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mote1985 [20]
3 years ago
12

How do I find an exactions for a graph

Mathematics
1 answer:
butalik [34]3 years ago
8 0
3x-6 this is the answer because it rose 3 and the Y intercept is -6
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Does the point (0, 0) satisfy the equation y = 9x?
miskamm [114]

Answer:

yes it does

Step-by-step explanation:

because the equation y=9x does not have a y-intercept (all slopes come in the form y=mx+b -- it can be written differently though) and since there is no 'b' that means the y-intercept is 0. So whenever there is no y-intercept, the slope starts at 0.

6 0
2 years ago
Which answer choice ?
Snezhnost [94]

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the fourth one

Step-by-step explanation:

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I need to find the exterior angle
Zina [86]

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Step-by-step explanation:

8 0
3 years ago
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

8 0
3 years ago
Can u pls help with this question ​
Alinara [238K]

Answer:

noooooooooooooooooooooooooooooooooooooooooooo

Step-by-step explanation:

7 0
2 years ago
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