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VashaNatasha [74]
3 years ago
10

A pair of fair dice is rolled.

Mathematics
2 answers:
mezya [45]3 years ago
5 0
A) 4/36, or 1/9
b) 1-1/9=8/9
Kobotan [32]3 years ago
5 0
Sample space={6 x 6} = 36 possible outcomes:

Favorable outcome {(3+6),(4+5),(5+4),(6+3) = 4 favorable outcome:

P(getting a sum = 9) = 4/36 = 1/9 = 0.1111

the odds  AGAINST = the number of failure/number of success

the odds  AGAINST = 1/9
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2.34 per pound is unit price

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(2x - y + 3) (2x - y - 3)using identities ​
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x(y) = y(x)

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3 years ago
What is x2+5x=0? Please show all work!
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7 0
3 years ago
Read 2 more answers
a shopkeeper sold a certain number ( a two-digit number)of toys all priced at a certain value (also a two-digit number when expr
Makovka662 [10]

The answer is 91 toys sold, make the number ab where a is the 10th digit and b is the first digit. The value is 10a + b that can expressed as 10 (3) + 4 = 34

Let the price of each item: xy

10x + y

He accidentally reversed the digits to: 10b + a toys sold at 10y + x rupees per toy. To get use the formula, he sold 10a + b toys but thought he sold 10b + a toys. The number of toys that he thought he left over was 72 items more than the actual amount of toys left over. So he sold 72 more toys than he thought:

10a + b =10b + a +72

9a = 9b + 72

a = b + 8

The only numbers that could work are a = 9 and b = 1 since a and b each have to be 1 digit numbers. He reversed the digits and thought he sold 19 toys. So the actual number of toys sold was 10a + b = 10 (9) + 1 = 91 toys sold. By checking, he sold 91 – 19 = 72 toys more than the amount that he though the sold. As a result, the number of toys he thought he left over was 72 more than the actual amount left over as was stated in the question.

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3 0
3 years ago
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