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3 years ago

Solve for x. 3X - 4 3X - 8

Mathematics

1 answer:

Tems11 [23]3 years ago

8 0

Answer:

x = 2

As x = 12/6 = 2

Step-by-step explanation:

3x-4+3x-8

3x+3x-4-8

= 6x-12

FACTORISING for check and set to zero

6(x-2)

6x -12 = 0

Cancel out left by adding 12 to both sides then applying /6 to both sides to cancel 6 left side.

6x--12 + 12 = 0 + 12

6x / 6 = 12 / 6

x = 12/6

x = 2

or balance as grouping by separation = multiply at end by -1

3x+3x = --4 - 8

6x = - 12

x = - 12/6

x = - 2

x = - 2 ( -1 )

x = 2

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3 years ago

6x-14+10y+22x Simplify by combining like terms I NEED HELP ASAP

Kitty [74]

Hello there!

Answer: ⇒⇒⇒=28x+10y-14

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Step-by-step explanation:

First you had to used group like terms.

$6x-14+10y+22x$

Group like terms:

↓↓↓↓↓↓↓

$6x+22x+10y-14$

Then you add similar elements.

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Hope this helps!

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7 0

3 years ago

Read 2 more answers

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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