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ANEK [815]
3 years ago
13

Bob rolls a fair six-sided die each morning. If Bob rolls a composite number, he eats sweetened cereal. If he rolls a prime numb

er, he eats unsweetened cereal. If he rolls a 1, then he rolls again. In a non-leap year, what is the expected number of times Bob will roll his die?
Mathematics
1 answer:
iragen [17]3 years ago
8 0

Answer:

He will roll his dice 365 days,

The expected number of days he will roll composite numbers (and eat sweetened cereal) is 146 days.

The expected number of days he will roll prime numbers (and eat unsweetened cereal) is 219 numbers.

Step-by-step explanation:

Bob rolls a fair six-sided dice each morning.

If he rolls a composite number, he eats sweetened cereal, if he rolls a prime number, he eats unsweetened cereal. If he rolls a 1, he rolls again.

We can conclude that when he gets 1 it doesn't really matter for the calculation of the probabilities since he will roll again. Therefore, our possible outcomes are 2, 3, 4, 5 and 6

  • The composite numbers we have are: 4 and 6.
  • The prime numbers we have are: 2, 3 and 5.

So out of the 5 outcomes, 40% (2/5) are composite numbers, and 60% (3/5) are prime numbers.

So in general, we would expect that Bob rolls composite numbers 40% of the days and prime numbers 60% of the days.

A non-leap year has 365 days, we expect him to roll composite numbers 40% of the time this will give us:

365 x 0.40 = 146 days.

We expect him to roll prime numbers 60% of the days:

365 x 0.60 = 219 numbers.

Therefore, he will roll his dice 365 dies and the expected number of days he will roll composite numbers (and eat sweetened cereal) is 146 days. The expected number of days he will roll prime numbers (and eat unsweetened cereal) is 219 numbers.

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Suppose that y varies directly as the square of x, and that y=567 when x=9. What is y when x=4
Gwar [14]

Answer:

y  = 112

Step-by-step explanation:

If  y varies directly as the square of x

mathematically;

y ∝ x²

y = kx² where 'k' is the constant of proportionality

let u know the value of k for this equation by making k the subject of the formular from y = kx²

divide both sides by x²

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6 0
3 years ago
Consider an urn containing 8 white balls, 7 red balls and 5 black balls.
weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053

2) the probability of getting 2 white balls is equal to:

=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465

3) The probability of getting at least 72 white balls is:

=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72}  \times \left( \frac{7}{20} \right)^{78}  +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73}  \times \left( \frac{7}{20} \right)^{77}  + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149}  \times \left( \frac{7}{20} \right)^{1}  +\left( \frac{8}{20} \right)^{150}

=\sum^{150}_{k=72} [C^{k}_{150}\times  \left( \frac{8}{15} \right)^{k}  \times \left( \frac{7}{15} \right)^{150-k}]

5 0
1 year ago
Read 2 more answers
PLS HELP ASAP!!!!!
zmey [24]
B. (7,-2) is the answer
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2 years ago
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