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STALIN [3.7K]
3 years ago
14

riverside elementary school is holding a school wide election to choose a school color. five-eighths of the votes were for blue

five-ninth of the remaining votes were for green
Mathematics
1 answer:
katen-ka-za [31]3 years ago
8 0
Blue won that election
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John is looking into a cell phone plan that will cost him $60 per month+$3.50 per gigabyte of data that is used. write an expres
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The answer is 60x+3.50
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X= a) 50 b) 100 c) 25
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Answer:

x=a)50 so that would be the answer

would be e correct anwser meep meep

Step-by-step explanation:

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3 years ago
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I keep on getting the wrong answer
Anon25 [30]

1) Area = (b*h)/2 = (9*5)/2 = 45/2 = 22.5 (Letter A)

2) Area = b*h = 8*14 = 112 (Letter D)

3) Surface area of a prism is SA=2B+ph (B = area of the base, p = perimeter of the base, h = height)

B = 15 * 5 = 75 cm^2

p = 15 + 5 = 20 cm

SA = 2*75 + 20*7 = 150 + 140 = 290 (G)

4) V = (B*H*L)/2 = (15*7*5)/2  = 525/2 = 262.5 cm^3 (G)

5) V = 9^3 = 81 cm^3 worth of wrapping (A)

6) V = (B*H*L)/2 = (13*6*8)/2 = 312 cube feet (J)

It gave me ADGGAJ. I don't know if this is right, but I atleast tried to do something, right?

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A 2-column table has 4 rows. The first column is labeled x with entries negative 4, negative 2, 3, 3. The second column is label
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Answer:

y ≤ –2x + 3

Step-by-step explanation: i got it right on edge 2020

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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
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