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VMariaS [17]
3 years ago
15

A box contains 12 balls of which 5 are red, and 7 are blue. If 4 balls are drawn from the box, in how many ways can this be done

if 1 is red and 3 are blue?
Mathematics
1 answer:
mash [69]3 years ago
6 0

Answer:

175 ways

Step-by-step explanation:

5 balls are red, and 7 balls  are blue making a total of 12 balls.

4 balls are drawn from the box,

1 ball is red and 3  balls are blue out of the 4 drawn from the box.

We have to choose 1 ball from 5 red balls and 3 blue balls from 7 blue balls.

Since the order is not essential we use combinations to find out

5C1 * 7C3= 5*35 = 175 ways

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Umnica [9.8K]

Answer:

(a)

Shortest Side of the trapezoid =6 yds

Longest Side =8 yds

(b)42 Square Yards

Step-by-step explanation:

<u>Part A</u>

In the Trapezoid, Width =6 yd

Since the shortest side of the playground and its width have the same dimension, then:

  • Shortest Side of the trapezoid =6 yd
  • Longest Side =6+1+1=8 yd

<u>Part B</u>

Area of the Playground

In the trapezoid

a=6 yd, b=8 yd, h=Width=6 yd

Area of a trapezoid=\frac{1}{2}(a+b)h<u />

<u />Therefore: Area=\frac{1}{2}(6+8)*6\\=14*3\\=42$ Square Yards<u />

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Step-by-step explanation:

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Solve 5x + 3x - 6 = 42​
xxMikexx [17]

Answer:

The answer to your question is x = 6

Step-by-step explanation:

Equation                   5x + 3x - 6 = 42

Process

Step 1. Add 6 units in both sides

                                 5x + 3x - 6 + 6 = 42 + 6

Step 2. Simplify

                                  5x + 3x = 48

Step 3. Simplify like terms

                                         8x = 48

Step 4. Divide by 8 both sides

                                        8x/8 = 48/8

Step 5. Simplify

                                          x = 6

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