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Karolina [17]
2 years ago
11

A batch of 500 machined parts contains 10 that do not conform to customer requirements. Parts are selected successively, without

replacement, until a nonconforming part is obtained. The random variable is the number of parts selected.
Mathematics
1 answer:
sveticcg [70]2 years ago
7 0

Question:

For the following exercise, determine the range (possible values) of the random variable. A batch of 500 machined parts contains 10 that do not conform to customer requirements. Parts are selected successively, without replacement, until a nonconforming part is obtained. The random variable is the number of parts selected.

Answer:

Possible values are all integers from 1 to 491.

{1,2,...........491}

Step-by-step explanation:

Given:

Total number of machined parts = 500

Number that do not conform to customers requirements = 10

Number that conforms to customers requirements = 500 - 10 = 490

Since, parts are selected successively, without replacement, until a nonconforming part is obtained, the range of the random variable (number of parts selected) here will be all integers from 1 to 491 ie {1,2,...........,491}.

Here, we have a total of 490 conforming parts, and 10 non conforming parts, there must be a non-conforming part among 491 selections, considering the fact that total number of parts are 500

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Answer:

1.5 mph

Step-by-step explanation:

Let speed of boat be x

let speed of current be c

Also, note D = RT

D is distance

R is rate

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Now, for first leg, we can write:

(x+c)3 = D

And for second leg , we can write:

(x-c)4.8 = D  [note 4 hour 48 minutes is 4.8 hours]

We can equate both D's to get:

(x+c)3 = (x-c)4.8

3x + 3c = 4.8x - 4.8c

7.8c = 1.8x

We know x = 6.5 [given], plugging it in and solving for c:

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c = 1.5

Speed of Current = 1.5 miles per hour

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2 years ago
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joyce is helping her aunt create craft kits. her aunt has 138 pipe cleaners, and each kit will include 6 pipe cleaners. what is
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Dave’s Automatic Door, referred to in Exercise 29, installs automatic garage door openers. Based on a sample, following are the
HACTEHA [7]

The question is not complete and the full question says;

Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

Answer:

A) Range = 30 minutes

B) Mean = 38

C) Mean Deviation = 7.2

D) This is well written in the explanation.

Step-by-step explanation:

A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.

Thus; Range = 54 - 24 = 30 minutes

B) In statistics,

Mean = Σx/n

Where n is the number of times occurring and Σx is the sum of all the times occurring

Thus,

Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380

n = 10

Thus, Mean(x') = 380/10 = 38

C) Mean deviation is given as;

M.D = [Σ(x-x')]/n

Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72

So, M.D = 72/10 = 7.2

D) The range of the times is 30 minutes.

The average time required to open one door is 38 minutes.

The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes

4 0
2 years ago
A print shop purchases a new printer for $25,000. The equipment depreciates at a rate of 5% each year. The relationship between
Pani-rosa [81]

Answer:

The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.

Step-by-step explanation:

Since the printer depreciates at a rate of 5% per year, I believe the stated equation is miss typed. Therefore I'll answer this with the correct equation that would represent that setting:

y(x) = 25,000*0.95^x

In the first year the value of the printer is:

y(1) = 25,000*0.95^1 = 23,750

On the second year the value of the printer is:

y(2) = 25,000*0.95^2 = 22,562.5\\

On the third year the value of the printer is:

y(3) = 25,000*0.95^3 = 21,434.38\\

The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.

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