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sergeinik [125]
3 years ago
9

Given the area of a rectangular pool is 6x^3+19x^2+2x-20 and the length of one side of the pool is 2x+5, find the measure of the

missing side.
The missing side length is

Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0

Answer:

The missing side length is 3x^2 + 2x -4

Step-by-step explanation:

Here, we are interested in calculating the length of the missing side.

Mathematically, the area of a rectangle = L * B

So since we have the area and one side, to get the other side, we shall need to divide the area by the side we have ;

Thus, the missing side length will be;

6x^3+19x^2+2x-20/2x+ 5

Please check attachment for the polynomial long division

The value of the other factor after the polynomial long division is 3x*2 + 2x -4

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galben [10]

Answer:

\\ \mu = 118\;grams\;and\;\sigma=30\;grams

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find \\ \mu\;and\;\sigma.

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}

\\ -0.6 = \frac{100-\mu}{\sigma} (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}

\\ 2.4 = \frac{190-\mu}{\sigma} (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

\\ -0.6 = \frac{100-\mu}{\sigma} (1)

\\ 2.4 = \frac{190-\mu}{\sigma} (2)

Rearranging these two equations:

\\ -0.6*\sigma = 100-\mu (1)

\\ 2.4*\sigma = 190-\mu (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

\\ 0.6*\sigma = -100+\mu (1)

\\ 2.4*\sigma = 190-\mu (2)

Summing both equations, we obtain the following equation:

\\ 3.0*\sigma = 90

Then

\\ \sigma = \frac{90}{3.0} = 30

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

\\ 2.4*30 = 190-\mu (2)

\\ 2.4*30 - 190 = -\mu

\\ -2.4*30 + 190 = \mu

\\ \mu = 118

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