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Hunter-Best [27]
2 years ago
8

elect the correct answer. graph A graph B graph C graph D A credit card company uses these rules to calculate the minimum amount

owed: For a bill of less than $100, the entire amount is due. For a bill of at least $100 but less than $500, the minimum due is $100. For a bill of at least $500 but less than $1,000, the minimum due is $300. For a bill of $1,000 or more, the minimum due is $500. Which graph shows the minimum amount due for a credit amount of x (given that the credit limit is $2,000). A. graph A B. graph B C. graph C D. graph D
Mathematics
1 answer:
laila [671]2 years ago
5 0

The graph that correctly shows the minimum amount due for a credit amount (x)  is graph B.

<h3>How to determine the graph with the minimum amount due?</h3>

First of all, we would assign a variable to the different amount of money as follows:

  • Let x be the credit amount.
  • Let y be the minimum amount due.

Translating the word problem into an algebraic expression, we have;

For a bill of less than $100, the entire amount is due:

0 ≤ x < $100, therefore, y = x

For a bill of at least $100 but less than $500, the minimum due is $100: $100 ≤ x < $500, therefore, y = 100

For a bill of at least $500 but less than $1,000, the minimum due is $300:

$500 ≤ x < $1000, therefore, y = $300

For a bill of $1,000 or more, the minimum due is $500:

Since the credit limit is equal to $2000, we have $1000 ≤ x < $2000 and y = $500

In conclusion, the graph that correctly shows the minimum amount due for a credit amount (x)  is graph B.

Read more on graphs here: brainly.com/question/16883029

#SPJ1

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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

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#V=27.72×1000 (V= volume)

# r = 42/2=21 (r = radius)

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