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kondor19780726 [428]
3 years ago
12

(PLEASE HELP 30 POINTS)

Mathematics
1 answer:
san4es73 [151]3 years ago
8 0

Answer:

We have to:

"All of the plots are the same length, x"

L = x

"and the width of each plot is 5 yards less than the length"

W = x-5

"The total number of plots Liam owns is 20 more than the length of a plot"

20 + x

"the total area of all the plots Liam owns is 2,688 square yards"

A = (20 + x) * (x) * (x-5)

A = (20x - 100 + x ^ 2 -5x) * (x)

A = (x ^ 2 + 15x - 100) * (x)

2688 = (x ^ 3 + 15x ^ 2 - 100x)

x ^ 3 + 15x ^ 2 - 100x = 2688

x ^ 3 + 15x ^ 2 - 100x - 2688 = 0

Answer:

*** The equation x3 + 15x2 - 100x - 2.688 = 0 can be used to find the length of each plot.

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Basile [38]

Answer:

Since M1 has the higher probability of being in the desired range, we choose M1.

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:

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Two machines M1, M2 are used to manufacture resistors with a design specification of 1000 ohm with 10% tolerance.

So we need the machines to be within 1000 - 0.1*1000 = 900 ohms and 1000 + 0.1*1000 = 1100 ohms.

For each machine, we need to find the probabilty of the machine being in this range. We choose the one with the higher probability.

M1:

Resistors of M1 are found to follow normal distribution with mean 1050 ohm and standard deviation of 100 ohm. This means that \mu = 1050, \sigma = 100

The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So

X = 1100

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

Z = \frac{1100 - 1050}{100}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

X = 900

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

Z = \frac{900 - 1050}{100}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.6915 - 0.0668 = 0.6247.

M1 has a 62.47% probability of being in the desired range.

M2:

M2 are found to follow normal distribution with mean 1000 ohm and standard deviation of 120 ohm. This means that \mu = 1000, \sigma = 120

X = 1100

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

Z = \frac{1100 - 1000}{120}

Z = 0.83

Z = 0.83 has a pvalue of 0.7967.

X = 900

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

Z = \frac{900 - 1000}{120}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

0.7967 - 0.2033 = 0.5934

M2 has a 59.34% probability of being in the desired range.

Since M1 has the higher probability of being in the desired range, we choose M1.

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