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Burka [1]
3 years ago
6

How many different combinations are possible if a number cube is tossed once and a coin is tossed twice?

Mathematics
1 answer:
MAVERICK [17]3 years ago
8 0
Cube has 6 sides and coin has 2

6*2*2=24 combinations
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3. About how much of the earth's surface is deserts?
Murrr4er [49]

Answer:

B- One-Quarter

Step-by-step explanation:

8 0
3 years ago
Solve the following system of equations please ASAP
swat32

Given:

The system of equation:

x+y = -4 -------- (1)

y = x^{2} -6x --------- (2)

To find the values of x and y.

We will use substitution method.

From (1) we get,

y = -4-x

We will put the value of y in (1) and we get,

-x-4 = x^{2} -6x

or, x^{2} -5x+4= 0

Now we will apply middle term factor method.

  x^{2} -(4+1)x+4 = 0

     x^{2} -4x-x +4 = 0

x(x-4)-1(x-4) = 0

       (x-4)(x-1)= 0

so, x = 4 and 1

Now,

Substitute x = 4 in (1) we get,

y = -4-4 = -8

And putting x = 1 in (1) we get,

y = -4-1 = -5

Hence, the solution of the given system of equation is (4,-8) and (1,-5)

Thus, Option A is the correct answer.

3 0
3 years ago
An experiment consists of ranomly drawing three cards in succession without replacement from a standard deck of 52 cards. Let A
nadya68 [22]

Answer:

0.0045248 ;

0.1312218 ;

0.0001809 ;

0.1659729

Step-by-step explanation:

Number of Kings in deck = 4

Total number of cards in deck = 52

Picking without replacement :

A = King on first draw :

P(A) = 4 / 52

A = King on 2nd draw :

P(B) = 3 / 51

A = King on 3rd draw :

P(C) = 2 / 50

1.) P(A n B) = P(A) * P(B)

P(A n B) = 4/52 * 3/51 = 12 / 2652 = 0.0045248

2.) P(A u B) = P(A) + P(B) - P(AnB)

P(AuB) = 4/52 + 3/51 - 0.0045248 = 0.1312218

3.) P(A ∩ B ∩ C) = P(A) * P(B) * P(C)

P(A ∩ B ∩ C) = 4/52 * 3/51 * 2/50 = 0.0001809

4.) P(A U B U C) =

P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) - P(AnBnC)

P(AnC) = P(A) * P(C) = 4/52 * 2/50 = 0.0030769

P(BnC) = P(B) * P(C) = 3/51 * 2/50 = 0.0023529

4/52 + 3/51 + 2/50 - 0.0045248 - 0.0030769 - 0.0023529 + 0.0001809 = 0.1659729

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