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tigry1 [53]
3 years ago
6

The midpoint of a segment is (−6,−5) and one endpoint is (1,3). Find the coordinates of the other endpoint.

Mathematics
2 answers:
Evgen [1.6K]3 years ago
6 0
C. (-13,-13)
You take the x of the midpoint and equal it to the formula of the midpoint so it will be -6 = 1+x/2 so 1st you times 2 by 6 because you wanna get rid of 2 so it will -12 Then when you move 1 so x is alone it will be -1 so -12-1= -13
Same thing for y
larisa86 [58]3 years ago
5 0

Answer: C. (-13, -13)

Step-by-step explanation:

The midpoint (x,y) of a line segment having two end points (a,b) and (c,d) is given by :-

x=\dfrac{a+c}{2}\ ;\ y=\dfrac{b+d}{2}

Given : The midpoint of a segment is (-6,-5) and one endpoint is (1,3).

Let the coordinates of other end point be (a,b) then , we have

-6=\dfrac{a+1}{2}\ ;\ -5=\dfrac{b+3}{2}\\\\\Rightarrow\ a+1=2\times-6\ ;\ b+3=2\times-5\\\\\Rightharrow\  a+1=-12\ ;\ b+3=-10\\\\\Rightarrow\ a=-12-1\ ;\ b=-10-3\\\\\Rightarrow\ a=-13,\ ;\ b=-13

Hence, the coordinates of the other endpoint = (-13,-13)

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A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio
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Answer:

The probability is P(|p-\^{p}| >  0.03)  =   0.0040

Step-by-step explanation:

From the question we are told that

   The population proportion is p =  0.07

   The mean of the sampling distribution is   \mu_p =  0.07

   The sample size is n = 600

Generally the standard deviation is mathematically represented as

     \sigma_p  =  \sqrt{\frac{p (1 -p)}{n} }

=>    \sigma_p  =  \sqrt{\frac{0.07(1 -0.07)}{600} }    

=>    \sigma_p  =  0.010416    

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

      P(|p-\^{p}| >  0.03) =  1 - P(|p -\^{p}| \le 0.03)

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(-0.03 \le p -\^{p} \le 0.03 )

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=>   P(|p-\^{p}| >  0.03)  =  1 -  P( 0.07-0.03  \le \^{p} \le 0.03+ 0.07 )

=>   P(|p-\^{p}| >  0.03)  =  1 -  P(0.04 \le \^{p} \le 0.10 )

Now  converting the probabilities to their respective standardized score

=> P(|p-\^{p}| >  0.03)  =  1 -  P(\frac{0.04 - 0.07}{0.010416}  \le Z \le \frac{0.10 -0.07}{0.010416}  )

=> P(|p-\^{p}| >  0.03)  =  1 -  P(-2.88  \le Z \le 2.88 )

=>   P(|p-\^{p}| >  0.03)  =   1 - [P(Z \le 2.88) - P(Z \le -2.88)]

From the z-table  

       P(Z \le 2.88)  =  0.9980

and

       P(Z \le -2.88)  = 0.0020

So

     P(|p-\^{p}| >  0.03)  =   1 - [0.9980 - 0.0020]

=>   P(|p-\^{p}| >  0.03)  =   0.0040

     

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