85 EN PHYSIQUE Fonctionnement de l'ail

kicyunya [14]

Answer:

8.69% probability that at least 285 rooms will be occupied.

Step-by-step explanation:

For each booked hotel room, there are only two possible outcomes. Either there is a cancelation, or there is not. So we use concepts of the binomial probability distribution to solve this question.

However, we are working with a big sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

A popular resort hotel has 300 rooms and is usually fully booked. This means that $n = 300$

About 7% of the time a reservation is canceled before the 6:00 p.m. deadline with no pen-alty. What is the probability that at least 285 rooms will be occupied?

Here a success is a reservation not being canceled. There is a 7% probability that a reservation is canceled, and a 100 - 7 = 93% probability that a reservation is not canceled, that is, a room is occupied. So we use $p = 0.93$

Approximating the binomial to the normal.

$E(X) = np = 300*0.93 = 279$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.93*0.07} = 4.42$

The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{285- 279}{4.42}$

$Z = 1.36$

$Z = 1.36$ has a pvalue of 0.9131.

So there is a 1-0.9131 = 0.0869 = 8.69% probability that at least 285 rooms will be occupied.

8 0

2 years ago

What is the value of this expression when a=4, b=-5, and c=-7?

nasty-shy [4]

Answer: C. 6 1/6

Step-by-step explanation:

2 * -5 * -7 = -10 * -7 = 70

70 + 4 = 74

3 * 4 = 12

74/12 = 6 1/6

4 0

3 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$