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3 years ago

How do I solve the absolute value of -5x=45

1 answer:

7 0

Well, You divide both sides by -5.

−5x/−5=45/−5

x=−9

Answer:

x=−9

You might be interested in

"A day care program has an average daily expense of $75.00. The standard deviation is $5.00. The owner takes a sample of 64 bill

kolbaska11 [484]

Answer:

p=1

Step-by-step explanation:

Given that a day care program has an average daily expense of $75.00. The standard deviation is $5.00.

Sample size = 64

Std error of sample = $\frac{5}{\sqrt{64} } \\=0.625$

Thus sample mean will follow normal with

(75, 0.625)

Required probability

= $P(70$

It is almost a certain event with p =1

8 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

A 12 ft. rope is attached to the wall of a garage. A dog’s collar is attached to the end of the rope to allow him to safely stay

Zolol [24]

Since it's a wall, I'm guessing it will only have access to half the circle? Therefore:

Circle area formula: πr²

Radius: 12/2 = 6ft

Our case:

(π * 6²)/2 =

36π/2 =

18π = 56.55ft²

3 0

3 years ago

Ninety-eight and seventy-three hundredths which place are tens , ones , tenths , hundredths?

Naily [24]

8 is the hundreds, 7 in the tens, and 3 in the ones

4 0

3 years ago

Read 2 more answers

In the data set below, what is the interquartile range? 7 1 5 5 9 Submit

Marizza181 [45]

Answer:

Step-by-step explanation:

The interquartile range is the difference between the upper quartile and the lower quartile.

First find the median.

The median is the middle value of the data set arranged in ascending order

1 5 5 7 9 ← data in ascending order

↑ median

The lower quartile is the middle value of the data to the left of the median. If there is not an exact middle then it the average of the values on either side of the middle.

1 5

↑ lower quartile = $\frac{1+5}{2}$ = 3

The upper quartile is the middle value of the data to the right of the median.

7 9

↑ upper quartile = $\frac{7+9}{2}$ = 8

Thus

interquartile range = 8 - 3 = 5

3 0

3 years ago