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kifflom [539]
2 years ago
11

Y=-5/8x+3/2 in standard form

Mathematics
2 answers:
Pani-rosa [81]2 years ago
8 0

Answer:

y=0.625

Step-by-step explanation:

ladessa [460]2 years ago
8 0

Answer:

Sorry about the way it is, I don't know how to fix it But I hope you can try and understand it.

Step-by-step explanation:

y

=

3

8

x

−

3

4

in standard form is

3

x

−

8

y

=

6

.

Explanation:

Standard form for a linear equation is

A

x

+

B

y

=

C

.

y

=

3

8

x

−

3

4

Simplify

3

8

x

to

3

x

8

.

Subtract

3

x

8

from both sides of the equation.

−

3

x

8

+

y

=

−

3

4

Multiply both sides by

−

1

.

−

3

x

8

(

−

1

)

+

y

(

−

1

)

=

−

3

4

(

−

1

)

=

3

x

8

−

y

=

3

4

Multiply both sides by

8

.

(

3

x

)

⋅

8

8

−

y

(

8

)

=

(

3

)

⋅

8

4

=

(

3

x

)

⋅

8

1

8

1

−

y

(

8

)

=

(

3

)

⋅

8

2

4

1

=

3

x

−

8

y

=

6

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Step-by-step explanation:

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Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number o
Ksenya-84 [330]

Answer:

Step-by-step explanation:

Step1:

We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t

Step2:

Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “”

The probability mass function of poisson distribution is given by

P(X) = , x = 0,1,2,3,...,n.

Where, μ(mean of the poisson distribution)

a).

Given that time period t = 1hr.

Then,μ = 8t

             = 8(1)

             = 8

Now,

The probability that exactly 6 small aircraft arrive during a 1-hour period is given by

P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6)

Consider,

P(X = 6) =  

              =  

              =  

              = 0.1219.

Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.

1).P(At least 6) = P(X 6)

Consider,

P(X 6) = 1 - P(X5)

                = 1 - {+++++}

                = 1 - (){+++++}

                = 1 - (0.000335){+++++}

                = 1 - (0.000335){1+8+32+85.34+170.67+273.07}

                = 1 - (0.000335){570.08}

                = 1 - 0.1909

                = 0.8090.

Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090.

2).P(At least 10) = P(X 10)

Consider,

P(X 10) = 1 - P(X9)

                 = 1 - {+++++

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3 years ago
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Arisa [49]

Answer:

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Step-by-step explanation:

Since the winner gets the larger amount of the two slips picked, random variable 'w' which is the amount awarded could be $1 or $10 or $25.

The probability distribution of 'w' is the values that the statistic takes on. Which could be: $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $10,$1  $10,$1  $10,$10  $10,$25  $10,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25

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