Make a conjecture about the diagram below. Do you think you can conclude that △JKL ≅ △XYZ? Explain your reasoning.

Mathematics

2 answers:

BlackZzzverrR [31]3 years ago

7 0

Answer:

Sample Response: Since these are right triangles, the unmarked sides can be determined to be congruent because of the Pythagorean theorem. Once the third set of congruent sides is determined, the SSS or SAS theorem can be used to prove triangle congruency.

Step-by-step explanation:

sample response

konstantin123 [22]3 years ago

6 0

Answer:

Yes, you can conclude that ΔJKL ≅ XYZ.

Step-by-step explanation:

I forget how to describe it in geometric terms but, if you read the similarity statement again, and follow each angle, you'll notice that the shapes match, and they correspond to each other, it would be different if it said something like ΔLKJ ≅ XYZ because it does not match.

If you're looking for a specific reason, we'd say that they're congruent because of SAS (Side-Angle-Side)

If this still confuses you, hit me up.

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Answer:

The 95% confidence interval for the standard deviation of the duration times of game play is (279.76, 670.50).

Step-by-step explanation:

The data for the twelve different video games showing substance use were observed and the duration of times of game play (in seconds) are listed below:

S = {4049, 3884, 3859, 4027, 4318, 4813, 4657, 4033, 5004, 4823, 4334, 4317}

It is assumed that the sample was obtained from a population with a normal distribution.

The confidence interval for population standard deviation is given by,

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

Compute the sample variance s² as follows:

$\text{s}^{2} = \dfrac{1}{n - 1}\sum\limits_{i=1}^{n}(x_i - \overline{x})^{2}\\$

$=\frac{1}{12-1}\times 1715527.67\\\\$

$=155957.061\\$

Compute the critical values of Ci-square for 95% confidence level and (n - 1) degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.025, (12-1)}=\chi^{2}_{0.025, 11}=21.92\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.025, (12-1)}=\chi^{2}_{0.975, 11}=3.816$