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wel
3 years ago
5

.)A bag has 7 red marbles, 8 blue marbles and 3 black marbles. What is the probability of picking a blue marble, replacing it, a

nd then picking a red marble? Answer a. 14/81 b. 7/108 c. 2/27 d. 1/9 2.)A bag has 7 red marbles, 8 blue marbles and 3 black marbles. What is the probability of picking a black marble, replacing it, and then picking a red marble? Answer a. 2/27 b. 7/108 c. 14/81 d. 1/9
Mathematics
1 answer:
topjm [15]3 years ago
3 0
Choice A. <span>14/81 
</span>Possible outcomes = total number of marbles = 18 
P (1st blue) = 8/18 = 4/9 
P (2nd red with replacement of blue) = 7/18 
P (both) = (4/9) * (7/18) = 28/162 = 14/81 

Choice B. 7/18
Possible outcomes = total number of marbles = 18 
<span>P (1st black) = 3/18 = 1/6 </span>
<span>P (2nd red with replacement of black) = 7/18 </span>
<span>P (both) = (1/6) * (7/18) = 7/108</span>
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Answer for part A: Definition of perpendicular
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Explanations:

Part A:

We are given that \overline{RX} \perp \overline{ST} which means, in english, "line segment RX is perpendicular to line segment ST"

By the very definition of perpendicular, this means that the two line segments form a right angle. This is visually shown as the red square angle marker for angle RXT. Angle RXS is also a right angle as well.

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Part B:

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Part C:

Any line segment is congruent to itself. This is because any line segment will have the same length as itself. It seems silly to even mention something so trivial but it helps establish what we need for the proof. 

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Part D:

We are given "X is the midpoint of segment ST" so by definition, X is in the very exact middle of ST. Midpoints cut segments exactly in half. SX is one half while TX is the other half. The two halves are congruent which is why SX = TX

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Part E:

Writing \triangle SXR \cong \triangle TXR means "triangle SXR is congruent to triangle TXR". These two triangles are the smaller triangles that form when you draw in segment RX

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Part F:

CPCTC stands for "Corresponding Parts of Congruent Triangles are Congruent"

It means that if we have two congruent triangles, then the corresponding parts are congruent. Back in part E, we proved the triangles congruent. For this part, we look at the pieces RS and RT (which correspond to one another; they are the hypotenuse of each triangle). They are proven congruent by CPCTC

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