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nadya68 [22]
2 years ago
10

At a hockey game, a vendor sold a combined total of 148 sodas and hot dogs. The number of sodas sold was three times the number

of hot dogs sold. Find the number of sodas and the number of hot dogs sold.
Mathematics
2 answers:
tatyana61 [14]2 years ago
5 0
111 sodas
37 hotdogs
148 = 3x (sodas) + x (hotdogs)
AleksandrR [38]2 years ago
3 0

Answer:

37 hot dogs and 111 sodas

Step-by-step explanation:

148 divided by 4= 37

148 - 37= 111

if the number of sodas was three times it would have to be broken up into three groups making the hot dogs 37 and sodas three times the amout would be 111. hope this helped sorry for any spelling mistakes

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A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o
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Answer:

a) 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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Step-by-step explanation:

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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:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial probability distribution

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.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

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This probability is:

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

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(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that n = 6

Each of these days, 0.6592 probability of no accident on this stretch, which means that p = 0.6592.

This probability is P(X = 2). So

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This means that n = 80. So

\mu = 0.01(80) = 0.8

The probability of no damaging accidents is P(X = 0). So

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

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

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