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Mathematics

2 answers:

nataly862011 [7]3 years ago

4 0

Answer:

the anwer is D: all of the above

Step-by-step explanation:

erastova [34]3 years ago

3 0

Answer:

all of the above

Step-by-step explanation:

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How many times greater is the value if the digit 2 in 23876 than the value of the digit 2 in 3254

mixas84 [53]

In 23,876 the 2 represents 20,000 while in 3,254 the 2 represents 2,000. So, 2 is ten times greater in 23,876 than in 3,254 (20,000÷2,000=10).

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3 years ago

Find the value of f(-4).

Alex

Answer:

y = f(x) = f(-4) =-4 because the value of f is equal to -4

4 0

2 years ago

A manufacturing company uses an acceptance scheme on items from a production line before they are shipped. The plan is a two-st

pishuonlain [190]

Answer:

a) The probability that a box containing 3 defectives will be shipped is $51.74\%$

b) The probability that a box containing only 1 defective will be sent back for screening is $17.39\%$

Step-by-step explanation:

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 20 items of the box (discounting the 3 defectives) as the possible ways to succeed, so $\left[\begin{array}{ccc}20\\4\end{array}\right] =20C4=4845$

Finally we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{4845}{8855}=0.5471=P(S)$ , therefore the probability that a box containing 3 defectives will be shipped is $P(S)=54.71\%$

a) The first step is to count the number of total possible random sets of taking a sample size of 4 items over 23 items of the box, so $\left[\begin{array}{ccc}23\\4\end{array}\right] =23C4=8855$

The second step is to count the number of total possible random sets of taking a sample size of 4 items over 22 items of the box (discounting the defective 1) as the possible ways to succeed, so $\left[\begin{array}{ccc}22\\4\end{array}\right] =22C4=7315$

Then we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 4} =\frac{7315}{8855}=0.8260=P(S)$ , therefore the probability that a box containing 1 defective will be shipped is $P(S)=82.60\%$