What is negative one half plus two?

Lelu [443]

Answer: Phillip is correct. The triangles are <u>not </u>congruent.

How do we know this? Because triangle ABC has the 15 inch side between the two angles 50 and 60 degrees. The other triangle must have the same set up (just with different letters XYZ). This isn't the case. The 15 inch side for triangle XYZ is between the 50 and 70 degree angle.

This mismatch means we cannot use the "S" in the ASA or AAS simply because we don't have a proper corresponding pair of sides. If we knew AB, BC, XZ or YZ, then we might be able to use ASA or AAS.

At this point, there isn't enough information. So that means John and Mary are incorrect, leaving Phillip to be correct by default.

Note: Phillip may be wrong and the triangles could be congruent, but again, we don't have enough info. If there was an answer choice simply saying "there isn't enough info to say either if the triangles are congruent or not", then this would be the best answer. Unfortunately, it looks like this answer is missing. So what I bolded above is the next best thing.

7 0

3 years ago

Read 2 more answers

The area of a rectangle is represented by the expression 6x³y5. Which of the following could be the

oee [108]

The length and width of the rectangle are: $2x^2y^2 , 3xy^3$

<h3>What is the area of the rectangle?</h3>

The area of the rectangle is the product of the length and width of a given rectangle.

The area of the rectangle = length × Width

The area of a rectangle is represented by the expression $6x^3y^5$ .

The area of the rectangle = $2x^2y^2 \times 3xy^3$

= $6x^3y^5$

So, the length and width of the rectangle are: $2x^2y^2 , 3xy^3$

Learn more about the area;

brainly.com/question/1658516

#SPJ1

7 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

What is the following product?

Leona [35]

Answer:

First option: 6y^2sqrt(10) + 12sqrt(5y)

Step-by-step explanation:

3sqrt(10) * (2y^2 + 2sqrt(2y)

= 6y^2sqrt(10) + 6sqrt(20y)

= 6y^2sqrt(10) + 12sqrt(5y)

3 0

3 years ago

Given two dice, what is the probability of rolling a four on the second die, given that you already rolled a four on the first d

Sholpan [36]

Answer: 1/6

Reason:

There's 1 side labeled "4" out of 6 sides total. So that's where the 1/6 comes from.

The "given that you already rolled a four on the first die" is unneeded info in my opinion, because each die is separate or independent from one another.

8 0

3 years ago