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enot [183]
3 years ago
12

What is CPCTC in analyzing triangle relationships?

Mathematics
1 answer:
monitta3 years ago
7 0

Answer:

Exercises Use the Plan for Proof to write a two-column proof.

Step-by-step explanation:

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The scatter plot below shows the relationship between latitude of cities and their average January temperature. Which is the bes
natka813 [3]

Answer:

-2.5

Step-by-step explanation:

it is what it is

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3 years ago
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Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi
aalyn [17]

Answer:

a) t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222  

df=12+15-2=25  

p_v =P(t_{25}>6.222) =8.26x10^{-7}

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

b) (91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309

(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691

Step-by-step explanation:

Notation and hypothesis

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}  

This last one is an unbiased estimator of the common variance \sigma^2  

Part a

The system of hypothesis on this case are:  

Null hypothesis: \mu_2 \leq \mu_1  

Alternative hypothesis: \mu_2 > \mu_1  

Or equivalently:  

Null hypothesis: \mu_2 - \mu_1 \leq 0  

Alternative hypothesis: \mu_2 -\mu_1 > 0  

Our notation on this case :  

n_1 =12 represent the sample size for group 1  

n_2 =15 represent the sample size for group 2  

\bar X_1 =85 represent the sample mean for the group 1  

\bar X_2 =91 represent the sample mean for the group 2  

s_1=3 represent the sample standard deviation for group 1  

s_2=2 represent the sample standard deviation for group 2  

First we can begin finding the pooled variance:  

\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2  

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

S_p=2.490  

Calculate the statistic

And now we can calculate the statistic:  

t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222  

Now we can calculate the degrees of freedom given by:  

df=12+15-2=25  

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:  

p_v =P(t_{25}>6.222) =8.26x10^{-7}

Conclusion

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

For the 99% of confidence we have \alpha=1-0.99 = 0.01 and \alpha/2 =0.005 and the critical value with 25 degrees of freedom on the t distribution is t_{\alpha/2}= 2.79

And replacing we got:

(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309

(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691

7 0
3 years ago
PLEASE HELP DUE IN 10 MINS!!!
Jet001 [13]

Answer:

yes, all answers are correct

Step-by-step explanation:

8 0
2 years ago
40. Find UW.<br> T<br> 6<br> 5<br> U +-<br> The measure of UW is<br> type your answer...
Ahat [919]

Answer:

9

Step-by-step explanation:

UT^2 = UV * UW

6^2 = (x - 7)(x - 7 + 5)

36 = (x - 7)(x - 2)

36 = x^2 - 2x - 7x + 14

x^2 - 9x - 22 = 0

(x - 11)(x + 2) = 0

x - 11 = 0  or  x + 2 = 0

x = 11

UW = x - 7 + 5

UW = 11 - 7 + 5

UW = 9

7 0
3 years ago
PLEASE HELP ME!!!!!!!!!
vitfil [10]
<h2>1)</h2>

(x - 4) {}^{2}  - 28 = 8 \\ (x - 4) {}^{2}  = 8 + 28 \\ (x - 4) {}^{2}  = 36

This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.

\sqrt{(x - 4) {}^{2} }  =  ± \sqrt{36}  \\x - 4 = ±6 \\ x _{1}- 4 = 6 \:  \:  \:  \:  \:  \:  \: \:  \:  \:  x _{2}- 4 =  - 6 \\ x_{1} = 6 + 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2} =  - 6 + 4 \\ x_{1} = 10 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: \:  \: x_{2} =  - 2

<h2>2)</h2>

Well he started off correct to the point of completing the square.

(x - 3)  {}^{2}  = 16 \\ x - 3 = ±4 \\  \: x_{1}  - 3 = 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2}  - 3 =  - 4 \\ x_{1}  = 7 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2}  =  - 1

8 0
1 year ago
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