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omeli [17]
4 years ago
9

Simplify the expression state any excluded values 2a^2-4a+2 --------------- 3a^2-3

Mathematics
1 answer:
chubhunter [2.5K]4 years ago
8 0

Answer:

The simplified form is \dfrac{2(x-1)}{3(x+1)}.

x =1 is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

\dfrac{2a^2-4a+2}{3a^2-3}

Let us simplify the numerator and denominator separately.

The numerator is given as 2a^2-4a+2

2 is a common factor in all the three terms. So, we factor it out. This gives,

=2(a^2-2a+1)

Now, a^2-2a+1=(a-1)(a-1)

Therefore, the numerator becomes 2(a-1)(a-1)

The denominator is given as: 3a^2-3

Factoring out 3, we get

3(a^2-1)

Now, a^2-1 is of the form a^2-b^2=(a-b)(a+b)

So, a^2-1=(a-1)(a+1)

Therefore, the denominator becomes 3(a-1)(a+1)

Now, the given expression is simplified to:

\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}

There is (x-1) in the numerator and denominator. We can cancel them only if x\ne1 as for x=1, the given expression is undefined.

Now, cancelling the like terms considering x\ne1, we get:

\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}

Therefore, the simplified form is \dfrac{2(x-1)}{3(x+1)}

The simplification is true only if  x\ne1. So, x =1 is the excluded value for the given expression.

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Answer:

The first quartile of the data set represented by the box plot is 12

The median of the data set is 16

Step-by-step explanation:

The computation is shown below:

Quartile means the third digit form the starting

While on the other hand the median represent the mid value of the data set that is shown in the inner side of the box plot as on the vertical line

So, the median of the data set is 16

The same is to be considered

Kindly find the attachment below:

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Step-by-step explanation:

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3 years ago
The lengths of lumber a machine cuts are normally distributed with a mean of 106 inches and a standard deviation of 0.3 inch. ​(
Vinvika [58]

Answer:

a) P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)

And we can find this probability with the complement rule:

P(z>0.37)=1-P(z

b) z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}=  2.431

And if we use the z score we got:

P(z>2.431) =1-P(z

Step-by-step explanation:

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

X \sim N(106,0.3)  

Where \mu=106 and \sigma=0.3

Part a

We are interested on this probability

P(X>106.11)

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And using this formula we got:

P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)

And we can find this probability with the complement rule:

P(z>0.37)=1-P(z

Part b

For this case we select a sample of n =44 and the new z score formula is given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score we got:

z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}=  2.431

And if we use the z score we got:

P(z>2.431) =1-P(z

6 0
3 years ago
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