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yaroslaw [1]
3 years ago
5

A grab bag contains 4 football cards and 6 basketball cards. An experiment consists of taking one card out of the bag, replacing

it, and then selecting another card. What is the probability of selecting a football card and then a basketball card? Express your answer as a decimal.
Mathematics
2 answers:
Zinaida [17]3 years ago
5 0
Total Number of cards = 4 + 6 = 10
Number of Football cards = 4
Number of Basketball cards = 6

Probability of choosing a football card = 4/10 = 0.4
Since this card is replaced, the total number of cards will remain the same.

Probability of selecting a basketball card = 6/10 = 0.6

Since, the two events are independent, the probability of selecting a football and a basketball card will be the product of two probabilities we calculated. 

Thus, probability of selecting a football card and then a basketball card = 0.4 x 0.6 = 0.24
Sonja [21]3 years ago
4 0

Answer:

0.24

Step-by-step explanation:

6/10 - baseball cards

4/10 - football cards

6/10 = 0.6

4/10 = 0.4

0.6 X 0.4 = 0.24

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(a) The product of (2x-4)\ and\ (3x^2-x+4)  is calculated to be =6x^3-14x^2+12x-16

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<u>Step-by-step explanation:</u>

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(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16

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(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8

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3 years ago
Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5
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Answer:

J (6,2) K (2,2)

Step-by-step explanation:

Now distance of AB = [(y2-y1)^2 + (x2-x1)^2]^(1/2)

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CB =  [(y2-y1)^2 + (x2-x1)^2]^(1/2)

= 6 units by using same formula

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Similarly K is (2,2) for the same reason.

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