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antiseptic1488 [7]
3 years ago
13

What is the equation of line l? y = -x - 1 y = x + 1 y = 3x - 3 y = -3x - 3

Mathematics
2 answers:
ivolga24 [154]3 years ago
5 0

Answer:

y = 3x-3

Step-by-step explanation:

The line has a y intercept of -3   (where it crosses the y axis)

We know the slope is positive since it goes up from left to right

The only answer with a positive slope and a y intercept of -3 i

y = 3x-3

ruslelena [56]3 years ago
3 0

To find the equation of this line, let's first find the y-intercept of the line which is the point where the line crosses the y-axis.

To find the y-intercept, we start at the origin.

We go 0 units to the left or right and 3 units down.

So that's (0, -3).

Now we can confirm that the y-intercept is -3.

So for any of the equations that have a y-intercept that's not -3, we can immediately eliminate them.

So we can eliminate y = -x - 1 and y = x + 1.

Now we have two equations left, y = 3x - 3 and y = -3x - 3.

Now we have to look at our line.

Remember that positive slopes go up and negative slopes go down so we can confirm that the equation won't have a slope of -3.

So therefore, the equation is y = 3x - 3.

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Answer:

1= 75 2= 60 3=180 4=45 5= vertical angles 6=45

Step-by-step explanation:

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1. The probability of telesales representative making a sale on a customer call is 0.15.
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Answer:

1c

 n = 33

1d

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Step-by-step explanation:

From the question we are told that

   The  probability of telesales representative making a sale on a customer call is  p = 0.15

     The mean is  \mu  =  5

Generally the distribution of sales call  made by a  telesales representative follows a binomial distribution  

i.e  

         X  \~ \ \ \  B(n , p)

and the probability distribution function for binomial  distribution is  

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Generally the mean is mathematically represented as

     \mu =  n*  p

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Generally the least number of calls that need to be made by a representative for the  probability of at least 1 sale to exceed 0.95 is mathematically represented as

      P( X \ge 1) = 1 - P( X < 1 ) > 0.95

=>    P( X \ge 1) = 1 - P( X =0 ) > 0.95

=>    P( X \ge 1) = 1 - [ ^{n}C_0 *  (0.15 )^0 *  (1- 0.15)^{n-0}] > 0.95

=>    1 - [1  *  1*  (0.85)^{n}] > 0.95

=>    [(0.85)^{n}] > 0.05

taking natural  log of both sides

n = \frac{ln(0.05)}{ln(0.85)}

=>  n = 19

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