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Dima020 [189]
2 years ago
5

Evaluate the function at x = -3.

Mathematics
2 answers:
trapecia [35]2 years ago
6 0

Answer:

hold on

Step-by-step explanation:

lilavasa [31]2 years ago
5 0
Given the function, p(x) = 4x^3 + 2x^2 - 3x + 2:

We can substitute the value of x = -3 as input into the given function:

p(-3) = 4x^3 + 2x^2 - 3x + 2

p(-3) = 4(-3)^3 + 2(-3)^2 - 3(-3) + 2

p(-3) = 4(-27) + 2(9) - 3(-3) + 2

p(-3) = -108+ 18 - (-9) + 2

p(-3) = -90 + 9 + 2

p(-3) = -81 + 2

p(-3) = -79

Please mark my answers as the Brainliest if you find this solution helpful :)
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Aleksandr-060686 [28]
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3 years ago
5) Two machines M1, M2 are used to manufacture resistors with a design
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:

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.

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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Answer:

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