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3 years ago

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Tess works at a sandwich shop. When assembling their turkey sandwich, they are instructed to use 5 pieces of sliced turkey and 2

pieces of cheese. If Tess is assembling a party platter of these turkey sandwiches and uses 12 pieces of cheese, how many slices of turkey will she use?

1 answer:

Romashka-Z-Leto [24]3 years ago

6 0

12/2x5=30 30 slices of turkey.

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We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

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$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

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$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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$\mu = 0.6\frac{n}{60} = 0.01n$

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This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

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$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

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