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enyata [817]
3 years ago
11

Marbles Selected Color Time Selected Purple 2 Orange 5 Black 4 White 6 Green 3 Using these results, what is the probability that

a black marble will be the next color selected?
Mathematics
1 answer:
notka56 [123]3 years ago
6 0
The probability is 5/20.

Explanation:

All together there are 20 marbles, if there are 5 black marbles then that means you would have 5 chances of picking a black marble.
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A

Step-by-step explanation:

theare are 5 part from 0 to -1

since, the value from 0 to -1 is -1 then if we break it to 5 part, each part has value of -1/5

so, to be -4/5, -1/5 must be multiplied by 4, so we need to go to the 4th part from 0, so it A

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Step-by-step explanation:

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Any figure to this question?

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savannah has $5 in her wallet. she owes her brother 20 and owes her sister 18 . she earned 25 babysitting. write an algebraic ex
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The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>

7 0
3 years ago
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