(07.05 MC)
1 answer:
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Answer:
1,082
Step-by-step explanation:
The sample size n in Simple Random Sampling is given by
where
z = 1.645 is the critical value for a 90% confidence level (*)
s = 2 is the estimated population standard deviation
e = 0.1 mm points is the margin of error
so
rounded up to the nearest integer.
So the manufacturer should test 1,083 parts.
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(*)This is a point z such that the area under the Normal curve N(0,1) outside the interval [-z, z] equals 10% = 0.1
It can be obtained in Excel with
NORMINV(1-0.05,0,1)
and in OpenOffice Calc with
NORMINV(1-0.05;0;1)
Answer:
Step-by-step explanation:
-0.4(3i + 4i) - 0.3(-2i + 5j) + 0.2(- +
j)
-1.2i - 1.6j + 0.6i - 1.5j - i + 0.5j
Converting to fraction form;
<u>Solving the i part;</u>
=
<u>Solving the j part;</u>
j
So -0.4a - o.3b + 0.2d = -
2734375 employee ID codes can be generated
<em><u>Solution:</u></em>
Given that Each code will have one digit followed by four letters
The letter V and the digits 3, 4, and 5 will not be used
So, there are 25 letters and 7 digits that will be used
Assume that the letters can be repeated
Each code = 1 digit + four letters
digits 3 , 4 , 5 cannot be used
Therefore, the remaining digits are 1, 2, 6, 7, 8 , 9, 0
7 digits that will be used
One digit can be chosen from 7 digits in 7 ways
The first letter can be chosen from 25 letters in 25 ways
Since given that letters can be repeated
The second letter can be chosen from 25 letters in 25 ways
Similarly, the third letter can be chosen from 25 letters in 25 ways
Similarly, the fourth letter can be chosen from 25 letters in 25 ways
<u><em>Therefore, the total number of ways are:</em></u>
Thus 2734375 employee ID codes can be generated