What is the answer to this problem? 2 1/3 + 3 1/3 = ?

Tommy bought 30 gallons of gas for 60.50. How much would 12 gallons of gas cost?

bogdanovich [222]

Answer:

24.2

Step-by-step explanation:

divide 60.50 by 30 then multiply that by 12

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

2 years ago

Asap help.! Circle measurements

Margaret [11]

Answer:

D

Step-by-step explanation:

Here, we want to get the value of x;

To get this, we simply use one of the circle angle and arc properties

Mathematically;

arc MN = 2 * arc MLN

3x + 22 = 134/2

That will be;

3x + 22 = 67

3x = 67-22

3x = 45

x = 45/3

x = 15

5 0

3 years ago

PLZ HELP !!! THIS TEST IS VERY IMPORTANT!!!!

andrew11 [14]

Answer:

E=20H

Step-by-step explanation:

E =20 when H=1

there fore the amount he receives per working time will be 20*H

8 0

3 years ago

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In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

2 years ago