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devlian [24]
3 years ago
7

HELP!!!!!!!!!!!!!!!!!!!!!!!!!! EXPLAIN TOO PLZ!

Mathematics
2 answers:
user100 [1]3 years ago
8 0

Answer:B) \frac{14}{95}

Step-by-step explanation:

Step 1: list down the probability of the marbles,

you will get the probability

of \frac{3}{5} for the blue marbles and \frac{2}{5} for the red ones .

Step 2: <em>Forget</em><em> about the blue marbles and  focus on the red marbles </em>

On the first pick of marbles  the probability of getting a red marble is \frac{2}{5} but then you didn't replace the marble so now you have only 19 marbles as your total.

Step 3: lets say you picked  a red marble on the first pick now you only have 7 red marbles out of 19 marbles (minus the red marble you picked -> step 2)

so the probability of getting a red marble on the second pick is \frac{7}{19}

Step 4: multiply the probabilities of both the first and second pick

it will look like this

\frac{2}{5}  × \frac{7}{19}

  ^                                              ^

first pick                                 second pick

Note : You will only minus your probability and number of "marbles "( in this case ) if the question says "without replacing "

final answer : \frac{14}{95}

 

asambeis [7]3 years ago
7 0

Ok so you would just take the two fractions and simply if needed and that the answer!

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{\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.

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★ 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 ]

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• Perimeter = 400m

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

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

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

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\longrightarrow \tt S = 180m

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\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} }

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As both the triangles have same sides

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

★ Figure in attachment.

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

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