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Aleksandr-060686 [28]
2 years ago
13

PLS HELP ME Draw the image of ∆JKL under the dilation with scale factor ½ and center of dilation at the origin. Label the image

∆J’K’L’.

Mathematics
1 answer:
Vsevolod [243]2 years ago
6 0

Answer:

See explanation

Step-by-step explanation:

Given triangle JKL with vertices at J(-2,4), K(-8,4), L(-4,6).

The dilation with scale factor ½ and center of dilation at the origin has the rule

(x,y)\rightarrow \left(\dfrac{1}{2}x,\dfrac{1}{2}y\right)

Thus,

  • J(-2,4)\rightarrow J'(-1,2);
  • K(-8,4)\rightarrow K'(-4,2);
  • L(-4,6)\rightarrow L'(-2,3).

See attached diagram.

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Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

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Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

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Now, for each case, we have to find the probability that the last marble is red. So

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}

P_{1} is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is \frac{30}{50} = \frac{3}{5}.

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is \frac{33}{53}

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is \frac{36}{56}

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Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is \frac{39}{59}. So

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P_{2} is the probability that we go R - R - B - R

P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788

P_{3} is the probability that we go R - B - R - R

P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076

P_{4} is the probability that we go R - B - B - R

P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511

P_{5} is the probability that we go B - R - R - R

P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733

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P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493

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P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078

There is a 20.78% probability that we actually drew the same marble all four times

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