What equation results from completing the square and then factoring?

Mathematics

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:

$\large\boxed{D.\ (x-4)^2=55}$

Step-by-step explanation:

$(a-b)^2=a^2-2ab+b^2\qquad(*)\\\\\\x^2-8x=39\\\\x^2-2(x)(4)=39\qquad\text{add}\ 4^2=16\ \text{to both sides}\\\\\underbrace{x^2-2(x)(4)+4^2}_{(*)}=39+16\\\\(x-4)^2=55$

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A teacher gives a test to a large group of students. The results are closely approximated by a normal curve. Teh mean is 73, wit

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The bottom for an A is 75. Round to the nearest whole number as needed.

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the scores on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=73,\sigma=7)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

2) The bottom for an A is _____. Round to the nearest whole number as needed.

And we want the top 5.5% of the scores, so we need a value a such that:

$P(X>a)=0.055$ or $P(X$

We need on the right tail of the distribution a value a that gives to us 94.5% of the area below and 5.5% of the area above. Both conditions are equivalent.

Let's use the condition $P(X$ , the best way to solve this problem is using the z score with the following formula:

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

So we need a value from the normal standard distribution that accumulates 0.945 of the area on the left and 0.055 on the right. This value on this case is 1.598 and we can founded with the following code in excel:

"=NORM.INV(0.945,0,1)"

If we apply the z score formula to our case we have this:

$P(X$