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kiruha [24]
3 years ago
5

Please help please and thanks

Mathematics
1 answer:
Bingel [31]3 years ago
7 0
This will help a lot if you go to desmos! Just graph the equation on there! It’s much easier to see on computer

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It's #25 guys. Show your work!
Ronch [10]
y+8 = \frac{1}{2}(x-7) is equivalent to <span>y-(-8) = \frac{1}{2}(x-7)

That is in the form </span><span>y-y_{1} = m(x-x_{1}) which is point slope form

In this case,

m = 1/2 is the slope
(x1,y1) = (7,-8) is the point the line goes through</span>
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decide if 30x-12 could be the result of using the distributive property. If it is, possible combonation of factors whose product
sergij07 [2.7K]

Answer:

89

Step-by-step explanation:

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3 years ago
Justin needs to earn a score of more than 80 on his next math test to earn a B for the semester. Which inequality shows the grad
zvonat [6]

Answer:

b

Step-by-step explanation:

Justin's score has to be greater than 80. the appropriate inequality sign is >

Note that

> means greater than

< means less than

≥ means greater than or equal to  

≤ less than or equal to  

On the number line, since the score has to be greater than 80, the line would start from 80 and end on 95

> or < is represented by unfilled circle

≤ or ≥  is represented by filled circle

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3 years ago
Long division 5th grade I’m sorry if I’m asking a lot but I’m really new to this
Nuetrik [128]

Answer:

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Step-by-step explanation:

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2 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
2 years ago
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