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kati45 [8]
3 years ago
7

Which is a counterexample of the following conjecture: "If two angles are supplementary, they have the same measure?"

Mathematics
1 answer:
Marrrta [24]3 years ago
4 0

Answer:i dont know but i think its D.

Step-by-step explanation:

Pls mark meh Brainlessed                  

My friend justen here hes tacken and hes cracked at fornight my guy AHHHH A BREAD DOG

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Simplify the picture below
podryga [215]
((x^2-3x-18)/(x+3))
((x-6)(x+3))/(x+3)
The x+3 on top cancels out the x+3 on the bottom so all you have left is

x-6


Hope I didn't mess up for your sake!
3 0
3 years ago
In a bag there are 4 purple marbles, 3 green marbles, and 10 yellow marbles. If a student draws one marble what is the sample sp
guajiro [1.7K]
The correct answer Is D. Purple , Green , Yellow 

Hope I helped! ( Smiles )
6 0
3 years ago
Can someone please help !!!!!
Katen [24]
I'm I am so sorry I haven't done this hey sorry
8 0
3 years ago
Someone plssss helppppp i need it finished in like 20 min plsss help
denis23 [38]

Answer:

number of guest tickets, g, is 100 tickets

number of student tickets, s, is 125 tickets

Step-by-step explanation:

Given;

total number of tickets sold, t = 225 tickets

cost of each guest ticket,  = $5.5

cost of each student ticket, = $3.5

total cost of the tickets, Tc = $987.5

number of guest tickets = g

number of student tickets, = s

g(5.5) + s(3.5) = 987.5 ------- (1)

g + s = 225 ----------- (2)

From equation (2); s = 225 - g

Substitute the value of g in equation (1)

5.5g + 3.5(225 - g) = 987.5

5.5g + 787.5 - 3.5g = 987.5

5.5g - 3.5g = 987.5 - 787.5

2g = 200

g = 200 / 2

g = 100 tickets

Then, s = 225 - g

s = 225 - 100

s = 125 tickets

3 0
3 years ago
The probability distribution of the number of students absent on Mondays, is as follows: X 0 1 2 3 4 5 6 7 f(x) 0.02 0.03 0.26 0
alexgriva [62]

a) Add up all the probabilities f(x) where x>3:

f(4)+f(5)+f(6)+f(7)=0.35

b) The expected value is

E[X]=\displaystyle\sum_xx\,f(x)=3.16

Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.

c) The variance is

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58

So the variance is

V[X]=11.58-3.16^2\approx1.59

The standard deviation is the square root of the variance:

\sqrt{V[X]}\approx1.26

d) Since Y=7X+3 is a linear combination of X, computing the expectation and variance of Y is easy:

E[Y]=E[7X+3]=7E[X]+3=25.12

V[Y]=V[7X+3]=7^2V[X]\approx78.13

e) The covariance of X and Y is

\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]

We have

XY=X(7X+3)=7X^2+3X

so

E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54

Then the covariance is

\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16

f) Dividing the covariance by the variance of X gives

\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638

5 0
3 years ago
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