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Vesnalui [34]
3 years ago
11

In a set of 1,000 integers from 1 to 1,000, an integer chosen at random on a single trial should be an integer from 1 to 25 abou

t 25 out of every 1,000 trials, or one out of every 40 integers selected. A sample of 5 integers selected is shown. Does this sample represent the general rule for picking an integer from 1 to 25 in the population of integers from 1 to 1,000?
A.Yes - the sample is representative of the expected number of integers from 1 to 25 in a sample of 5 integers, which would be none or zero.
B. No - the sample is not representative of the expected number of integers from 1 to 25 in a sample of 5 integers, which would be at least 10.

Reminder: Please do not just say the answer, make sure to include your work on how you got the answer!
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0

Answer: hi

Step-by-step explanation:

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3 1/2 • 4 3/4 please help
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Answer: 16 5/8

Step-by-step explanation:

3 1/2 = 7/2

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Help!!!!!!!!!!!!!!!!!!!!!!!!!
torisob [31]

Answer:

53

Step-by-step explanation:

\frac{sin(56)}{78.6}  =  \frac{sin(90)}{ab}  \\  \\ absin(56) = sin(90) \times 78.6 \\  \\ ab =  \frac{sin(90) \times 78.6}{sin(56)}  \\ ab = 94.8 \\  \\  {x}^{2}   +  {78.6}^{2}  =  {94.8}^{2}  \\  {x}^{2}  =  {94.8}^{2}  -  {78.6}^{2}  \\  {x}^{2}  = 8987.04 - 6177.96 \\  {x}^{2}  = 2809.08 \\ x =  \sqrt{2809.08}  \\ x = 53

6 0
2 years ago
What is the probability that no 1s will be in the output for an input sequence of 10 digits??
emmasim [6.3K]
The total numbers with 10 digits are 9*10*10*10*10*10*10*10*10*10 = 9*(10^9)

If you are restricted to use no 1s, the numbers are 8*(9^9)

Then, the probability is : numbers with no 1s /  total numbers

 = 8*(9^9) / [9*(10^9)]= 0.344
5 0
3 years ago
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