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alukav5142 [94]
3 years ago
11

Use the given information and the figures below to find m Given: BCDE = WPHY A: 225 B: 135 C: 180 D: 200

E: 168

Mathematics
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

the answer is B

Step-by-step explanation:

the total measure of all the angles of any quadrilateral is 360, so all you have to do is subtract all the known angles from 360 to find the unknown angle

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For their twentieth wedding anniversary, Richard and Sylvia had a party for several of their best friends. Things were going alo
kompoz [17]

Answer:

Richard is older

Step-by-step explanation:

We can set up an inequality for both statements.

Let r equal Richard's age and s equal Sylvia's age.

"I am older than my wife."

Since Richard is speaking, the inequality would look like this:

r > s

This means Richard is older than Sylvia.

"I am younger than my husband."

Since Sylvia is speaking, the inequality would look like this:

s < r

This means that Sylvia is younger than Richard.

We can flip one inequality to "see" them from the same perspective.

Let's use s < r

To make it so that we can see the relationship from Richard's perspective, flip the entire inequality.

s < r

to

r > s

The inequality from the first quote is identical to this one!

Therefore, Richard is older than Sylvia.

8 0
3 years ago
What property is occurring in the following example? 11.5 + (- 11.5) = 0
kramer

Answer: A property occurring in the example 11.5 + (-11.5) = 0, is additive inverse.

Step-by-step explanation:

A property where sum of any number and its inverse is equal to zero is called additive inverse property.

For example, 11.5 + (-11.5) = 0

Here, 11.5 is the number and its inverse is (-11.5). The sum of both these is equal to zero. Hence, it shows a property of additive inverse.

Thus, we can conclude that property occurring in the example 11.5 + (-11.5) = 0, is additive inverse.

5 0
3 years ago
Please hellppp i will mark brainliest ​
castortr0y [4]
30 degrees it's right
8 0
3 years ago
For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics;
umka2103 [35]

Answer:

a) The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Step-by-step explanation:

For any distribution,

The median is the variable at the middle when all the variables in the dsitribution are arranged in ascending or descending order.

The median is the (n+1)/2 th variabke.

where n = size of the distribution or number of variables. For these questions, the sample size is 5, hence, the median will always be the 3rd variable when the variables are arranged in ascending or descending order.

The IQR, known as the inter quartile range is a measure of dispersion for the distribution. Although, it isn't as effective as other measures of dispersion such as the standard deviation because the IQR unlike the the standard deviation isn't responsive to changes in the variables, especially ones at the end of the distribution.

The IQR is simply given mathematically as the third quartile minus the first quartile.

IQR = (Third quartile) - (First quartile)

Third quartile is the 3(n+1)/4 th variable. For a sample of n=5, the third quartile is the 4.5th variable, that is the average of the 4th and 5th variable.

First quartile is the (n+1)/4 th variable. For a sample of n=5, the first quartile is the 1.5th variable, that is the average of the 1st and 2nd variable.

Taking the questions one at a time

a) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,6,7,20

Median = 3rd variable = 6

Third quartile = (7+20)/2 = 13.5

First quartile = (3+5)/2 = 4

IQR = 13.5 - 4 = 9.5

The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) (1) 1,2,3,4,5

Median = 3rd variable = 3

Third quartile = (4+5)/2 = 4.5

First quartile = (1+2)/2 = 1.5

IQR = 4.5 - 1.5 = 3

(2) 6,7,8,9,10

Median = 3rd variable = 8

Third quartile = (9+10)/2 = 9.5

First quartile = (6+7)/2 = 6.5

IQR = 9.5 - 6.5 = 3

The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,7,8,9

Median = 3rd variable = 7

Third quartile = (8+9)/2 = 8.5

First quartile = (3+5)/2 = 4

IQR = 8.5 - 4 = 4.5

The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) (1) 0,10,50,60,100

Median = 3rd variable = 50

Third quartile = (60+100)/2 = 80

First quartile = (0+10)/2 = 5

IQR = 80 - 5 = 75

(2) 0,100,500,600,1000

Median = 3rd variable = 500

Third quartile = (600+1000)/2 =800

First quartile = (0+100)/2 = 50

IQR = 800 - 50 = 750

The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Hope this Helps!!!

5 0
3 years ago
June will spin the arrow on the spinner 5 times. What is the probability that the arrow will stop on A, then B, then C, then D,
inessss [21]
Ok so assuming the board only has 4 spaces to land on (A,B,C,D) all we need to do is weight the probability, 

1/4 x 1/4 x 1/4 x 1/4 x 1/4 = 1/1024 

To solve I put the number of favorable outcomes over the number of total outcomes, in this case we had 1 favorable outcome each time and a constant of 4 possible outcomes. 
4 0
3 years ago
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