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2 years ago

9

A bag contains 3 red marbles, 2 blue marbles and 4 green marbles. If two marbles are drawn out of the bag, what is the probabili

ty, to the nearest 10th of a percent, that both marbles drawn will be red?

1 answer:

Basile [38]2 years ago

3 0

Answer:

9

Step-by-step explanation:

Well we

do 3 + 4 + 2

= 9

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

$2489ft^{2}$

Step-by-step explanation:

The pool are is divided into 4 separated shapes: 2 circular sections and 2 isosceles triangles. Basically, to calculate the whole area, we need to find the area of each section. Due to its symmetry, both triangles are equal, and both circular sections are also the same, so it would be enough to calculate 1 circular section and 1 triangle, then multiply it by 2.

<h3>Area of each triangle:</h3>

From the figure, we know that <em>b = 20ft </em>and <em>h = 25ft. </em>So, the area would be:

$A_{t}=\frac{b.h}{2}=\frac{(20ft)(25ft)}{2}=250ft^{2}$

<h3>Area of each circular section:</h3>

From the figure, we know that $\alpha =2.21 radians$ and the radius is $R=30ft$ . So, the are would be calculated with this formula:

$A_{cs}=\frac{\pi R^{2}\alpha}{360\°}$

Replacing all values:

$A_{cs}=\frac{(3.14)(30ft)^{2}(2.21radians)}{6.28radians}$

Remember that $360\°=6.28radians$

Therefore, $A_{cs}=994.5ft^{2}$

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$A_{total}=2A_{t}+2A{cs}=2(250ft^{2} )+2(994.5ft^{2})\\A_{total}=500ft^{2} + 1989ft^{2}=2489ft^{2}$

Therefore the area of the symmetrical pool is $2489ft^{2}$

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3 years ago

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