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earnstyle [38]
2 years ago
10

the markup on a restaurant meal is 250%. A meal costs $7.30 to produce. How much will the customer be charged before tax and tip

?
Mathematics
2 answers:
monitta2 years ago
5 0
2.5 times 7.3=18.25
it costs $18.25
Simora [160]2 years ago
4 0

Answer:  The required amount of money that a customer be charged is $25.55.

Step-by-step explanation:  Given that the markup on a restaurant meal is 250% and a meal costs $7.30 to produce.

We are to find the amount of money that a customer be charged before tax and tip.

The markup price on the meal is given by

m_p\\\\=250\%\times7.30\\\\=\dfrac{250}{100}\times7.30\\\\=\dfrac{5}{2}\times7.30\\\\=18.25.

Therefore, the amount of money that a customer be charged before tax and tip is

\$(7.30+18.25)=\$25.55.

Thus, the required amount of money that a customer be charged is $25.55.

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Answer:

a) p=0.2

b) probability of passing is 0.01696

.

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}.

Recall that \binom{n}{k}= \frac{n!}{k!(n-k)!} In this case, the student passes if X is at least four correct questions, then

P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696

c)The expected value of a binomial random variable with parameters n and p is E[X] = np. IN our case, n=6 and p =0.2. Then the expected value of correct answers is 6\cdot 0.2 = 1.2

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3 years ago
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