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vova2212 [387]
3 years ago
10

Ten years ago in Louisiana, schools averaged 182 pupils for every 10 teachers. Write a ratio for this problem in 3 different way

s.
Mathematics
1 answer:
Anettt [7]3 years ago
6 0
182:10

182/10


182 to 10
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Support requests arrive at a software company at the rate of 1 every 30 minutes. Assume that the requests arrive as events in a
rewona [7]

Answer:

a. 0.5413

b. 20

c. 0.3724

d. 4.4721

Step-by-step explanation:

Solution:-

- We will start by defining a random variable X.

           

                     X : The number of support requests arrived

- The event defined by the random variable ( X ) is assumed to follow Poisson distribution. This means the number of request in two distinct time intervals are independent from one another. Also the probability of success is linear within a time interval.

- The time interval is basically the time required for a poisson event to occur. Consequently, each distributions is defined by its parameter(s).

- Poisson distribution is defined by " Rate at which the event occurs " - ( λ ). So in our case the rate at which a support request arrives in a defined time interval. We define our distributions as follows:

                                   X ~ Po ( λ )

                                 

Where,                        λ = 1 / 30 mins

Hence,

                                   X ~ Po ( 1/30 )

a)

- We see that the time interval for events has been expanded from 30 minutes to 1 hour. However, the rate ( λ ) is given per 30 mins. In such cases we utilize the second property of Poisson distribution i.e the probability of occurrence is proportional within a time interval. Then we scale the given rate to a larger time interval as follows:

                                   λ* =  \frac{1}{\frac{1}{2} hr} = \frac{2}{1hr}

- We redefine our distribution as follows:

                                   X ~ Po ( 2/1 hr )

- Next we utilize the probability density function for poisson process and accumulate the probability for 2 to 4 request in an hour.

                           P ( X = x ) = \frac{e^-^l^a^m^b^d^a . lambda^x}{x!}

- The required probability is:

                   P ( 2 \leq X \leq 4 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )\\\\P ( 2 \leq X \leq 4 ) = \frac{e^-^2 . 2^2}{2!} +  \frac{e^-^2 . 2^3}{3!} + \frac{e^-^2 . 2^4}{4!}\\\\P ( 2 \leq X \leq 4 ) = 0.27067 + 0.18044 + 0.09022\\\\P ( 2 \leq X \leq 4 ) = 0.5413            Answer    

b)

We will repeat the process we did in the previous part and scale the poisson parameter ( λ ) to a 10 hour work interval as follows:

                               λ* = \frac{2}{1 hr} * \frac{10}{10} = \frac{20}{10 hr}

- The expected value of the poisson distribution is given as:

                             E ( X ) = λ

Hence,

                            E ( X ) = 20  (10 hour work day)    .... Answer

c)

- We redefine our distribution as follows:

                                   X ~ Po ( 20/10 hr )

- Next we utilize the probability density function for poisson process and accumulate the probability for 20 to 24 request in an 10 hour work day.

                           P ( X = x ) = \frac{e^-^l^a^m^b^d^a . lambda^x}{x!}

- The required probability is:

                   P ( 20 \leq X \leq 24 ) = P ( X = 20 ) + P ( X = 21 ) + P ( X = 22 )+P ( X = 23 ) + P ( X = 24 )\\\\P ( 20 \leq X \leq 24 ) = \frac{e^-^2^0 . 20^2^0}{20!} +  \frac{e^-^2^0 . 20^2^1}{21!} + \frac{e^-^2^0 . 20^2^2}{22!} + \frac{e^-^2^0 . 20^2^3}{23!} + \frac{e^-^2^0 . 20^2^4}{24!} \\\\P ( 20 \leq X \leq 24 ) = 0.0883 +0.08460 +0.07691 +0.06688+0.05573\\\\P ( 20 \leq X \leq 24 ) = 0.3724            Answer  

c)

The standard deviation of the poisson process is determined from the application of Poisson Limit theorem. I.e Normal approximation of Poisson distribution. The results are:

                                σ = √λ

                                σ = √20

                                σ = 4.4721 ... Answer

8 0
3 years ago
Will give brainlist if right! Number 13 and 14 plz
Gelneren [198K]

Answer:

13 - 14

14 - base area 30 and height 19

base area 285 and height 2

Step-by-step explanation:

1428/17/6=14

5 0
3 years ago
PLEASE HELP! Geometry
inysia [295]
The answer is b and I also need 20 characters so here are random words
7 0
3 years ago
The Doe family had a rectangular garden with a 58-foot perimeter. They reduced the length by 7 feet and decreased the width to b
Vaselesa [24]

Answer:

60 square foot

Step-by-step explanation:

Perimeter of a rectangular garden = 58 foot

Let x and y denote length and width of the rectangular garden.

2 (Length + Width) = Perimeter of the rectangular garden

2(x+y)=58\\x+y=\frac{58}{2}\\ x+y=29

So,

length = x

width = y = 29-x

The length is reduced by 7 feet and the width becomes half.

New length = x-7

New width = \frac{1}{2}(29-x)

New perimeter = 34 foot

2[(x-7)+\frac{1}{2}(29-x)]=34\\ 2[2(x-7)+(29-x)]=2(34)\\2(x-7)+(29-x)=34\\2x-14+29-x=34\\2x-x-14+29=34\\x+15=34\\x=34-15\\x=19

So,

New length =x-7=19-7=12\,\,foot

New width = \frac{1}{2}(29-19)=\frac{1}{2}(10)=5\,\,foot

Area of the new smaller garden = New length × New Width

= 12 × 5

= 60 square foot

6 0
4 years ago
If 4+2=26 and 8+1=79 and 6+5=111 what is 7+3=?<br> *Please Show Work*
Natali5045456 [20]
410 ?
Explanation : 4 comes from what you subtract and 10 comes from what the question would be if it was added
7 0
4 years ago
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