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3 years ago

The value 5 is an upper bound for the zeros of the function shown below. f(x)=x^4+x^3 -11x^2-9x+18

2 answers:

7 0

Answer: True

Explanation:
According to the rational zeros theorem, if x=a is a zero of the function f(x), then f(a) = 0.

Given: f(x) = x⁴ + x³ - 11x² - 9x + 18
From the calculator, obtain
f(5) = 448
f(4) = 126
f(3) = 0
f(2) = -20
f(1) = 0
f(0) = 18
f(-1) = 16
f(-2) = 0
f(-3) = 0

The polynomial is of degree 4, so it has at most 4 real zeros.
From the calculations, we found all 4 zeros as x = -3, -2, 1, and 3.
Therefore
f(x) = (x+3)(x+2)(x-1)(x-3).

For x>3, f(x)increases rapidly. Therefore there are no zeros for x>3.
The statement that x=5 is an upper bound for the zeros of f(x) is true.

yawa3891 [41]3 years ago

4 0

Answer:

true

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Since L means the number of lessons taken, we will have:

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3 years ago

Pleaseeee help me solve this

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Answer:

3/5 is the answer!

Step-by-step explanation:

y = 3x/5 also means y = 3/5 * x

Which means in y = mx form, the 3/5 = m

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3 years ago

I’m I right? I kinda but kinda don’t get this question I lost my notes and I just learnt this

belka [17]

Yes, you are correct

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3 years ago

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1. Write the equation for a line in point slope form that had a slope of 4 and passes through (1,1)

Dahasolnce [82]

Solution:

Slope = m = 4
Given point: (x₁,y₁) = (1,1) [x₁ = 1; y₁ = 1]
Formula: y - y₁ = m(x - x₁)

<u>Use the formula to convert the line in point slope form.</u>

y - y₁ = m(x - x₁)
y - 1 = 4(x - 1)

The equation of the line (In point slope form) is y - 1 = 4(x - 1).

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3 years ago

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The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0

3 years ago