Which is the best name for the quadrilateral with vertices at (2,2) (5,-2) (1,-5) (-2,-1)

What is the greatest common factor of 8 and 18?

SCORPION-xisa [38]

Answer: 2

Step-by-step explanation: To find the greatest common factor or GCF of 8 and 18, we begin by finding all of the factors of each number.

To find the factors of 8, we know that 8 ÷ 1 is 8 so 1 and 8 are factors.

8 ÷ 2 is 4 so 2 and 4 are factors.

However, if we continue to divide 8 by 3, 4, 5, and so on, we won't find any new factors. So the factors of 8 are 1, 2, 4, and 8.

Now let's find the factors of 18.

18 ÷ 1 is 18 so 1 and 18 are factors.

18 ÷ 2 is 9 so 2 and 9 are factors.

18 ÷ 3 is 6 so 3 and 6 are factors.

However, if we continue to divide by 4, 5, 6, and so on, we won't find any new factors. So the factors of 18 are 1, 2, 3, 6, 9, and 18.

Now that we have our list of factors, to find the greatest common factor, we simply find the largest number that is shared by the two lists.

Notice that our lists share a 1 but the largest number they share is a <em>2</em><em>.</em>

<em />

So the greatest common factor or <em>gcf</em> of 8 and 18 is 2.

I have also shown my work on the whiteboard.

4 0

3 years ago

Solve the inequality |* +31 < |2x + 1.<br> |x+3|<|2x+1|

Kisachek [45]

Answer:

Step-by-step explanation:

Here are the steps to follow when solving absolute value inequalities:

Isolate the absolute value expression on the left side of the inequality.

If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions.

If your problem has a greater than sign (your problem now says that an absolute value is greater than a number), then set up an "or" compound inequality that looks like this:

(quantity inside absolute value) < -(number on other side)

OR

(quantity inside absolute value) > (number on other side)

The same setup is used for a ³ sign.

If your absolute value is less than a number, then set up a three-part compound inequality that looks like this:

-(number on other side) < (quantity inside absolute value) < (number on other side)

The same setup is used for a £ sign

7 0

4 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

The rule for the number of fish in a home aquarium is 1 gallon of water for each inch of fish length. Marta's aquarium holds 33

tigry1 [53]

<h3>Answer:</h3>

A: 7 inches
B: 6 inches

<h3>Step-by-step explanation:</h3>

We can let the variables A and B stand for the length in inches of fish A and fish B, respectively.

If we assume each person bought fish having a total length of 1 inch per gallon of aquarium, then we can write equations ...

... 3A +2B = 33 . . . . . total length of Marta's fish

... 3A +4B = 45 . . . . . total length of Hank's fish

Subtracting the first equation from the second, we get ...

... 2B = 12

... B = 6 . . . . . divide by 2

Using this value in the first equation, we have ...

... 3A + 2·6 = 33

... 3A = 21 . . . . . . . . subtract 12

... A = 7 . . . . . . . . . . divide by 3

Fish A is 7 inches long; fish B is 6 inches long.

6 0

3 years ago

Please help this is due in about three minutes

mihalych1998 [28]

Answer:

13) Constant

14) Variable

15) Coefficient

16) Exponent

6 0

3 years ago