All categories

3 years ago

Consider the two triangles.

Mathematics

1 answer:

GuDViN [60]3 years ago

4 0

Answer:

(A)Show that the ratios StartFraction U V Over X Y EndFraction , StartFraction W U Over Z X EndFraction , and StartFraction W V Over Z Y EndFraction are equivalent.

$\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}$

Step-by-step explanation:

In Triangles WUV and XZY:

$\angle VUW$ and \angle YXZ$ are congruent. \\\angle U W V$ and \angle X Z Y$ are congruent.\\ \angle U V W$ and \angle Z Y X$ are congruent.$

Therefore:

$\triangle UWV \cong \triangle XZY$

To show that the triangles are similar by the SSS similarity theorem, we have:

$\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}$

As a check:

$\dfrac{UW}{XZ}=\dfrac{40}{32}=1.25\\\\\dfrac{WV}{ZY}=\dfrac{60}{48}=1.25\\\\\dfrac{UV}{XY}=\dfrac{50}{40}=1.25$

The correct option is A.

You might be interested in

A european business has offered to donate to the Salvation Army, but they are requiring a detailed report on expenditures. Chase

dedylja [7]

Answer:

Both processes are correct.

Step-by-step explanation:

Let the mile taken by Chase be 10miles .

He doubles it so it will be 20 and

20% of 20 will be 4

20-4 = 16km

Now Alex operations

(10/5 )*8 = 2*8= 16 km

In both cases the answer will be the same

1 mile = 1.60034 km

Now 10 miles = 16.00 34 km

which is the same as above.

Both the processes are correct.

7 0

3 years ago

Finding Distance in the Coordinate Plane

GuDViN [60]

The distance between the two points (-1, 4) and (5, 4) is 6 units.

Step-by-step explanation:

The given Coordinate points are (-1, 4) and (5, 4).

The points are considered as (x1,y1) and (x2,y2).

Here,

x1 = -1 and x2 = 5

y1 = 4 and y2 = 4

The distance formula is given by :

Distance = $\sqrt{(x2-x1)^2+(y2-y1)^2}$

To find the distance between these two points :

The distance formula is used,

Distance = $\sqrt{(5+1)^2+(4-4)^2}$

⇒ √(6)²

⇒ √36

⇒ 6

Therefore, the between the points is 6 units.

5 0

3 years ago

The amount of money in a bank account with an initial balance of $15,000 earns 3% per year. In the exponential model of the form

oksian1 [2.3K]

1.03=b this makes sense because it increases 3 percent so you add 1 to .03

8 0

3 years ago

Read 2 more answers

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA ba

PSYCHO15rus [73]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA basketball tournament (not all private schools report their coach's salaries). Consider these 62 salaries to be a sample from the population of salaries of all 346 NCAA Division I basketball coaches.

Question 1. Use the 62 salaries from the TOTAL PAY column to construct a 95% confidence interval for the mean salary of all basketball coaches in NCAA Division I.

xbar = $1,465,752

SD = $1,346,046.2

lower bound of confidence interval ________

upper bound of confidence interval _______

Question 2. Coach Mike Krzyzewski's high salary is an outlier and could be significantly affecting the confidence interval results. Remove Coach Krzyzewski's salary from the data and recalculate the 95% confidence interval using the remaining 61 salaries.

xbar = $1,371,191

SD = $1,130,666.5

lower bound of confidence interval _________

upper bound of confidence interval. ________

Answer:

Question 1:

lower bound of confidence interval = $1,124,027

upper bound of confidence interval = $1,807,477

Question 2:

lower bound of confidence interval = $1,081,512

upper bound of confidence interval = $1,660,870

Step-by-step explanation:

Question 1:

The sample mean salary of 62 couches is

$\bar{x} = 1,465,752$

The standard deviation of mean salary is

$s = 1,346,046.2$

The confidence interval for the mean salary of all basketball coaches is given by

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } )$

Where $\bar{x}$ is the sample mean, n is the sample size, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 62 - 1 = 61

From the t-table at α = 0.025 and DoF = 61

t-score = 1.999

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (\frac{1,346,046.2}{\sqrt{62} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (170948.04 ) \\\\CI = 1,465,752 \pm 341,725 \\\\LCI = 1,465,752 - 341,725 = 1,124,027 \\\\UCI = 1,465,752 + 341,725 = 1,807,477\\\\$

Question 2:

After removing the Coach Krzyzewski's salary from the data

The sample mean salary of 61 couches is

$\bar{x} = 1,371,191$

The standard deviation of the mean salary is

$s = 1,130,666.5$

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 61 - 1 = 60

From the t-table at α = 0.025 and DoF = 60

t-score = 2.001

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (\frac{1,130,666.5}{\sqrt{61} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (144767 ) \\\\CI = 1,371,191 \pm 289,678.8 \\\\LCI = 1,371,191 - 289,678.8 = 1,081,512 \\\\UCI = 1,371,191 + 289,678.8 = 1,660,870\\\\$

6 0

3 years ago

If a house is 15m long and 12m wide how long is the diagonal of the house

Nitella [24]

If this talks about the diagonal of the flooring then it is 9m

3 0

3 years ago