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pochemuha
1 year ago
13

Quadrilateral A'B'C'D' is the image of quadrilateral ABC D under a dilation with a scale factor of 4. c What is the length of se

gment C" D'?
Mathematics
1 answer:
DENIUS [597]1 year ago
4 0

a dilation with a scale factor of 4, is multiplication by 4 each side, so we need the measure of the side CD and then multiply by 4 to find C'D'

we can notice tht length of CD, is 3 squares,multiplying by 4 the new length is 3x4= 1 2 squares

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➷ We know this:

3/7 = 27

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To do this, divide by 3

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Multiply this by 7 to get the total number of balls:

9 x 7 = 63

There are 63 balls.

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3 years ago
How many solutions does 2x^2-3x+6=0 have?
Ipatiy [6.2K]

Answer:

i beileve in this answer its a question of more than one salution, one salution and no solutions ................ so more than one salutions

Step-by-step explanation:

8 0
3 years ago
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What is the equation of this graph
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Answer:

y-1=x^2

Step-by-step explanation:

That is the equation of a parabola with vertex at (0,1). The equation is y-1=x^2.

3 0
3 years ago
Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

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