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3 years ago

In the triangle below, angle B measures 60° and BC is 18. What is the length of AC

Mathematics

2 answers:

TEA [102]3 years ago

6 0

Ac is 42 i think is i didnt help sorry ._. ...................................

hichkok12 [17]3 years ago

3 0

Answer:

$AC=9\sqrt{3}\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

In the right triangle ABC

$sin(60\°)=\frac{AC}{BC}$

$sin(60\°)=\frac{\sqrt{3}}{2}$

$\frac{AC}{BC}=\frac{\sqrt{3}}{2}$

we have

$BC=18\ units$

substitute

$\frac{AC}{18}=\frac{\sqrt{3}}{2}$

$AC=18\frac{\sqrt{3}}{2}=9\sqrt{3}\ units$

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The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

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Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

Step-by-step explanation:

Let the random variable X denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

X = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

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The formula to compute the sample variance for the daily price change is:

$s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}$

The sample mean is computed using the formula:

$\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

$\bar X$ =AVERAGE(A2:A13)

$s^{2}$ =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

$s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}$

Consider the Excel output attached below.

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$s$ =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

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The (1 - α)% confidence interval for population variance is:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

Compute the critical value of Chi-square for α = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816$

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

$=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]$

$=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]$

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