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4 years ago

Factor. x^2−9x+20 Enter your answer in the box.

2 answers:

5 0

First, you’d look at what can multiply together to make 20. You can multiply 10x2, 1x20, or 4x5 (or change them so each number is a negative) to make a positive 20. You then must look at which one of these can be added or subtracted together to create -9. Negative 4 and 5 can be subtracted to make -9, so you would factor them out, as well as the x^2, and get the answer (x-4)(x-5).

Answer: (x-4)(x-5)

disa [49]4 years ago

5 0

Answer:

(x-5) (x-4)

Step-by-step explanation:

x^2−9x+20

Using AC method

4 × 5 = 20 and 4 + 5 = 9

=(x-5) (x-4)

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ali wants to buy a piano the piano measures 19/4 feet long she has a space 5 feet long for the paino in her house. does she have

IceJOKER [234]

Yes. This is because 19/4 as a mixed number is 4 3/4. 4 3/4 < 5. It might be a tight fit but she has enough room

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3 years ago

Solve using the pathagorean theorem

seropon [69]

Answer:

2 sqrt(13)

Step-by-step explanation:

We can find the hypotenuse using the Pythagorean theorem

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

4^2 + 6^2 = c^2

16+36 = c^2

52 = c^2

Taking the square root of each side

sqrt (52) = c

sqrt(4*13) = c

2 sqrt(13) = c

4 0

2 years ago

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

Triangle XYZ is rotated 90° counterclockwise using the origin as the center of rotation.

lina2011 [118]

Answer:

im pretty sure its 270 clockwise.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Evaluate the limit. lim x ? 5 (x-5)^2/x-5

slega [8]

As x approaches 5
hmm, we can divide the (x-5) from top and bottom to get x-5
if we input 5 for x we get
5-5=0
it approaches 0 as x approaches 5

7 0

3 years ago