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3 years ago

a parabola opening up and down has vertex (-1, 0) and passes through (-8, -49/8). Write its equation in vertex form.

Mathematics

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

Step-by-step explanation:

hello :

here is an solution

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What are effective elements to include in introductions? Check all that apply.

dlinn [17]

The effective elements to include in the introductions are statement of the topic, a relevant anecdote and a surprising fact about the topic. Therefore the option 1st, 2nd and 6th options are correct.

<h3>What is Introduction? </h3>

Introduction is the starting line or the opening hook or portion that attracts the readers attention which carries the relevant background of the information.

It is the statement which represents the main point of the following argument or the paragraphs.

There are many elements that included in the introductory part some of them are as follows:-

a statement of the topic
a relevant anecdote or tale
a surprising fact about the topic

Therefore the option 1st, 2nd and 6th options are correct.

Learn more about Introduction here:

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2 years ago

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I need help please help me

Leviafan [203]

Is that the Algebra work/practice book? If so, go into algebra nation and they have videos where they go step by step through the questions. Hope this helps?

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3 years ago

What the heck is this?

max2010maxim [7]

Answer:

(2,-1)

Step-by-step explanation:

(-2+6)/2 (2+4)/2

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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.

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

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3 years ago

2nd grade represent even numbers

julsineya [31]

Can you please tell me your question?

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2 years ago

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