Giving brainly for correct answer dont spam if you dont get right away

max2010maxim [7]

Step-by-step explanation:

Adjacent angles are two angles that have a common vertex and common side but do not overlap. They share the same vertex and same common side.

Vertical angles are opposite angles made by two intersecting lines.

Hope this helps!

7 0

3 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

<u>Hours of TV per week (X)</u>: 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

<em>Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the true average, </u> $\mu$ <u> is ;</u>

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

<u>80% confidence interval for</u> $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0

3 years ago

Write a system of inequalities that defines a shaded region that looks like a<br> right triangle.

eduard

$(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y$

<h2>Explanation:</h2>

A right triangle is a special triangle that has a right angle. In this case, we have to write a system of inequalities that defines a shaded region that looks like a right triangle. First of all, let's say that at the origin the triangle will have a right angle. To do so, we'd need to set:

$x>0 \\ \\ y>0$

So the shaded region in this first part will be the First Quadrant as indicated in the first figure below. So if the opposite side lies on the x-axis the adjacent side will lie on the y-axis or if the adjacent side lies on the x-axis the opposite side will lie on the y-axis. Everything is ok up to this point. We just need to define the hypotenuse, so we'd need to define a line. In order to have a right triangle, we need a line with negative slope and positive y-intercept shaded under the line. So, this inequality could be:

$y$