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3 years ago

If BD bisects ABC, ABD = 2x + 3, and DBC = 3x - 13, find ABD.

Mathematics

1 answer:

8090 [49]3 years ago

5 0

Answer:

Step-by-step explanation:

Since its an angle bisector, both angles would be equal to each other.

3x-13=2x+3

Now you need to get x alone.

Subtract 2x from each side.

x-13=3

Still need x alone so add 13 to each side (since it negative addition)

x=16

You have x, but you need ABD

Plug 16 in for x

2(16)+3

32+3

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Following are the published weights (in pounds) of all of the team members of Football Team A from a previous year.178; 203; 212

mylen [45]

Answer:

a. 241; b. 206.0; c. 272.0; d. 174 to 232; other answers (see below).

Step-by-step explanation:

First, at all, we need to organize the data from the smallest to the largest value:

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

This step is important to find the median, the first quartile, and the third quartile. There are numerous methods to find the median and the other quartiles, but in this case, we are going to use a method described by Tukey, and it does not need calculators or software to estimate it.

<h3>Part a: Median</h3>

In this case, we have 53 values (an odd number of values), the median is the value that has the same number of values below and above it, so what is the value that has 26 values below and above it? Well, in the organized data above this value is the 27th value, because it has 26 values above and below it:

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

The median is 241.

<h3>Part b: First Quartile</h3>

For the first quartile, we need to calculate the median for the lower half of the values from the median previously obtained. Since we have an odd number of values (53), we have to include the median in this calculation. We have 27 values (including the median), so the "median" for these values is the value with 13 values below and above it.

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 <em>241</em>

Since we are asking to round the answer to one decimal place, the first quartile is 206.0

<h3>Part c: Third Quartile</h3>

We use the same procedure used to find the first quartile, but in this case, using the upper half of the values.

<em>241</em> 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

So, the third quartile is 272.0

<h3>Part d: The middle 50% </h3>

The second quartile is the median and "50% of the data lies below this point" (Quartile (2020), in Wikipedia). Having this information into account, 50% of the weights lies below the median 241.

Thus, the middle 50% of the weights are from 174 to 232.

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 <em>241</em>

<h3>Part e: Sample or Population 1</h3>

If the population were all professional football players, the right option is:

<em>The above data would be a sample of weights because they represent a subset of the population of all football players. </em>

It represents a sample. Supposing Football Team A are all professional, they can be considered a sample from all professional football players.

<h3>Part f: Sample or Population 2</h3>

If the population were Football Team A, the right option is:

<em>The data would be a population of weights because they represent all of the players on Football Team A. </em>

<h3>Part g: population mean and more</h3><h3>Part i</h3>

The mean for the population of weights of Football Team A is the sum of all weights (12529 pounds) divided by the number of cases (53).

$\\ \mu = \frac{12529}{53} = 236.39622641509433$

$\\ \mu = \frac{12529}{53} = 236.40$

The standard deviation for the population is:

$\\ \sigma = \sqrt(\frac{(x_{1}-\mu)^2 + (x_{2}-\mu)^2 + ... + (x_{n}-\mu)^2}{n})$

In words, we need to take either value, subtract it from the population mean, square the resulting value, sum all the values for the 53 cases (in this case, the value is 74332.67924), divide the value by 53 (1402.50338) and take the square root of it (37.45001).

Then, the population standard deviation is 37.45.

The weight that is 3 standard deviations below the mean can be obtained using the following formula:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ \mu= 236.40\;and\;\sigma=37.45$

Then, in the case of three standard deviations below the mean, z = -3.

$\\ -3 = \frac{x - 236.40}{37.45}$

$\\ x = -3*37.45 + 236.40$

$\\ x = 124.05$

For the player that weights 209 pounds:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{229 - 236.40}{37.45}$

$\\ z = -0.19$

<h3>Part h: Comparing Weights of Team A and Team B</h3>

For Team B, we have a <em>mean</em> and a <em>standard deviation</em> of:

$\\ \mu=240.08\;and\;\sigma=44.38$

And Player B weighed 229 pounds.

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{229 - 240.08}{44.38}$

$\\ z = -0.70$

This value says that Player B is lighter with respect to his team than Player A, because his weight is 0.70 below the mean of Football Team B, whereas Player A has a weight that is closer to the mean of his team. So, the answer is:

<em>Player B, because he is more standard deviations away from his team's mean weight.</em>

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4 0

3 years ago

There are 1493 men and 1438 women in the audience at a concert. What is the percentage of the audience that are men

Advocard [28]

To calculate this answer you have to sum up the amount of women and amount of men at the concert: 1493 + 1438 = 2931.

