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kolezko [41]
3 years ago
9

there are 11 days until Richards's concert. Susie's concert is 7 days after Richard's. How many days are there until Susie's con

cert? A. 77÷11=7. B. 11-7=4. C. 11 x 7=77. D. 11+7=18
Mathematics
1 answer:
matrenka [14]3 years ago
8 0
There are 11 days until Richards's concert.
Susie's concert is 7 days after Richard's.

so Susie's concert: 11+7=18
answer
D. 11+7=18
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Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


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Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

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3 years ago
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He doesn’t have enough for both to be on at a time but if the cars are on without the train the cars will take up 2.4 meters leaving them with an extra .1 meters of length
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3 years ago
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3 years ago
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Vedmedyk [2.9K]

\\ \sf\longmapsto 2:3=1:n

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Answer:

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