Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

11

Which statement describes what these four powers have in common? 4 Superscript 0 (negative 2) Superscript 0 (one-third) Superscr

ipt 0 9 Superscript 0 All the powers have a value of 0 because the exponent is zero. All the powers have a value of 1 because the exponent is zero. All the powers have a value of –1 because the exponent is zero. All the powers have a fractional value because the exponent is zero.

2 answers:

kiruha [24]3 years ago

8 0

Answer:

The answer would be that All the powers have a value if 1 because the exponent is zero.

Step-by-step explanation:

Use the exponent rule to help you

stich3 [128]3 years ago

5 0

Answer:

the answer is A.

Step-by-step explanation:

You might be interested in

Rachel made 28 hats. Rachel made 7 times as many hats as Greg. Let n be the number of hats that Greg made.

il63 [147K]

Answer:

28/7=n is what I believe would be the answer

5 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

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

<u>The area of rectangle is equal to</u>

$A=14*5=70\ in^{2}$

<u>The area of the small semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

<u>The area of the larger semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

Read 2 more answers

Solve the quadratic equation for x.What is one of the roots?

zubka84 [21]

-13 because 6 x 2 = 12 plus x = -13 x represents 1 btw

7 0

3 years ago

Read 2 more answers

A new automobile cost 11300 which is 100 more than 25 times a certain number what is the number

olga_2 [115]

Answer:

The number is 448.

Step-by-step explanation:

Hope it helps

3 0

2 years ago