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makvit [3.9K]
3 years ago
5

Three cards are drawn sequentially from a shuffled deck without replacement. What is the approximate probability all three drawn

cards have NUMBERS on them?
Mathematics
1 answer:
bekas [8.4K]3 years ago
8 0

Answer:

Step-by-step explanation:

There are 36 numbered cards out of 52 total cards .

so in the first attempt probability of numbered cards being drawn

= 36 / 52

After first draw , numbered cards remaining = 35 , total no of card remaining = 51

probability of numbered card  in 2nd attempt = 35 / 51

Probability in third attempt

= 34 / 50

Total probability  of all cards being numbers = (36 / 52) x (35 / 51) x (34/50)

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The length is 12, because if it's 9 inches wide it has two sides of 9 inches which is 18. Then subtract that from the intial value 42 to get 24. After, you divide 24 by 2 due to there being 2 sides of the box and 24 is the sum of both sides. Hopethis helps!
8 0
3 years ago
Solve (x − 2) = 3<br> Give your solutions correct to 3 significant figures.
Pachacha [2.7K]

Answer:

(x − 2) = 3

Remove the bracket

x - 2 = 3

Group the constants at one side of the equation

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x = 5

Therefore x = 5.00 to 3 significant figures.

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6 0
3 years ago
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
wolverine [178]

Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

b^3-4b^2=0

Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

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