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vitfil [10]
2 years ago
10

A bowl contains 7 red balls and 10 blue balls. A woman selects 4 balls at random from the bowl. How many different selections ar

e possible if at least 3 balls must be blue?
Physics
1 answer:
denis-greek [22]2 years ago
8 0

Answer:

1050 possible selections

Explanation:

<em>Number of red balls = 7</em>

<em>Number of blue balls = 10</em>

<em>Let blue balls be B and red balls be R</em>

<em>four balls are to be selected. At least three must be blue.</em>

the first combination is 4blue 0red

second combination is 3blue 1red

(10C4*7C0 + 10C3*7C1)

210 *1 +120 *7

210+840

=1050 possibilities

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3 years ago
Okay i'm totally stuck and nobody I know really gets it either, so i've turned to Yahoo for help :)
OlgaM077 [116]

Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:

                     (weight) x (distance from the pivot) <u>on one side</u>
is equal to
                     (weight) x (distance from the pivot) <u>on the other side</u>.

That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.

       (Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).


So now we come to the strange beings on the alien planet.
There are three choices right away that both work:

<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.

       (400) x (1)  =    (200) x (2)

<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.

       (200) x (1)  =    (100) x (2)

<u>#3).</u>

On one side:  (300 N) in the end-seat       (300) x (2) = <u>600</u>

On the other side:
                      (400 N) in the middle-seat  (400) x (1) = 400
           and     (100 N) in the end-seat      (100) x (2) = 200
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These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.


5 0
3 years ago
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