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mamaluj [8]
4 years ago
11

Two six-sided fair dice are rolled. The probability that at least one number is odd and the sum of the two numbers is even is .

The probability that exactly one number is 6 and the product of the two numbers is at most 15 is .
Mathematics
1 answer:
Oxana [17]4 years ago
5 0
The easiest way to figure out probability problem with small data sets is to write out your entire sample space then divide by the total:

Sample size = 6 * 6 = 36

S = {[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[2,1],[2,2],[2,3],[2,4],[2,5],[2,6],[3,1],[3,2],[3,3],[3,4],[3,5],[3,6],[4,1],[4,2],[4,3],[4,4],[4,5],[4,6],[5,1],[5,2],[5,3],[5,4],[5,5],[5,6],[6,1],[6,2],[6,3],[6,4],[6,5],[6,6]}

The only way to make a number combination that's even while 1 die is odd is to have 2 odd numbers.

{[1,1],[1,3],[1,5],[3,1],[3,3],[3,5],[5,1],[5,3],[5,5]}

This gives us 9 results.

The probability of this happening is 9/36 = 1/4 = 0.25

Now if we have to get a 6 with the product being at most 15 we know that the biggest number that 6 can be multiplied by is 2 which gives us 12.

We are left with 4 options:

{[1,6],[2,6],[6,1],[6,2]}

The probability of this happening is 4/36 = 1/9 = 0.1111...
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Multiply r2+7r+10/3 by 3r-30/ r2-5r-50
kvv77 [185]
((r²+7r+10)/3) × (3r-30)/(r²-5r=50)
= Factoring the expression (r²+7r+10) we get  (r+2)(r+5)
Then multiply it with the other numerator, we get
= (r+2)(r+5)(3(r-10)
Then we factorise
= (r²-5r-50) = (r+5)(r-10)
Multiplying it with the other denominator, we have;
= 3(r+5)(r-10)
Therefore dividing

(r+2)(r+5)(3(r-10)/(3(r+5)(r-10) we get;
= r+2
Therefore, the correct answer is r+2

7 0
4 years ago
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Nick handed in 2 paper to his professor, each paper is worth 5.5%. How much will both paper increase his grade?
Jobisdone [24]
Grade increase would be 11%
5 0
3 years ago
What is the slope of the line that contains the points (-2, 7) and (2, 3)?
charle [14.2K]
Slope = rise/run
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3 0
3 years ago
a freight elevator CAN safely hold no more than 2000 Pound. an elevator operator must take 55lb boxes to a storage área. if he w
BARSIC [14]

Answer:

33 boxes.

Step-by-step explanation:

We have been given that a freight elevator can safely hold no more than 2000 pound. Weight of elevator operator is 165 pound. Weight of each box is 55 pound.

Let n be number of boxes. We will write an inequality as weight of elevator operator and weight of all boxes should be no more than 2000 pound.

55*n+165\leq 2000

Now we will solve our inequality to find number of boxes that could be moved safely at one time.

55*n\leq 2000-165

55*n\leq 1835

n\leq \frac{1835}{55}

n\leq 33.363636

Therefore, 33 boxes can be moved safely at one time.  


8 0
4 years ago
Helpppppppp here plz
miv72 [106K]

Answer:

x + x+12 + 3(x+12) = 123

x = 15

Step-by-step explanation:

Jamil : x

Kiera : x+12

Luther : 3 ( x+12)

x + x+12 + 3(x+12) = 123

Distribute

x + x+12 + 3x+36 = 123

Combine like terms

5x+ 48 = 123

Subtract 48 from each side

5x+48-48 = 123-48

5x =75

Divide by 5

5x/5 = 75/5

x = 15

7 0
3 years ago
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