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2 years ago

Sorry it’s so large u can zoom in

Mathematics

1 answer:

nataly862011 [7]2 years ago

6 0

Answer:

15 feet

Step-by-step explanation:

dyscshdududhsgs. don't mind that

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The vertical _____ of a function secant are determined by the points that are not in the domain.

shusha [124]

The vertical asymptotes of a function secant are determined by the points that are not in domain.
Thank you.

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2 years ago

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Answer choices: 16/3 40/9 4/9 4/3

son4ous [18]

Answer:

0.44444444 or 4/9

Step-by-step explanation:

Since we know the value of $x$ , we can rewrite the equation replacing the variables.

$\frac{2}{3}^{2} + 3*\frac{2}{3} - 2$

Now solve. While solving, remember the order of operations.

$\frac{2}{3}^{2} + 3*\frac{2}{3} - 2 = 0.44444444$ or $\frac{4}{9}$

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2 years ago

Find the x-intercept of the graph of the linear equation y = 2x + 4.

Tamiku [17]

Answer:

The x-intercept is -2

Step-by-step explanation:

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3 years ago

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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.

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

5 0

3 years ago

The diagonals of a quadrilateral QRST intersect at P(-1,3). QRST has vertices at Q(3,6) and R(-4,5). What must be the coordinate

erica [24]

S = (-5,0)

T = (2,1)

Step-by-step explanation:

Step 1 :

Given

Q = (3,6) and R = (-4,5). P = (-1,3)

Let S be (a,b) and T be (c,d)

The diagonals of a parallelogram bisect each other. so in order to ensure that QRST is a parallelogram, P must be the mid point of the diagonals QS and RT.

Step 2 :

P is the midpoint of QS

So we have (3+a) ÷ 2 = -1 and (6 + b) ÷ 2 = 3

=> 3 + a = -2 and 6 + b = 6

=> a = -5 and b =0

So S should be (-5,0)

Step 3 :

P is the midpoint of RT

So we have (-4+c) ÷ 2 = -1 and (5 + d) ÷ 2 = 3

=> -4+ c = -2 and 5 + d = 6

=> c = 2 and d =1

So T should be (2,1)

Step 4 :

Answer :

S = (-5,0)

T = (2,1)

6 0

2 years ago