Add and subtract positive and negative decimals
1 answer:
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Answer:
535,095 different samples of size six that contain exactly 2 defective parts.
0.0337 = 3.37% probability that a sample contains exactly 2 defective parts.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
As the order of the parts is not important, the combinations formula is used to solve this question.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
How many different samples are there of size six that contain exactly 2 defective parts?
2 defective from a set of 3, and 4 non-defective from a set of 47. So
535,095 different samples of size six that contain exactly 2 defective parts.
What is the probability that a sample contains exactly 2 defective parts?
The total number of samples is:
Then...
0.0337 = 3.37% probability that a sample contains exactly 2 defective parts.
The coefficients of the binomial expansion 
, where n is the row number, is given in the Pascal's triangle shown below.
First, to find the <span>coefficient of the fourth term of (x+2)^5 we look at row 5, term 4. The coefficient there is 10.
But, we must also remember that the term 2 also is taken to a certain power here. Mainly , for each term, the power of 2 is as follows:
2^0, 2^1, 2^2, 2^3=8.
So, in total we have: 10*8=80.
Second, to find </span><span> the coefficient of the third term of (3x-1)^5 we again go to the row 5, this time term 3 and we have 10 there. Now we must check how each of (3x) and 1 expand, now being careful about the sign as well.
we have:
(3x)^5 (1) -(3x)^4 (1) (3x)^3(1)=27x^3.
Thus, the coefficient of the third term is 27*10=270.
Third, we want to find </span><span>the coefficient of the third term of (a+5b^2)^4. We look at row 4, term 3. There we have 6.
The terms a and 5b^2 are as follows:
a^4 (5b^2)^0 </span> a^3 (5b^2)^1 a^2 (5b^2)^2=25a^2b^4
Thus, the coefficient is 25*6=150.
Answer:
80; 270; 150
First, to find the <span>coefficient of the fourth term of (x+2)^5 we look at row 5, term 4. The coefficient there is 10.
But, we must also remember that the term 2 also is taken to a certain power here. Mainly , for each term, the power of 2 is as follows:
2^0, 2^1, 2^2, 2^3=8.
So, in total we have: 10*8=80.
Second, to find </span><span> the coefficient of the third term of (3x-1)^5 we again go to the row 5, this time term 3 and we have 10 there. Now we must check how each of (3x) and 1 expand, now being careful about the sign as well.
we have:
(3x)^5 (1) -(3x)^4 (1) (3x)^3(1)=27x^3.
Thus, the coefficient of the third term is 27*10=270.
Third, we want to find </span><span>the coefficient of the third term of (a+5b^2)^4. We look at row 4, term 3. There we have 6.
The terms a and 5b^2 are as follows:
a^4 (5b^2)^0 </span> a^3 (5b^2)^1 a^2 (5b^2)^2=25a^2b^4
Thus, the coefficient is 25*6=150.
Answer:
80; 270; 150