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Alex73 [517]
3 years ago
6

Find the derivative dy, dx for y equals the quotient of the quantity x cubed minus 5 times x and the quantity x squared minus 1.

. Did I do this right.

Mathematics
1 answer:
Elza [17]3 years ago
7 0
Using the quotient rule  

dy/dx  =  (x^-1) * (3x^2 -5 ) - (x^3 - 5x) * 2x
              ----------------------------------------------
                              (x^2 - 1)^2

You are correct
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Which equation represents this statement?
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2n - 10 = 12

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Use the provided ruler to measure the segment shown. Find the scale of the drawing.
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3 years ago
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds,What perc
soldier1979 [14.2K]

Answer:

57.62% of players weigh between 180 and 220 pounds

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

\mu = 200, \sigma = 25

What percent of players weigh between 180 and 220 pounds

We have to find the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 180.

X = 220

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

Z = \frac{220 - 200}{25}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881

X = 180

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

Z = \frac{180 - 200}{25}

Z = -0.8

Z = -0.8 has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% of players weigh between 180 and 220 pounds

4 0
3 years ago
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