Solve by elimination: (x + 7y = 12) (3x

Simplify<br><br> √12m • √15m<br><br> = [1] m √[2]

Elodia [21]

Answer:

6m√5

Step-by-step explanation:

√(12m)·√(15m) = √(180m²) = √(5·(36m²)) = √5·√(6m)²

= 6m√5 . . . . . . for m ≥ 0

8 0

3 years ago

I NEED HELP RN PLSSS!!! (check whole picture) i’ll give brainiest if u don’t give a link just pls help immediately!!!)

zysi [14]

Answer:

So I would say your answer is..............A

Step-by-step explanation:

For most places, global warming will result in more frequent hot days and fewer cool days, with the greatest warming occurring over land. Longer, more intense heat waves will become more common. Storms, floods, and droughts will generally be more severe as precipitation patterns change

Also, A is correct because lets say you put an ice cube in the sun...okay well if it keeps getting hotter and hotter multiply ice cubes into ice caps. Well same thing will happen. It will melt causeing water from the ice caps. Which will result into more water. Which will cause water to go above sea level!

Hope this helps!!!

6 0

2 years ago

Which is the simplified form of the expression (6^-2•6^5)^-3

Arisa [49]

Answer:

1

Step-by-step explanation:

$( {6}^{ - 2} \times {6}^{5} ) ^{ - 3} \\ = ( {6}^{3} )^{ - 3} \\ = {6}^{0} \\ = 1$

6 0

3 years ago

Read 2 more answers

The owner expects you to hand wash each coffee pot before the coffee shop opens. It takes 10 minutes for all of the coffee pots

ankoles [38]

Pretty sure it is 16

7 0

3 years ago

Read 2 more answers

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

Solve by elimination: (x + 7y = 12) (3x - 5y = 10) *