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Tems11 [23]
3 years ago
6

Park Hyatt Philadelphia at the Bellevue, located at Walnut and Broad in downtown Philadelphia has a capacity of 240 king rooms.

Customers of Hyatt are typically either leisure travelers or business customers. Hyatt charges a discount fare of $125 for a midweek stay (but requires booking a week in advance) which contrast the regular fare of $275. Typically, business customers book in the last minute, and are willing to pay the regular fare, if they can be guaranteed accommodation. Suppose we are interested in the bookings in Park Hyatt on August 6th (the day of our final exam). Hyatt knows that there are plenty of leisure travelers, willing to pay the low fares. However, all else being equal, Hyatt would like to fill those rooms with the high-fare travelers. The objective of Hyatt is to maximize the sum of revenue from both sections of the travelers. If Hyatt followed the ‘booking limit policy’, by reserving some rooms for last-minute business customers, how many rooms should it reserve? Assume that there is ample demand of leisure customers willing to pay the discount fare, and the number of business customers is normally distributed, with mean 50 and standard deviation 26. (6 points)
Mathematics
1 answer:
Reika [66]3 years ago
7 0

Answer:

X=53

Step-by-step explanation:

From the question we are told that:

Regular fare R= $275

Discount fare of $125

Mean \=x =50

Standard deviation \sigma= 26.

Generally, the equation for Critical Fraction is mathematically given by

C=\frac{Pf-Pd}{Pf}

C=\frac{275-125}{275}

C=0.5

From Z Distribution Table

Z=0.1131

Therefore

Reservation  made is give as for High fare travellers is

X = \=x+(z* \sigma)

X = 50 + (0.1131 * 26)

X=53

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Answer:

16 and 7

Step-by-step explanation:

48/3 = 16.   16 = 16/1

56/8 = 7.     56/8 = 7/1

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3 years ago
How do I give an example of a positive and negative tempature that has a different of 5 degrees
love history [14]

Answer:

You could say +3 and -2 or +4 and -1

Step-by-step explanation:For example if you draw a number line

count 1 hump between each line.   it equals 5.

     |_______|_______|_______|_______|_______|

    -2            -1               0              1              2              3

4 0
3 years ago
The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine
salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

                                              f(x)=\frac{e^{\frac{-x}{1000} }}{1000}

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

              b. P(1000<x<2000) = 0.2325

              c. P(x<1000) = 0.6321

              d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = \int\limits^b_a {P(x)} \, dx

Then, for the electronic component, probability will be:

P(a<x<b) = \int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx

P(a<x<b) = \frac{1000}{1000}.e^{\frac{-x}{1000} }

P(a<x<b) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

a. For a component to last more than 3000 hours:

P(3000<x<∞) = e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = e^{-3}

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}

P(1000<x<2000) = e^{-2}-e^{-1}

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}

P(0<x<1000) = e^{-1}-1

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

0.1 = 1-e^\frac{-x}{1000}

-e^{\frac{-x}{1000} }=-0.9

{\frac{-x}{1000} }=ln0.9

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

5 0
4 years ago
Please answer this question, will give brainliest!
Andrej [43]

Answer:

6.63 cm

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The segments within the circle form a right angle. Triangle CRS as a right triangle must follow the Pythagorean theorem which says the square of each leg adds to the square of the hypotenuse.

a² + b² = c²

Here a is unknown, b = 10 and c = 12.

a² + 10² = 12²

a² + 100 = 144

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8 0
3 years ago
Bill reads 1/5 of a book on Monday he reads 2/3 of the book on Tuesday if he finishes reading the book on Wednesday or fraction
ratelena [41]

Answer:

\frac{2}{15}

Step-by-step explanation:      

We have been given that Bill reads 1/5 of a book on Monday he reads 2/3 of the book on Tuesday.

Let x be the part of book that Bill read on Wednesday. We can find x by equating the sum of parts of book read on each day by 1.  

\frac{1}{5}+\frac{2}{3}+x=1

First of all we will find a common denominator.

\frac{1\times 3}{5\times 3}+\frac{2\times 5}{3\times 5}+x=1

\frac{3}{15}+\frac{10}{15}+x=1

\frac{3+10}{15}+x=1

\frac{13}{15}+x=1

x=1-\frac{13}{15}

x=\frac{15-13}{15}  

x=\frac{2}{15}

Therefore, Bill read \frac{2}{15} of the book on Wednesday.  

7 0
3 years ago
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