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4 years ago

9

The cost of a party is $550. The price per person depends on how many people attend the party. Write an expression for the price

per person if p people attend the party. Then find the price per person if 25, 50, an 55 people attend the party.

1 answer:

Alexxandr [17]4 years ago

8 0

x= 550/p 25p=22 50p=11 55p=10

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HELP!!!!! <br> answer asap! this is due in two hours!!

lapo4ka [179]

Answer:

it is 56

Step-by-step explanation:

there are 4 sides and they are all 10 and if you add them they are 40 and if you do the bottom it is 16 and 40 + 16 is 56 hope it helps

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3 years ago

第 5 个问题 a multiple choice exam has 10 questions. each question has 3 possible answers, of which one is correct. a student knows

tatuchka [14]

The chances that the student was merely guessing is 1/3.

Bayes Theorem determines the conditional probability of an event A given that event B has already occurred.

denoted by

$P(A/B)=\frac{P(A)*P(B/A)}{P(B)}$

let A be the event that the student knows the answer .

B be the event that the student does not knows the answer .

and

E be the event he gets answer correct .

According to the given question

$P(A)=\frac{4}{10} \\\\ P(B)=1-\frac{4}{10} =\frac{6}{10}$

Probability that the answer is correct ,given that he knows the answer is

$P(E/A)=1$

Probability that the answer is correct ,given that he guesses it is

$P(E/B)=\frac{1}{3}$ [as the MCQ has 3 options and only one is correct]

We need to find the probability that he guesses the answer given that it is correct.

Required probability $P(B/E)=\frac{P(B)*P(E/B)}{P(A)*P(E/A)+P(B)*P(E/B)}$

Substituting the values we get

$P(B/E)=\frac{\frac{6}{10} *\frac{1}{3} }{\frac{4}{10} *1+\frac{6}{10} *\frac{1}{3} }$

$=\frac{6}{30}*\frac{30}{18} \\ \\ =\frac{6}{18} \\ \\ =\frac{1}{3}$

Therefore , the chances that the student was merely guessing is 1/3.

Learn more about Probability here brainly.com/question/13140147

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1 year ago

How do you do that I don’t understand

Nataliya [291]

I think the distance is 8 1/8 feet

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3 years ago

B) How much mg can the child receive per dose? A: 531.82 mg

mars1129 [50]

A.) 531.82
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4 0

3 years ago

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The

Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$ , z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$ , so $Z = 1.405$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

6 0

3 years ago