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aksik [14]
3 years ago
12

Consider a deck of cards:

Mathematics
1 answer:
Ganezh [65]3 years ago
3 0

Ohhhh nasty !  What a delightful little problem !

The first card can be any one of the 52 in the deck.  For each one ...
The second card can be any one of the 39 in the other 3 suits. For each one ...
The third card can be any one of the 26 in the other 2 suits.  For each one ...
The fourth card can be any one of the 13 in the last suit.

Total possible ways to draw them = (52 x 39 x 26 x 13) = 685,464 ways.

But wait !  That's not the answer yet.

Once you have the 4 cards in your hand, you can arrange them
in (4 x 3 x 2 x 1) = 24 different arrangements.  That tells you that
the same hand could have been drawn in 24 different ways.  So
the number of different 4-card hands is only ...

                     (685,464) / (24) = <em>28,561 hands</em>.

I love it !


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Derrick earned 1472 during the 4 weeks he had his summer job. If he earned the same amount each week, how much did he earn each
Agata [3.3K]
368 he earned each week.

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3 years ago
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A magician’s stage has a rectangle trapdoor which is (x+ 1/2) by 2x ft. The outside rectangle length is (x+ 16) ft.
Katen [24]

Answer:

Part A:- x +11 ft     Part B:- (4x+54)ft   part C:- x= 2ft  

Part D:- Yes it satisfy his requirement

Step-by-step explanation:

Given that,

length and breadth of Trapdoor are (x+ \frac{1}{2} ) by 2x ft.

The Outside length of Rectangle is (x+16 ) ft.

Part A:- Total area (in square feet) of the stage can be represented by  

             x^{2} + 27x+ 176. Write an expression for the width of stage.

So,        Area of rectangle =  length \times breadth

                 x^{2} + 27x+ 176 = (x+16) \times breadth

                     Breadth = \frac{x^{2} + 27x+ 176}{x+16}

                     Breadth = x+11 ft.                       ................(1)

Part B:- Write an expression for the Perimeter of the stage.

Here,   Perimeter of Rectangle = 2(L+B)

                                                    = 2(x+16 + x +11)

                                                    = (4x + 54) ft

Part C:- The area of trapdoor is 10 ft^{2}. Find the value of x.

So,                        Area of trapdoor = 10 ft^{2}

                              (x +\frac{1}{2}) \times 2x =10 ft^{2}

                               2x^{2} +x - 10 = 0

                                  x = \frac{-1-9}{4} , \frac{-1+9}{4}

                                  x = \frac{-5}{2} , 2

Hence value of x is 2 ft             (Neglecting the negative value because                      

                                                   length cannot be negative).

Part D:-  The magician wishes to have the area of the stage be at least 20 times the area of the trapdoor. Does this stage satisfy his requirement? Explain.

So,   Length of trapdoor is (2+\frac{1}{2} ) = \frac{5}{2} ft

        breadth of trapdoor is 2\times 2 = 4ft

Now, outside length of Rectangle is = 18ft

And outside breadth of rectangle is = 13ft

Here, Area of trapdoor = \frac{5}{2} \times 4 = 10ft^{2}              ...................(2)

         Area of rectangle = 18 \times 13= 234 ft^{2}        ....................(3)

Thus comparing Equation (1) & (2) we found that

Area of rectangular stage is 20 times greater than the area of trapdoor.

7 0
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konstantin123 [22]

Answer:

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Step-by-step explanation:

Hope this helped! :)

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