Answer Perimeter = 33in
Step-by-step explanation:
AC = 20
BD = 20
AB = 13
O is the midpoint
OD = OC = 20/2 = 10
AB = DC = 13 in
P = DC + OD + OC = 13+10+10 = 33in
Answer:

Step-by-step explanation:
<u>The full question:</u>
<em>"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current positions?"</em>
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The permutation of choosing 3 members from a group of 11 would be:
P(n,r) = 
Where n would be the total [in this case n is 11] & r would be 3
Which is:
P(11,3) = 
So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:
1/990
Answer:
The answer is below
Step-by-step explanation:
Let S denote syntax errors and L denote logic errors.
Given that P(S) = 36% = 0.36, P(L) = 47% = 0.47, P(S ∪ L) = 56% = 0.56
a) The probability a program contains both error types = P(S ∩ L)
The probability that the programs contains only syntax error = P(S ∩ L') = P(S ∪ L) - P(L) = 56% - 47% = 9%
The probability that the programs contains only logic error = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
P(S ∩ L) = P(S ∪ L) - [P(S ∩ L') + P(S' ∩ L)] =56% - (9% + 20%) = 56% - 29% = 27%
b) Probability a program contains neither error type= P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
c) The probability a program has logic errors, but not syntax errors = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
d) The probability a program either has no syntax errors or has no logic errors = P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
Answer:

Step-by-step explanation:
"4 times the sum of 2 and y".
<h3>Hope it is helpful...</h3>