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3 years ago

(-1) + 4x + 2 + 2x + x Pls provide an explination.....

Mathematics

2 answers:

Elodia [21]3 years ago

7 0

Answer:

7x + 1

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Damm [24]3 years ago

5 0

Answer:

7x + 1

Step-by-step explanation:

Basically, we must solve the equation. Lets do it!

Apparently, we have constants and variables, so our answer will come in (variables + constants)

(-1) + 4x + 2 + 2x + x

=> 7x + 1

Our answer is 7x + 1.

I hoped this helped.

Please give me brainliest :D

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3 years ago

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2 years ago

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3 years ago

Triangle A″B″C″ is formed using the translation (x + 2, y + 0) and the dilation by a scale factor of one half from the origin. W

Travka [436]

Complete question:

Triangle A″B″C″ is formed using the translation (x + 2, y + 0) and the dilation by a scale factor of one half from the origin. Which equation explains the relationship between segment AB and segment A double prime B double prime?

A) segment a double prime b double prime = segment ab over 2

B) segment ab = segment a double prime b double prime over 2

C) segment ab over segment a double prime b double prime = one half

D) segment a double prime b double prime over segment ab = 2

Answer:

A) segment a double prime b double prime = segment ab over 2.

It can be rewritten as:

$A"B" = \frac{AB}{2}$

Step-by-step explanation:

Here, we are given triangle A″B″C which was formed using the translation (x + 2, y + 0) and the dilation by a scale factor of one half from the origin.

We know segment A"B" equals segment AB multiplied by the scale factor.

A"B" = AB * s.f.

Since we are given a scale factor of ½

Therefore,

$A"B" = AB * \frac{1}{2}$

$A"B" = \frac{AB}{2}$

The equation that explains the relationship between segment AB and segment A"B" is

$A"B" = \frac{AB}{2}$

Option A is correct

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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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4 years ago