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2 years ago

Select all the conditions for which it is possible to construct a triangle.

Mathematics

1 answer:

Margaret [11]2 years ago

8 0

The conditions for which it is possible to construct a triangle include:

A triangle with angles measures 60 ∘ , 80 ∘ , and 80 ∘

A triangle with side lengths of 4 cm, 5 cm, and 6 cm

A triangle with side lengths 4 cm and 5 cm and a 50 ∘ angle.

<h3>What is a Triangle?</h3>

This is defined as a three-sided polygon which has three edges and three vertices and a total internal angles of 180 degrees.

The triangle with side lengths 4 cm, 5 cm, and 15 cm can't be constructed due to the length of sides not being equal according to pythagoras theorem.

Read more about Triangle here brainly.com/question/17335144

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A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states.

umka21 [38]

Answer:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

$\bar X_1 = 28900$ the sample mean for the salaries of teachers in Indiana

$s_1 = 2300$ the sample deviation for the salary of teachers in Indiana

$n_1 =10$ the sample size from Indiana

$\bar X_2 = 30300$ the sample mean for the salaries of teachers in Michigan

$s_2 = 2100$ the sample deviation for the salary of teachers in Michigan

$n_2 =14$ the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom for this case are:

$df = n_1 +n_2 -2= 10+14-2 =22$

The confidence is 95%so then the significance is $\alpha=0.05$ and the $\alpha/2 =0.025$ , we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got:

$t_{\alpha/2}= 2.07$

And now replacing in the formula for the confidence interval we got:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

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3 years ago

11/12 times4/5 estimate

sveticcg [70]

Answer:

44/60

Step-by-step explanation:

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3 years ago

Which of the following is true about multi-collinearity? Group of answer choices The P-value reduces significantly, leading to r

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Answer:

It is measured using a measure called variance inflation factor (VIF).

Step-by-step explanation:

This is the statement that is true about multi-collinearity. Multi-collinearity is a situation that can sometimes arise when dealing with a multiple regression model. Multicollinearity refers to the occurrence of high intercorrelations among various independent variables. This can often lead to results that are skewed or misleading. The main way in which multi-collinearity can be measured is by using the variance inflation factor (IVF). This allows us to quantify the severity of multi-collinearity.

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Answer:

The 1st and 4th statements are true.

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