Now you have the total amount of people. To get the precentage of the audience that are men, you do this: 1493/2931. (you divide the amount of men with the amount of every person present).

1493/2931 = 0.5093

you multiply 0.5093 by 100

and you get 50.93%. And that is the final answer.

8 0

3 years ago

Read 2 more answers

What is the first step when constructing an angle bisector using only a compass and a straightedge

zloy xaker [14]

Step 1 is, "draw arcs through both legs of the angle, centered at the vertex of the angle."

_____

The arcs crossing the two legs have the same radius. One arc crossing both legs can be used, if you like.

3 0

3 years ago

Read 2 more answers

A subset $S \subseteq \mathbb{R}$ is called open if for every $x \in S$, there exists a real number $\epsilon > 0$ such that

const2013 [10]

Answer:

Step-by-step explanation:

REcall that given sets S,T if we want to prove that $S\subseteqT$ , then we need to prove that for all x that is in S, it is in T.

a) Let (a,b) be a non empty interval and $x\in (a,b)$ . Then a<x <b. Let $\varepsilon = \min{\min\{b-x, x-a\}}{2}$ Consider $y \in (x-\varepsilon,x+\varepsilon)$ , then

$y$ and

$y>x-\varepsilon>x-(x-a) = a$ .

Then $y\in (a,b)$ . Hence, (a,b) is open.

Consider the complement of [a,b] (i.e $(a,b)^c$ ).

Then, it is beyond the scope of this answer that

$(a,b)^c = (-\infty,a) \cup (b,\infty)$ .

Suppose that $x\in (a,b)^c$ and without loss of generality, suppose that x < a (The same technique applies when x>b). Take $\varepsilon = \frac{a-x}{2}$ and consider $y \in (x-\varepsilon,x+\varepsilon)$ . Then

$y$

Then y \in (-\infty,a). Applying the same argument when $x \in (b,\infty)$ we find that [a,b] is closed.

c) Let I be an arbitrary set of indexes and consider the family of open sets $\{A_i\}_{i\in I}. Let [tex]B = \bigcup_{i\in I}A_i$ . Let $x \in B$ . Then, by detinition there exists an index $i_0$ such that $x\in A_{i_0}$ . Since $A_{i_0}$ is open, there exists a positive epsilon such that $(x-\varepsilon,x+\varepsilon)\subseteq A_{i_0} \subseteq B$ . Hence, B is open.

d). Consider the following family of open intervals $A_n = (a-\frac{1}{n},b+\frac{1}{n})$ . Let $B = \bigcap_{n=1}^{\infty}A_n$ . It can be easily proven that

$B =[a,b]$ . Then, the intersection of open intervals doesn't need to be an open interval.

b) Note that for every $x \in \mathbb{R}$ and for every $\varepsilon>0$ we have that $(x-\varepsilon,x+\varepsilon)\subseteq \mathbb{R}$ . This means that $\mathbb{R}$ is open, and by definition, $\emptyset$ is closed.

Note that the definition of an open set is the following:

if for every $x \in S$ , there exists a real number $\epsilon > 0$ such that $(x-\epsilon,x \epsilon) \subseteq S$ . This means that if a set is not open, there exists an element x in the set S such that for a especific value of epsilon, the subset (x-epsilon, x + epsilon) is not a proper subset of S. Suppose that S is the empty set, and suppose that S is not open. This would imply, by the definition, that there exists an element in S that contradicts the definition of an open set. But, since S is the empty set, it is a contradiction that it has an element. Hence, it must be true that S (i.e the empty set) is open. Hence $\mathbb{R}$ is also closed, by definition. If you want to prove that this are the only sets that satisfy this property, you must prove that $\mathbb{R}$ is a connected set (this is a topic in topology)

6 0

3 years ago

Rachel's aunt gave her $85 to spend on

Orlov [11]

The amount Rachel has left of the amount given to her by her aunt after she makes the purchases is $22.15.

<h3>How much does Rachel have left?</h3>

In order to determine the amount she would have left, determine the total value of the purchases and subtract it from the amount she was given.

Total value of the purchases = (4 x 0.85) + 45.50 + (5 x 2.79) = $62.85

Change = $85 - $62.85 = $22.15

To learn more about addition, please check: brainly.com/question/19628082

7 0

2 years ago