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In-s [12.5K]
2 years ago
14

Explain how you can use the inscribed angle theorem to justify its second corollary, that an angle inscribed in a semicircle is

a right angle
Mathematics
1 answer:
mihalych1998 [28]2 years ago
7 0

Answer:

We find half the measure of the central angle of the circle. Where the given shape is a semi-circle, then the inscribed angle is half of 180, therefore, 90 degrees and a right angle is proven. We show this as Angle BAC is a RA 360 is the measure of circle 360/2 = 180 the semi circle Angle BAC is 180/2 = 90 degree.

Step-by-step explanation:

Hope it helps you!

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34/100
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Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
Harlamova29_29 [7]

Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

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df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

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Thtas what you must mean. 
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