Answer:
See explanation below
Step-by-step explanation:
Here a coin was tossed three times.
Let H = head & T = tail
Find the following:
a) The sample space:
Since a coin is tossed thrice, all possible outcome would be:
S = { HHH, HHT, HTH, HTT, TTT, TTH, THH, THT}
b) i) A = Exactly 2 tails: Here exactly 2 tails were recorded.
A = {HTT, TTH, THT}
ii) B = at least two tails: Here 2 or more tails were recorded.
B = {HTT, TTT, TTH, THT}
iii) C = the last two tosses are heads:
C = { HHH, THH}
c) List the elements of the following events:
i) A. This means all outcomes in A
= {HTT, TTH, THT}
ii) A∪B. A union B, means all possible outcomes present in A or B or in both
= {HTT, TTH, THT, TTT}
iii) A∩B. This means all possible outcomes of A that are present in B.
= {HTT, TTH, THT}
iv) A∩C. All outcomes A that are present in B
= {∅}
The sample space of tossing a coin three times consists of eight possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT. Events A, B, and C can be determined by listing the appropriate outcomes. The intersection and union of events A and B can also be determined.
(a) The sample space, Ω, of tossing a coin three times can be determined by listing all the possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT.
(b) i. A = {HHT, HTH, THH}
ii. B = {TTT, TTH, THT, HTT, HHT, HTH, THH}
iii. C = {HTH, TTH}
(c) i. A = {HHT, HTH, THH}
ii. A∪B = {HHT, HTH, THH, TTT, TTH, THT, HTT, HHT}
iii. A∩B = {HHT, HTH, THH}
iv. A∩C = {HHT, HTH}
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Answer: A
Step-by-step explanation:
For this problem, we can automatically eliminate B and D. We know this because of the inequality. The inequality is less than or equal to. anything that has an equal to sign is a solid line because any point on the line itself is a solution. If it was a < sign, it shows that the solution is less than, and not equal to. Therefore, the line would be dotted/dashed.
Now that we have figured out what type of line we have, we can see where the shading is located. To do so, we must make all the same variables go on the same side to see the equation.
y ≤ x²-4x+2
From this, we see that y is less than or equal to. This tells us that the shading is under the line. Under the line, y goes towards negative infinity, which makes y less than the line. Above the line, y goes towards positive infinity, which makes y greater than the line. That would make the inequality y≥x²-4x+2. Therefore, A is the correct answer.
12.5%
15%
20%
The percent error matching with its scenario :
1) 15%
2)20%
3)10%
4)12.5%
Here, from the given information we get:
the percent error are:
Leo = (11.5-10)/10 = 0.15
Candice = (13.2-12)/ 12 =0.1
Josie = (24-20)/20 = 0.2
Lenny = (18-16)/16 = 0.125
Now, we have,
For percentage, multiply each number by 100;
Leo = 15%
Candice = 10%
Josie = 20%
Lenny = 12.5%
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Answer:
1) 15%
2)20%
3)10%
4)12.5%
Answer:
Present value is $993.47
Step-by-step explanation:
PV = present value
Fv = future value = $1,600
Discount (i) = 10%
N = Years = 5
The formula for this is given by:
PV = FV/(1 + i)^N
PV = $1600/(1 + 0.10)^5
PV = $1600/1.1^5
PV = $1600/1.61051
PV = $993.47
Answer:
y = -7/5x - 6.7
Step-by-step
y = -7/5x + 6. A Line that is parallel will always have the same slope. The slope is m. m in this situation = -7/5x. A lines equation is y = mx + b. m = -7/5. y = -7/5x + b. Now to find b we can substitute the given point which goes through the new line, (2, -6). In this point x =2 and y = -6. Now substitute the x and y values into our equation. y = -7/5x + b is now -6 = -7/5(2) + b. -7/5(2) = -7/10. -6 = 7/10 + b. Subtract 7/10 from -6. Its -6 and 7/10 or -6.7 . -6.7 = b. b = -6.7. Now substitute the b value into the equation. y = -7/5x -6.7.