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4 years ago

Factor. 9x2−30xy+25y2

Mathematics

1 answer:

Mnenie [13.5K]4 years ago

5 0

Answer:

The factor form is: $\quad \left(3x-5y\right)^2$

Step-by-step explanation:

When the factor of the expression is required to be found, then take the common terms, or can be done by using the standard formulas, for example, $\left(a-b\right)^2=a^2-2ab+b^2$

Now, here we have to find the factor of the expression:

$9x^2-30xy+25y^2\\=\left(3x\right)^2-2\cdot \:3x\cdot \:5y+\left(5y\right)^2\\$

So now, we can see that this is the perfect square form of:

$\left(a-b\right)^2=a^2-2ab+b^2$

So the given expression is factorized as follows:

$\left(3x\right)^2-2\cdot \:3x\cdot \:5y+\left(5y\right)^2=\left(3x-5y\right)^2\\$

Thus the factored form is:

$9x^2-30xy+25y^2 = \quad \left(3x-5y\right)^2$

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The heights of twelve month old boys are normally distributed with a mean of 29.8 inches and a standard deviation of 1.2 inches.

Triss [41]

Answer:

About twenty-one percent of twelve month old boys are shorter than 28.8 inches.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 29.8, \sigma = 1.2$

About twenty-one percent of twelve month old boys are shorter than what height?

This is the value of X when Z has a pvalue of 0.21. So it is X when Z = -0.805.

$Z = \frac{X - \mu}{\sigma}$

$-0.805 = \frac{X - 29.8}{1.2}$

$X - 29.8 = -0.805*1.2$

$X = 28.8$

About twenty-one percent of twelve month old boys are shorter than 28.8 inches.

4 0

4 years ago

You and your friend race on a trail that is 10 miles long. In each situation, does your friend pass you before the end of the tr

Sliva [168]

For the first example given the answer would be No. you have a 4 mile head start. at the 4 mile line, your friend starts at 8 miles an hour and you start running at 6 miles an hour. basic addition should give the answer. In one hour time, you wouldve ran 6 miles plus the 4 you had as a head start, giving you the 10 miles you needed to reach the finish line. He on the other hand, Biked 8 miles in an hour time. By that time, you had just reached the finish line.

So the answer is no for the first example

For the second example the maths get bit harder. You start at the 5 mile point and you friend starts at the beginning point. You only need 5 miles to win, and your friend needs double (its actually more than double, because if it was perfectly doubled, you would tie the race. Your pace just has to be a bit more than half of his speed. his speed is 17mph. yours, by logic, needs to be even a tad bit more than 8.5mph. You need to have a faster speed than 8.5mph (8.51mph works perfectly) and you win by a hair. But when we se your example, you're only going at 7mph. A whole mile and a half behind pace. Sadly, he passes you short before winning.

The second example is YES he does pass you before the end of the trail.

Hope this helped. I spent around 20 minutes writing this answer, and a 5 star, like and noter as the "BRAINLIEST ANSWER" would be very much apreciated! Have a good day!!