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3 years ago

Please help me with this

Mathematics

1 answer:

3241004551 [841]3 years ago

7 0

120+x+16+x=180
2x+ 136=180
2x=44
X= 22

<B= 22

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In a game called Taxation and Evasion, a player rolls a pair of dice. If, on any turn, the sum is 7, 11, or 12, the player gets

AnnZ [28]

Answer:

$Probability = \frac{243}{1024}$

Step-by-step explanation:

Given

$Sum = \{7, 11, 12\}$

$Rolls = 5$

Required

Probability of not getting audited

If a pair of dice is rolled, the following are the observations of the sum

$Outcomes = 36$

$Sum\ of\ 7 = 6$

$Sum\ of\ 11 = 2$

$Sum\ of\ 12 = 1$

So, in a single roll; The probability of getting audited is:

$p= \frac{1 + 2 + 6}{36}$

$p= \frac{9}{36}$

$p= \frac{1}{4}$

The probability of not getting audited in a single roll is:

$q = 1 - p$ --- Complement rule

$q = 1 - \frac{1}{4}$

Take LCM

$q = \frac{4 - 1}{4}$

$q = \frac{3}{4}$

The probability of not getting audited in 5 rolls is:

$Probability = q^5$

$Probability = (\frac{3}{4})^5$

$Probability = \frac{3^5}{4^5}$

$Probability = \frac{243}{1024}$

7 0

3 years ago

What comes after this:

Lynna [10]

Answer:

16:20??

Lol I think it's wrong sorry.

Step-by-step explanation:

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3 years ago

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Point E is drawn on the graph so that the line EF is parallel to line CD

hoa [83]

What is the question though?? and what are the answer choices??

3 0

3 years ago

i need help with this equation please there are two more possible answers that were cut off they are 17,2% and 19,5%

Margarita [4]

Consider that the experimental probability of an event is based upon the previous trials and observations of the experiment.

The experimental probability of occurrence of an event is given by,

$\text{Probability of an event}=\frac{\text{ Number of outcomes that favoured the event}}{\text{ Total number of trials or outcomes}}$

As per the problem, there are a total of 1230 trials of rolling a dice.

And the favourable event is getting a 2.

The corresponding experimental probability is calculated as,

$\begin{gathered} P(\text{ getting a 2})=\frac{\text{ No. of times 2 occurred}}{\text{ Total no. of times the dice is thrown}} \\ P(\text{ getting a 2})=\frac{172}{1230} \\ P(\text{ getting a 2})\approx0.13984 \\ P(\text{ getting a 2})\approx13.98\text{ percent} \end{gathered}$

Thus, the required probability is 13.98% approximately.

Theref

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