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WINSTONCH [101]
3 years ago
7

Remy has $150 to spend on party supplies. She spends $75.50 on decorations and various other supplies. She spent the remaining m

oney to order invitations. If she invited 20 people to her party, how much was the cost to print each invitation?
Mathematics
1 answer:
love history [14]3 years ago
7 0

Answer:

the cost to price of each invitation is $3.725

Step-by-step explanation:

Given that

On party supplies she incurred $150

For decorations and various other supplies she incurred $75.50

The remaining amount i.e.

= $150 - $75.50

= $74.50

This amount would be incurred for the invitations

Now she invited 20 people

SO the cost to price of each invitation is

= $74.50 ÷ 20 people

= $3.725

Hence, the cost to price of each invitation is $3.725

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Answer:

<u><em>a) The probability that exactly 4 flights are on time is equal to 0.0313</em></u>

<u><em></em></u>

<u><em>b) The probability that at most 3 flights are on time is equal to 0.0293</em></u>

<u><em></em></u>

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Step-by-step explanation:

The question posted is incomplete. This is the complete question:

<em>United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures. </em>

<em>a) The probability that exactly 4 flights are on time is = </em>

<em>b) The probability that at most 3 flights are on time is = </em>

<em>c)The probability that at least 8 flights are on time is =</em>

<h2>Solution to the problem</h2>

<u><em>a) Probability that exactly 4 flights are on time</em></u>

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

           P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}

Where _nC_x is the number of different combinations of x success in n trials.

            _nC_x=\frac{x!}{n!(n-x)!}

Hence,

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                                _9C_4=\frac{4!}{9!(9-4)!}=126

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The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

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