Answer:
66 2/3 %
Step-by-step explanation:
First find the students not in the 8th grade
24 - 8 = 16
16 students are not in the 8th grade
Take the fraction of the students not in the 8th grade over the total
16/24 = 2/3
Change to a decimal
.66666666666
Multiply by 100 to change to a percent
66.666666%
66 2/3 %
Answer:
66.67% of students are not in eighth grade
Step-by-step explanation:
8/24=1/3
1/3=0.33333333333
1-0.33333333333=0.66666666667
0.66666666667=66.67%
Answer:
A, C, E.
Step-by-step explanation:
Just did it.
Answer:
A (square root 13, 146.3 degrees)
C (-squareroot13, -33.7 degrees)
E (-3,2)
Answer:
D. 30 feet per minute
Step-by-step explanation:
If you follow the x-axis (time) you see 1 minute as you go up the y-axis (distance) you see that 1 and 30 connect, therefore you get 30 feet per minute.
Answer:
30 feet per minute
Step-by-step explanation:
its very slow therefore it takes a lot of time
Answer:
The answer is 2.25
Step-by-step explanation:
Steps:
1. 1/2 inch + 1 3/4 inch = 2.25 inch
Answer:
2.25
Step-by-step explanation:
Answer:
a
Step-by-step explanation:
b) What is the probability that the class hangs Wisconsin's flag on Monday, Michigan's flag on Tuesday, and California's flag on Wednesday.?
c) What is the probability that Wisconsin's flag will be hung at least two of the three days?
Answer:
a.) P(x = X) =
b.)
c.) 0.00118
Step-by-step explanation:
The sample space Ω = flags of all 50 states
a.) Any one of the flags is randomly chosen. Therefore we can write the
probability measure as P(x = X) = , for all the elements of the sample
space, that is for all x ∈ Ω.
b.) the probability that the class hangs Wisconsin's flag on Monday,
Michigan's flag on Tuesday, and California's flag on Wednesday
=
c.) the probability that Wisconsin's flag will be hung at least two of the three days
= Probability that Wisconsin's flag will be hung on two days + Probability that Wisconsin's flag will be hung on three days
= P(x = 2) + P(x = 3)
=
=
=
= 0.00118
The sample space for this experiment is all the possible combinations of flags from the 50 U.S. states for the three days. The probability of hanging Wisconsin's flag on Monday, Michigan's on Tuesday, and California's on Wednesday is 1/125,000. The probability of hanging Wisconsin's flag at least two of the three days is 294/125,000.
a) The sample space Ω for this experiment comprises of all possible combinations of flags from the 50 U.S. states for the three days. Hence, the total number of outcomes in the sample space Ω would be 50*50*50 = 125,000. Every outcome in this space is equally likely, so the probability measure P would assign a probability of 1/125,000 to each outcome.
b) As each day's choice is independent of the others and each state's flag is equally likely to be chosen, the probability that Wisconsin's flag is hung on Monday, Michigan's flag is hung on Tuesday, and California's flag is hung on Wednesday would be (1/50) * (1/50) * (1/50) = 1/125,000.
c) To find the probability that Wisconsin's flag will be hung at least two of the three days, we have to add the probabilities for the three situations where Wisconsin's flag is hung exactly twice plus the situation where Wisconsin's flag is hung all three days. The final probability would be [(3 * (1/50)² * (49/50)) + (1/50)³] = 294/125,000.
#SPJ3
12×10 to the power of zero is 1.2×10 to the power of one and 12 times one is 12