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Galina-37 [17]
3 years ago
5

1. *

Mathematics
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

I am not so clear about this question

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-x+y=4 6x+y=-3<br> what are the coordinates
goldenfox [79]

Answer:

6x-y-4=0 is the linear coordinates.

Step-by-step explanation:

hope this helps.

8 0
2 years ago
A mobile auto-detailing company pays you $12.75 an hour. Last summer you worked 35.25 hours a week. What was your total straight
masya89 [10]

multiply the hourly rate by number of hours worked

12.75 x 35.25 = 449.44 per week

449.44 x 2 = 898.88 for two weeks

5 0
3 years ago
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A certain Bookstore that sells A Million books wants to hire you, and you get the job if you can answer these problems correctly
Maslowich

Answer:

a) P ( X = 2 ) = 0.23028

b) P ( X < 4 ) = 0.95689

c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292

Step-by-step explanation:

Given:-

- The parameter for the poisson distribution is given, λ = 1.3.

- Declare a random variable (X) which is the number of books sold in the next minute:

                               X ~ Po (1.3)

Find:-

a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

Solution:-

a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:

                       P(X = x ) =\frac{ (lambda)^k e^(^-^l^a^m^b^d^a^)}{k!}\\\\P(X = x )  =\frac{ (1.3)^k e^(^-^1^.^3^)}{k!}

Where, "k" is the number of books sold in next minute.

- Now compute P ( X = 2 ) :

                       P(X = 2 )  =\frac{ (1.3)^2 e^(^-^1^.^3^)}{2!}\\\\P(X = 2 )  = 0.23028    

b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:

                      P(X < 4 )  = P ( X = 0) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )\\\\P(X < 4 )  = \frac{ (1.3)^0 e^(^-^1^.^3^)}{0!}+ \frac{ (1.3)^1 e^(^-^1^.^3^)}{1!}+ \frac{ (1.3)^2 e^(^-^1^.^3^)}{2!} + \frac{ (1.3)^3 e^(^-^1^.^3^)}{3!}\\\\P(X < 4 )  = 0.27253 + 0.35429 + 0.23028 + 0.09979 = 0.95689

c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:

                   P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )

                                               = P ( X ≥ 3 ) / P (X ≥ 2 )

                                               = P ( X > 2 ) / P ( X > 1 )

                                               = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]  

- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:

                     P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028

                                      = 0.8571

                     P ( X ≤ 1 ) = 0.27253 + 0.35429

                                      = 0.62682

- Use the expression developed for conditional probability to evaluate the required probability:

                     P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

                                                  = [ 1 - 0.8571 ] / [ 1 - 0.62682 ]

                                                  = 0.38292

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How to find the distance between these points?
irakobra [83]

2/5 (rise/run) is the distance. You move up by 2 and 5 to the right

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