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Assoli18 [71]
3 years ago
8

What is the length of line segment DC

Mathematics
2 answers:
Misha Larkins [42]3 years ago
4 0

The answer is 3 units.


12/4= 3

9/3= 3 units

Y_Kistochka [10]3 years ago
3 0

Answer:

The length of DC is 3 unit.

Step-by-step explanation:

Consider the provided figure.

DE is parallel to CB.

According to basic proportionality theorem:

"If a line is drawn parallel to one side of a triangle to pass the other two sides in separate points, the other two sides are divided into the same ratio."

Therefore

\frac{AE}{EB}=\frac{AD}{DC}

\frac{12}{4}=\frac{9}{x}

3=\frac{9}{x}

x=\frac{9}{3}

x=3

Thus, the length of DC is 3 unit.

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A beacon is flashing on top of a 50 foot tower. A 6 foot tall man walks constantly away from the tower at 5 feet/sec. At the ins
Viefleur [7K]

Answer:\frac{253}{44}

Step-by-step explanation:

ignore the "at the instant the man is 30 feet away" part, set it as X and the man's shadow as Y.

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7 0
2 years ago
7.4 Practice
pickupchik [31]

The simulation of the medicine and the bowler hat are illustrations of probability

  • The probability that the medicine is effective on at least two is 0.767
  • The probability that the medicine is effective on none is 0
  • The probability that the bowler hits a headpin 4 out of 5 times is 0.3281

<h3>The probability that the medicine is effective on at least two</h3>

From the question,

  • Numbers 1 to 7 represents the medicine being effective
  • 0, 8 and 9 represents the medicine not being effective

From the simulation, 23 of the 30 randomly generated numbers show that the medicine is effective on at least two

So, the probability is:

p = 23/30

p = 0.767

Hence, the probability that the medicine is effective on at least two is 0.767

<h3>The probability that the medicine is effective on none</h3>

From the simulation, 0 of the 30 randomly generated numbers show that the medicine is effective on none

So, the probability is:

p = 0/30

p = 0

Hence, the probability that the medicine is effective on none is 0

<h3>The probability a bowler hits a headpin</h3>

The probability of hitting a headpin is:

p = 90%

The probability a bowler hits a headpin 4 out of 5 times is:

P(x) = nCx * p^x * (1 - p)^(n - x)

So, we have:

P(4) = 5C4 * (90%)^4 * (1 - 90%)^1

P(4) = 0.3281

Hence, the probability that the bowler hits a headpin 4 out of 5 times is 0.3281

Read more about probabilities at:

brainly.com/question/25870256

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Step-by-step explanation:

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