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Olin [163]
3 years ago
10

A bag contains 10 red marbles, 15 yellow marbles, 5 green marbles, and 20 blue marbles. Two marbles are drawn from the

Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

<u />

  •     <u>        (₁₀C₁) × (₂₀C₁)       </u>

                           ₅₀C₂

Explanation:

There are a total number of 10 + 15 + 5 + 20 = 50 marbles.

The number of ways in which any two marbles can be drawn is the combination of 50 marbles taken two at a time: ₅₀C₂, since the order does not matter.

There are 10 red marbles and 20 blue marbles.

Thus, the combinations with one red marble are: ₁₀C₁.

The combinations with one blue marble are: ₂₀C₁.

The combinations with one red and one blue marble is the product of the previous two:

  • ₁₀C₁ × ₂₀C₁

Thus, the probability that one of the marbles is read and the other is blue is the quotient of the two sets:

<u />

  •     <u>        (₁₀C₁) × (₂₀C₁)       </u>

                           ₅₀C₂

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6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
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