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spin [16.1K]
3 years ago
11

A student pulls out a marble from a pag containing blue, green, and red marbles. He records the color and places it back into th

e bag The table below shows the frequency of
each color after 100 marbles are pulled out
Color of Marbles Blue Green Red
Number of Draws 13 29 58
How many green draws can you expect if the marbles are pulled out 1,000 times?

Mathematics
2 answers:
Mashutka [201]3 years ago
8 0

Answer:

Answer is B - 290.

Step-by-step explanation:

The idea here is that we're assuming the frequency each colour is drawn is a reflection of the proportion of marbles in the bag that are that colour. The more of a particular colour you have, the more likely you are to draw it, and vice versa.

Therefore we can assume the proportion of blue, gree, and red marbles is 13%, 29%, and 58% respectively based on the first 100 draws.

These percents should stay relatively similar as more draws are done, approaching the true proportions of each colour better the higher the number of draws. That means 29%, or 290, of the 1000 draws will be green.

katrin [286]3 years ago
4 0

Answer: 290

Step-by-step explanation:

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

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AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

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