Write in miles per hour. Round to the nearest tenth. 211 ft/s
1 answer:
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Answer:
1) The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Step-by-step explanation:
Given : Assume there are 365 days in a year.
To find : 1) What is the probability that ten students in a class have different birthdays?
2) What is the probability that among ten students in a class, at least two of them share a birthday?
Solution :
Total outcome = 365
1) Probability that ten students in a class have different birthdays is
The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...
The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday
P(2 born on same day) = 1- P( 2 not born on same day)
The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Answer:
32, 37, 42, 47, 52, 57, 62.
Step-by-step explanation:
It is given that Carissa counts by 5s from 32 to 62.
It means, we have to add 5 in the number to get the next number.
Add 5 in 32.
Now, add 5 in 37.
Similarly, continue the process.
Therefore, the numbers counted by carissa are 32, 37, 42, 47, 52, 57, 62.