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FinnZ [79.3K]
3 years ago
10

For the following exercise, determine the range (possible values) of the random variable, X. A batch of 300 machined parts conta

ins 10 that do not conform to customer requirements. The random variable is the number of parts in a sample of 5 parts that do not conform to customer requirements. Choose the correct answer.
a. The range of X is {0,1,2,3,4,5,6}
b. The range of X is {0,1,2,3,4,5,6}
c. The range of X is {0,1,2,3,4,5,6}
d. The range of X is {0,1,2,3,4,5,6}
e. The range of X is {0,1,2,3,4,5,6}
Mathematics
1 answer:
solmaris [256]3 years ago
5 0

Answer:

{0,1,2,3,4,5}

Step-by-step explanation:

We are given that

Total number of  machine parts=300

Number of defective machine parts=10

Total number of good machine parts=300-10=290

Sample contain parts that do not conform to customer requirement=5

X  is a random variable which is the number of parts in a sample of 5 parts that do not conform to customer requirements.

We have to find the correct answer.

The sample contain 5 parts

Therefore, the possible values of random variable X

0,1,2,3,4,5

Hence, the range of X is given by

{0,1,2,3,4,5}

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Can a common denominator of mixed numbers be 20?
gladu [14]
Yes, a common denominator is just the smallest multiple of all numbers in a group. 

So the common denominator of 4, 8 and 12 would be 4, because 4 goes into all those numbers.
3 0
3 years ago
Read 2 more answers
If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is $ and the a
Vlada [557]

Answer:

The amount Sam invested the first year = $2000

The amount Sally invested the last year = $1900

Complete question related to this was found at brainly (ID 4527784):

For three consecutive years, Sam invested some money at the start of the year. The first year, he invested x dollars. The second year, he invested $2,000 less than 5/2 times the amount he invested the first year. The third year, he invested $1,000 more than 1/5 of the amount he invested the first year.

During the same three years, Sally also invested some money at the start of every year. The first year, she invested $1,000 less than 3/2 times the amount Sam invested the first year. The second year, she invested $1,500 less than 2 times the amount Sam invested the first year. The third year, she invested $1,400 more than 1/4 of the amount Sam invested the first year.

If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is $ and the amount Sally invested the last year is $ .

Step-by-step explanation:

First we would represent the information given with mathematical expressions.

Sam investment for 3 consecutive years:

Year 1 = x dollars

Year 2 = $2,000 less than 5/2 times the amount he invested the first year

Year 2 = (5/2)(x) - 2000

Year 3 = $1,000 more than 1/5 of the amount he invested the first year

Year 3 = (1/5)(x) + 1000

Sally investment for 3 consecutive years:

Year 1 = $1,000 less than 3/2 times the amount Sam invested the first year

Year 1 = (3/2)(x) - 1000

Year 2 = $1,500 less than 2 times the amount Sam invested the first year

Year 2 = 2x - 1500

Year 3 = $1,400 more than 1/4 of the amount Sam invested the first year.

Year 3 = (1/4)(x) + 1400

Since Sam and Sally invested the same total amount at the end of three years, we would equate their sum:

Sum of Sam investment for the 3years = x + (5/2)(x) - 2000 + (1/5)(x) + 1000

= x + 5x/2 -2000 + x/5 + 1000

= (10x+25x+2x)/10 - 1000

= 37x/10 - 1000

Sum of Sally investment for the 3years = (3/2)(x) - 1000 + 2x - 1500 + (1/4)(x) + 1400

= 3x/2 - 1000 + 2x -1500 + x/4 + 1400

= (6x+8x+x)/4 - 1100

= 15x/4 - 1100

37x/10 - 1000 = 15x/4 - 1100

37x/10 - 15x/4 = -100

(148x - 150x)/40 = -100

-2x = -4000

x = 2000

Therefore the amount Sam invested the first year = x = $2000

The amount Sally invested the last year (3rd year) = (1/4)(x) + 1400

(1/4)(2000) + 1400 = 500+1400 = 1900

The amount Sally invested the last year = $1900

8 0
3 years ago
I don't understand this please help :( :/
Lena [83]
Okay.  The equation is the way to figure out how much can be put into the suitcase.

What you do is substitute the various weights for w, and see if the equation is true.
So, 42 + w ≤ 50.
If you use the wieght of the books, 10lb, it looks like this: 42 + 10 ≤ 50.  If you solve that, the equation isn't true, so you can't put the books into the suitcase.

7 0
3 years ago
Do subtraction with monetory values.<br><br> $49.68 -$7.89=?
adell [148]

Answer:

U have $41.79 left i hope it is helpful have a nice day and merry Christmas

5 0
3 years ago
Read 2 more answers
Find LCM of:<br><br>x2 - 9, 3x3 + 81​
VikaD [51]

Answer:

\boxed{\pink{\sf (x + 3) \ is\ the \ LCM . }}

Step-by-step explanation:

Given two expressions ,

\qquad (1)\: x^2 - 9 \\\\ \qquad (2)\:3x^3 + 81

And , we need to find the LCM , that is lowest common factor . So , let's factorise them seperately .

<u>Factorising</u><u> </u><u>x²</u><u> </u><u>-</u><u> </u><u>9</u><u> </u><u>:</u><u>-</u><u> </u>

= x^2 - 9 \\\\= x^2 - 3^2 \\\\ \red{= (x+3)(x-3)} \qquad\bf [Using \ a^2-b^2 = (a+b)(a-b) ]

<u>Factorising</u><u> </u><u>3x</u><u>³</u><u> </u><u>+</u><u> </u><u>8</u><u>1</u><u> </u>

= 3x^3 + 81\\\\= 3(x^3+27) \\\\= 3( x^3+27) \\\\ = 3(x^3+3^3) \\\\ \red{= 3 [ (x+3)(x^2+9-3x) ] } \qquad \bf{[ Using \ a^3+b^3=(a+b)(a^2+b^2-ab) ] }

Hence we can see that (x+3) is common factor in both expressions.

<u>Hence</u><u> </u><u>the</u><u> </u><u>LCM</u><u> </u><u>is</u><u> </u><u>(</u><u> </u><u>x</u><u>+</u><u>3</u><u> </u><u>)</u><u> </u><u>.</u>

8 0
3 years ago
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