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3 years ago

Which equation represents the line through the points (3,2) and (-9,6) ?

Mathematics

1 answer:

natima [27]3 years ago

7 0

Answer:

Step-by-step explanation:

(3,2),(-9,6)

slope = (6 - 2) / (-9 - 3) = -4/12 = -1/3

y = mx + b

slope(m) = -1/3

use either of ur points...(3,2)....x = 3 and y = 2

now we sub into the formula and find b, the y int

2 = -1/3(3) + b

2 = -1 + b

2 + 1 = b

3 = b

so ur equation is : y = -1/3x + 3 <==

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Hi I need to know the slope(m) before 11:00 I know the y-intercept(b) is -3

Alexxandr [17]

Answer:

slope is 2/3

Step-by-step explanation:

y=mx + b where m is the slope

6 0

3 years ago

PLEASE HELP!!

lidiya [134]

Divide f(x) over g(x). To simplify, factor everything as much as possible and cancel out any common terms. In this case "4x+3" is a common term that shows up in the numerator and denominator.

(f/g)(x) = [ f(x) ]/[ g(x) ]
(f/g)(x) = (16x + 12)/(4x + 3)
(f/g)(x) = (4(4x+3))/(4x+3)
(f/g)(x) = 4

So the answer is choice C

3 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$