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ANEK [815]
3 years ago
13

Which image is the dilation of the rectangle with the center of the dilation at the origin and a scale factor of 1/2 ?

Mathematics
2 answers:
yKpoI14uk [10]3 years ago
8 0

Answer:

The 3rd graph represents the dilated image of the rectangle.

Step-by-step explanation:

We are given the rectangle having vertices,

(1,-2), (1,2), (-1,2) and (-1,-2).

<em>Now, this rectangle is dilated about the origin by the scale factor of \frac{1}{2}</em>

That is, the size of the rectangle is reduced by the factor \frac{1}{2}.

Then, the vertices of the new rectangle will be,

(1,-2) changes to \frac{1}{2}\times (1,-2) = (\frac{1}{2},-1)

(1,2) changes to \frac{1}{2}\times (1,2) = (\frac{1}{2},1)

(-1,2) changes to \frac{1}{2}\times (-1,2) = (\frac{-1}{2},1)

(-1,-2) changes to \frac{1}{2}\times (-1,-2) = (\frac{-1}{2},-1)

So, the vertices of the dilated rectangle are (\frac{1}{2},-1), (\frac{1}{2},1), (\frac{-1}{2},1) and (\frac{-1}{2},-1).

Thus, the 3rd graph shown below represents the dilated image of the rectangle.

ElenaW [278]3 years ago
4 0
A <span>To find the Scale Factor divide= </span><span>a side in the image /<span>the same side in the figure</span></span>
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Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

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P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

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Only 0.4% of the people who don't have it test positive.

P(A|D^c) = probability that a person is tested positive given he don't have a disease = 0.004

P(D^c)=1-\frac{1}{10000}

Formula:P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}

P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}

P(D|A)=\frac{2470}{2471}=0.9995

P(D|A)=0.9995

A)The probability that someone who tests positive has the disease is 0.9995

(B)

P(D^c|A^c)=probability that someone does not have disease given that he tests negative

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=0.996

P(A^c|D)=probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}

P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}

P(D^c|A^c)=0.99999

B)The probability that someone who tests negative does not have the disease is 0.99999

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3 years ago
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lianna [129]

Answer:

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