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suter [353]
3 years ago
10

Pls help geometry please and thank you

Mathematics
2 answers:
viktelen [127]3 years ago
8 0
1. is given. 2. is reflective property. 4. is alternate interior angles. and 5. is definition of a midpoint. (the answer above has 2 and 4 swapped. the statement for 4 is giving you interior angles, hence the reason for four being alternate interior angles. this could be wrong but i don’t believe it is.)
Elena L [17]3 years ago
4 0
#one would be given, #2 is alternate interior, and number 4 would be reflexive property
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204 thousands equals
jeka57 [31]
204 thousand equals 204,000 or two hundred and four thousand. 
Hope this helps.
6 0
3 years ago
HELPPPPPPPPPPPPPPPPPPPPP
Rus_ich [418]

Answer:

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wa=d

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3=

= 2412512.1.4.21.5421.3.12.542.3.215.42.4.2143.51.3

Step-by-step explanation:

8 0
4 years ago
Solve for w, x, and y. Show your work.
sveta [45]

First of all, the left triangle is an equilateral triangle. Which means that all the interior angles are 60 degrees.

4w=60

w=15

Then all 3 sides are equal thus we can set up equation 3x-y=4y then simplify,

3x=5y

Going to the triangle on the right its isosceles we can set up equation 2x+50=180

2x=130

x=75

Substitute this back into equation above 3x=5y

5y=3*75

y=45

Therefore

x=75

y=45

w=15

Done!

5 0
3 years ago
What is another way to write this number? 300+70+ 5/10+8/100
Mademuasel [1]
Hi,
It would be 370.58
4 0
3 years ago
Read 2 more answers
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
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