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3 years ago

A group of rowdy teenagers near a wind turbine decide to place a pair of

Mathematics

1 answer:

tamaranim1 [39]3 years ago

3 0

Answer:

a) Sinusoidal functions are y = a sin [b(x-h)] + k (or)

y = a cos [b(x-h)] + k

Where a is amplitude a= (max-min)/2=(16-2)/2=7

period p= 2π/b

b=2π/30

Horizontal transformation to 10 units right h=10

k= (max+min)/2=(16+2)/2=9

h = 7 cos [π/15(t-10)]+ 9

b) t=10min=600 sec

substitue in the above equation

h=5.5m

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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

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3 years ago

Help ...............

sammy [17]

A straight line is 180° so....

9x+24 +6x - 24 = 180

15x=180. ( because you have to combine like terms.

X= 12

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3 years ago

No. Zachary should have drawn the first two arcs with the compass centered at B.

qaws [65]

Answer:

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3 years ago

I WILL GIVE BRAINLIEST TT

Vinil7 [7]

Answer:

Step-by-step explanation:

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Whats the difference between coplanar and collinear

Sergio [31]

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