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gogolik [260]
3 years ago
12

What is the period, in seconds, of the simple harmonic motion described by the equation x = 5sin(8πt)?

Mathematics
2 answers:
Zolol [24]3 years ago
5 0
\bf \qquad \qquad \qquad \qquad \textit{function transformations}

\bf \begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\

\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}

now, let's see your equation  \bf \begin{array}{llccclll}
x=&5sin(&8\pi t&+0)&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\qquad period\implies \cfrac{2\pi }{B}\iff \cfrac{2\pi }{8\pi }
mart [117]3 years ago
3 0

<span>t=0.007958x
</span>
Hope that helps!

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I need help on this
valentinak56 [21]

Answer:

Im pretty sure it's the 2nd one

8 0
3 years ago
Read 2 more answers
Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor
kherson [118]

Answer:

A) 95% confidence interval for the population mean PEF for children in biomass households = (3.314, 3.486)

95% confidence interval for the population mean PEF for children in LPG households

= (4.195, 4.365)

Simultaneous confidence interval for both = (3.314, 4.365)

B) The result of the hypothesis test is significant, hence, the true average PEF is lower for children in biomass households than it is for children in LPG households.

C) 95% confidence interval for the population mean FEY for children in biomass households = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.365)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Step-by-step explanation:

A) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the z-distribution. This is because although, there is no information provided for the population standard deviation, the sample sizes are large enough for the sample properties to approximate the population properties.

Finding the critical value from the z-tables,

z-critical value for 95% confidence level = 1.960 (from the z-tables)

For the children in the biomass households

Sample mean = 3.40

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.20

N = sample size = 756

σₓ = (1.20/√756) = 0.04364

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 3.40 ± (1.960 × 0.04364)

CI = 3.40 ± 0.08554

95% CI = (3.31446, 3.48554)

95% Confidence interval = (3.314, 3.486)

For the children in the LPG households

Sample mean = 4.28

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.19

N = sample size = 752

σₓ = (1.19/√752) = 0.043395

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 4.28 ± (1.960 × 0.043395)

CI = 4.28 ± 0.085054

95% CI = (4.1949, 4.3651)

95% Confidence interval = (4.195, 4.365)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The null hypothesis usually goes against the claim we are trying to test and would be that the true average PEF for children in biomass households is not lower than that of children in LPG households.

The alternative hypothesis confirms the claim we are testing and is that the true average PEF is lower for children in biomass households than it is for children in LPG households.

Mathematically, if the true average PEF for children in biomass households is μ₁, the true average PEF for children in LPG households is μ₂ and the difference is μ = μ₁ - μ₂

The null hypothesis is

H₀: μ ≥ 0 or μ₁ ≥ μ₂

The alternative hypothesis is

Hₐ: μ < 0 or μ₁ < μ₂

Test statistic for 2 sample mean data is given as

Test statistic = (μ₂ - μ₁)/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = 3.40

n₁ = 756

s₁ = 1.20

μ₂ = 4.28

n₂ = 752

s₂ = 1.19

σ = √[(1.20²/756) + (1.19²/752)] = 0.061546

z = (3.40 - 4.28) ÷ 0.061546 = -14.30

checking the tables for the p-value of this z-statistic

Significance level = 0.01

The hypothesis test uses a one-tailed condition because we're testing in only one direction.

p-value (for z = -14.30, at 0.01 significance level, with a one tailed condition) = 0.000000001

The interpretation of p-values is that

When the p-value > significance level, we fail to reject the null hypothesis and when the p-value < significance level, we reject the null hypothesis and accept the alternative hypothesis.

Significance level = 0.01

p-value = 0.000000001

0.000000001 < 0.01

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that true average PEF is lower for children in biomass households than it is for children in LPG households.

C) For FEY for biomass households,

Sample mean = 2.3 L/s

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation = 0.5

N = sample size = 756

σₓ = (0.5/√756) = 0.018185

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 2.30 ± (1.960 × 0.018185)

CI = 2.30 ± 0.03564

95% CI = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.365)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Hope this Helps!!!

3 0
3 years ago
An organic farmer buys a piece of land seed plants tomatoes on 5/9 of the land and green beans on 1/12 of the of the land she pl
tino4ka555 [31]
Find how much is reamaining
remaining=1-used
used=tomatoes+beens
used=5/9+1/12

5/9 times 12/12=60/108
1/12 times 9/9=9/108
used=60/108+9/108
used=69/108
used=23/36

1-23/36=
36/36-23/36=
13/36

answer is 13/36
5 0
3 years ago
Read 2 more answers
Please can someone help me?
Debora [2.8K]
If the perimeters are equal, you can form an equation to find x:
x + x + 12 + x + x + 12 = x + 11 + x + 11 + x + 16
4x + 24 = 3x + 38
- 3x
x + 24 = 38
- 24
x = 14
Now you can substitute the x value into the perimeter:
(4 × 14) + 24
56 + 24 = 80 cm

(3 × 14) + 38
42 + 38 = 80 cm

So the perimeter would be 80 cm.
I hope this helps!
6 0
3 years ago
The following data are the distances between a sample of 20 retail stores and a large distribution center. The distances are in
sesenic [268]

Answer:

Variance = 1,227.27

Standard deviation = 35.03

Step-by-step explanation:

To calculate these, we use the following formulas:

Mean = (sum of the values) / n

Variance = ((Σ(x - mean)^2) / (n - 1)

Standard deviation = Variance^0.5

Where;

n = number of values = 20

x = each value

Therefore, we have:

Sum of the values = 29 + 32 + 36 + 40 + 58 + 67 + 68 + 69 + 76 + 86 + 87 + 95 + 96 + 96 + 99 + 106 + 112 + 127 + 145 + 150 = 1,674

Mean = 1,674 / 20 = 83.70

Variance = ((29-83.70)^2 + (32-83.70)^2 + (36-83.70)^2 + (40-83.70)^2 + (58-83.70)^2 + (67-83.70)^2 + (68-83.70)^2 + (69-83.70)^2 + (76-83.70)^2 + (86-83.70)^2 + (87-83.70)^2 + (95-83.70)^2 + (96-83.70)^2 + (96-83.70)^2 + (99-83.70)^2 + (106-83.70)^2 + (112-83.70)^2 + (127-83.70)^2 + (145-83.70)^2 + (150-83.70)^2) / (20 - 1) = 23,318.20 / 19 = 1,227.27

Standard deviation = 1,227.27^0.5 = 35.03

5 0
3 years ago
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