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Rama09 [41]
2 years ago
9

In a board game, you must roll two 6-sided number cubes. You can only start the game if you roll a 3 on at least one of the numb

er cubes.
(a)Make a list of all the different possible outcomes when two number cubes are rolled.
(b)What fraction of the possible outcomes is favorable?
(c)Suppose you rolled the two number cubes 100 times, would you expect at least one 3 more or less than 34 times? Explain.
Mathematics
1 answer:
NeTakaya2 years ago
6 0
B is 2 out of 12

c is if you were talking probability then you would have put down one out of 6 because there is one 3 out of 6 sides but i believe i would expect to see two times 34 because it is two time the dice and it gives you the capability to earn three twice as much so you should see it at least 68 times    

HOPE THIS HELPS


im not clear on a tho

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Three consecutive odd integers have a sum of 27. Find the integers.
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\sf \bf {\boxed {\mathbb {GIVEN:}}}

Sum of three consecutive odd integers = 27

\sf \bf {\boxed {\mathbb {TO\:FIND:}}}

The values of the three integers.

\sf \bf {\boxed {\mathbb {SOLUTION:}}}

\sf\purple{The\:three\:consecutive \:odd\:integers\:are\:7,\:9\:and\:11.}

\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}

Let us assume the three consecutive odd integers to be x, (x+2) and (x+4).

As per the condition, we have

Sum \:  \:  of \:  \:  the  \:  \: three \:  \:  consecutive \:  \:  odd \:  \:  integers  = 27

➺ \: x + (x + 2) + (x + 4) = 27

➺ \: x + x + 2 + x + 4 = 27

Now, collect the like terms.

➺ \: (x + x + x) + (2 + 4) = 27

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➺ \: x =  \frac{21}{3} \\

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Therefore, the three consecutive odd integers whose sum is 27 are \boxed{  7  }, \boxed{ 9   } and \boxed{ 11   } respectively.

\sf \bf {\boxed {\mathbb {TO\:VERIFY :}}}

⇢ 7 + 9 + 11 = 27

⇢ 27 = 27

⇢ L. H. S. = R. H. S.

\sf\blue{Hence\:verified.}

\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}

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