Plz help I need it or I’m gonna fail this test

In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police

Lyrx [107]

Answer:

Option 3 - 4 miles

Step-by-step explanation:

Given : In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. The distance between the library and the police station is 5 miles.

To find : How far apart are the police station and the fire station?

Solution :

Let the distance between police station and fire station be 'x'.

The distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station.

i.e. $2x-3$

The distance between the library and the police station is 5 miles.

i.e. $2x-3=5$

$2x=5+3$

$2x=8$

$x=4$

Therefore, The police and fire stations is 4 miles apart.

So, Option 3 is correct.

3 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

WHAT IS THE ANSWER PLEASE

viva [34]

Answer:

.top the at is 71 ebecaus because one third the IsIs

Step-by-step explanation:

6 0

2 years ago

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Write the steps to solv

Mkey [24]

Answer:

The computer can perform 1051050 calculations in 10 seconds.

Step-by-step explanation:

1. Put the quantity of calculations the computer can perform in 1 second:

$105105\frac{calculations}{1second}$

2. Multiply the quantity of calculations by 10 seconds:

$105105\frac{calculations}{1second}*10seconds=1051050calculations$

Note that the units of seconds are cancelled as they appear on the numerator and the denominator, and the final response unit is given in calculations number.

The computer can perform 1051050 calculations in 10 seconds.

4 0

3 years ago

I need help with this problem

rewona [7]

I believe the answer is A

6 0

3 years ago