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Ostrovityanka [42]

3 years ago

11

Roxy purchased a block of wood for a craft. The block of wood is 5 inches, or about 127 millimeters, in length. What is the appr

oximate number of centimeters per inch?

1 answer:

arsen [322]3 years ago

7 0

1 inch is about 2.54 centimeters.

So 5 inches is 12.7 centimeters. To find the answer you have to multiply 12.7×5=12.7

You might be interested in

If a bank account doubles in size every 5 years, then by what percent does it grow after only 3 years? Round to the nearest tent

mr Goodwill [35]

Answer:

Step-by-step explanation:

Alright, let get started.

Suppose the initial deposit in account = x

After five years, this accounts get doubled means = 2x

Means in 5 years , the growth = $2x-x = x$

So, in 1 year, the growth will be = $\frac{x}{5}$

So, in 3 years, the growth will be = $\frac{x}{5}*3=\frac{3x}{5}$

So percent grwoth will be = $\frac{\frac{3x}{5}}{x}*100 = 60$

Hence percentage growth is 60 %. : Answer

Hope it will help :)

4 0

3 years ago

If George received $ 78.00 for his work at Weaver's Store, and he was paid $ 6.00 per hour, how many hours did he work?

mihalych1998 [28]

13 hours
78.00 ÷ 6.00 equals 13 hours
6×13 equals 78

3 0

3 years ago

Read 2 more answers

If x is increased by 20%.<br>then what<br>will be the increased percent if there will be x^2?<br>

nevsk [136]

Answer:

the area will be increased by 44%

Step-by-step explanation:

5 0

3 years ago

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

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 30 mm and standard d

skelet666 [1.2K]

Answer:

$z=-0.842$

And if we solve for a we got

$a=30 -0.842*7.8=23.432$

So the value of height that separates the bottom 20% of data from the top 80% is 23.432.

Step-by-step explanation:

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(30,7.8)$

Where $\mu=30$ and $\sigma=7.8$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.80$ (a)

$P(X$ (b)

As we can see on the figure attached the z value that satisfy the condition with 0.20 of the area on the left and 0.80 of the area on the right it's z=-0.842

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842$

And if we solve for a we got

$a=30 -0.842*7.8=23.432$

So the value of height that separates the bottom 20% of data from the top 80% is 23.432.

8 0

3 years ago