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Yuki888 [10]
3 years ago
9

What is the period of the function f(x)=sin(x/3) ?

Mathematics
1 answer:
vitfil [10]3 years ago
6 0
<h3>Answer: 6pi radians</h3>

(this is equivalent to 1080 degrees)

======================================

Explanation:

f(x) = sin(x/3)

is the same as

f(x) = 1*sin( (1/3)(x-0) )+0

and that is in the form

f(x) = A*sin( B(x-C) )+D

The letters A,B,C,D are explained below

A = helps find the amplitude

B = 2pi/T, where T is the period

C = determines phase shift (aka left/right shifting)

D = determines vertical shift = midline

All we care about is the value of B as that is the only thing that is connected to the period T

--------

Compare f(x) = 1*sin( (1/3)(x-0) )+0 with f(x) = A*sin( B(x-C) )+D and we see that B = 1/3, so,

B = 2pi/T

1/3 = 2pi/T

1*T = 3*2pi ... cross multiply

T = 6pi

The period is 6pi radians. This is equivalent to 1080 degrees. To convert from radians to degrees, you multiply by (180/pi).

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Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= \frac{1}{n-1}*[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= \frac{1}{11}*[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= \frac{1}{9}*[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10

The rejection region is then F ≥ 3.10

F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

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