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3 years ago

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A box contains 2 red marbles, 3 green marbles, and 3 blue marbles. if we choose a marble, then another marble without putting th

e first one back in the box, what is the probability that the first marble will be green and the second will be blue?

2 answers:

Bad White [126]3 years ago

7 0

There are 7 marbles, including 3 green ones, so at the start, there is a 3/7 chance of getting a green marble.
Assuming you did get a green one, there are now 6 marbles left, with 2 blue marbles, so there is a 2/6 chance of taking a blue marble.
Given that both have to happen, you must multiply each probability, hence the total probability is 3/7 x 2/6, or a 1/7 chance.
Hope this helped

Zepler [3.9K]3 years ago

4 0

Answer:

The answer is actually 9/56

Step-by-step explanation:

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Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro

Neporo4naja [7]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

★ Area each of them [son and daughter] will get.

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Let, ABCD be the rhombus shaped field and each side of the field be $x$

[ All sides of the rhombus are equal, therefore we will let the each side of the field be $x$ ]

Now,

• Perimeter = 400m

$\longrightarrow \tt AB+BC+CD+AD=400m$

$\longrightarrow \tt x + x + x + x=400$

$\longrightarrow \tt 4x=400$

$\longrightarrow \tt \: x = \dfrac{400}{4}$

$\longrightarrow \tt x= \red{100m}$

$\therefore$ Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For $\triangle$ ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

$\therefore \tt Simi \: perimeter \: [S] = \boxed{ \sf \dfrac{a + b + c}{2} }$

$\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}$

$\longrightarrow \tt S = \cancel{ \dfrac{360}{2}}$

$\longrightarrow \tt S = 180m$

Using <u>herons formula</u><u>,</u>

$\star \tt Area \: of \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star$

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{180(80)(80)(20) }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{180 \times 80 \times 80 \times 20 }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{ {3}^{2} \times {20}^{2} \times {80}^{2} }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = 3 \times 20 \times 80$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \red{ 4800 \: {m}^{2} }$

Thus, area of $\triangle$ ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of $\triangle$ BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

★ Figure in attachment.

${\underline{\rule{290pt}{2pt}}}$

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