answer:

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Step-by-step explanation:

multiple the first equation by 2

after subtract equation 2 from 1

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Studentka2010 [4]

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lorasvet [3.4K]

Answer:

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Step-by-step explanation:

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3tanx = -4
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⇒ x = 143°

In Quadrant IV,

270° + 53°
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2 years ago

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