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Firlakuza [10]
2 years ago
6

What is the measure or ENV

Mathematics
2 answers:
Vlad [161]2 years ago
8 0
First we need to get x
<RNA = <ENV [v.o.a]
8x+12=5x+57
8x-5x=57-12
3x=45
x=15

<ENV= (5×15+57)
<ENV =132
the answer is D
kirza4 [7]2 years ago
5 0

Answer: The answer is (D) 132^\circ.

Step-by-step explanation: We are given a figure where the line segments AE and RV are intersecting at the point N. We are to find the measure of \angle ENV.

Also, given that

\angle ANR=(8x+12)^\circ,\\\\\angle ENV=(5x+57)^\circ.

Since the line segments AE and RV intersect at N, so we have

\angle ANR=\angleENV~[\textup{vertically opposite angles}]\\\\5x+57=8x+12\\\\\Rightarrow8x-5x=57-12\\\\\Rightarrow3x=45\\\\\Rightarrow x=15.

Therefore,

\angle ENV=(5\times 15+57)^\circ=(75+57)^\circ=132^\circ.

Thus, (D) is the correct option.

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in circle o, points a and b lie on the circle such that OA=5x-11 and OB=4(x-1) what represents the length of the circle' s diame
AysviL [449]
We know that
if <span>points a and b lie on the circle  and O is the center
so
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3 0
2 years ago
Ashish is 175 cm tall his sister Annu is 8% shorter than him what is Annu's height<br><br>pls help​
mafiozo [28]

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2 years ago
One hundred teachers attended a seminar on mathematical problem solving. The attitudes of representative sample of 12 of the tea
andrew11 [14]

Answer:

a) 1.92

b) 3.25

c) 1.5

d) -5.23

Step-by-step explanation:

We are given the following in the question:

4, 7, -1, 1, 0, 5, -2, 2, -1, 6, 5, -3

a) mean of score change

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{23}{12} = 1.92

b) standard deviation for this population

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.

Sum of squares of differences = 126.92

\sigma = \sqrt{\frac{126.92}{12}} = 3.25

c) median change score

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Sorted data: -3, -2, -1, -1, 0, 1, 2, 4, 5, 5, 6, 7

Median =

\dfrac{6^{th} + 7^{th}}{2} = \dfrac{1+2}{2} = 1.5

d) change score that is 2.2 standard deviations below the mean.

x = \mu - 2.2(\sigma)\\x = 1.92-2.2(3.25)\\x = -5.23

4 0
2 years ago
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