Part (a)
The experimental or empirical probability is based on the results shown in the table. There are 13 instances of grey out of 50 spins total. Therefore, we end up with an experimental probability of 13/50. This converts to the decimal form 0.26
<h3>Answer: 0.26</h3>
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Part (b)
Since each slice is of equal size, this means theoretically each slice should have the same chance of being landed on. We have 3 grey slices out of 10 total. The probability of landing on a grey space is 3/10 = 0.3
<h3>Answer: 0.3</h3>
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Part (c)
<h3>Answer: Choice A) </h3>
As the number of spins increases, we expect the experimental and theoretical probabilities to become closer, though they might not be equal.
The theoretical probability is locked to 0.3 the whole time (only the experimental probability changes). This is according to the Law of Large Numbers.
Answer:
Exact Form:
72/5
Decimal Form:
14.4
Mixed Number Form:
14 2/5
Step-by-step explanation:
yea thats what i got
Answer:
The answer is below
Step-by-step explanation:
Natalee transferred water from a square pyramid to a cube. To calculate how many times she will need to dump the water from the pyramid into the cube to completely fill the cube, we divide the volume of the cube by the volume of the square pyramid. Hence:
Number of times = volume of cube / volume of pyramid
The perimeter of the pyramid base = 25 in, hence the length of one side of the bae = 25 / 4 = 6.25 in
Volume of square pyramid = base² × (height / 3) = (6.25 in)² * (5 in / 3) = 65.1 in³
Volume of cube = 125 in³
Number of times = 125 in³ / 65.1 in³ = 1.92
Answer:
a^3 (a^4) = a^7
(2a)(5ab) = 10a^2b
Step-by-step explanation:
Answer:
Check the explanation
Step-by-step explanation:
Ans=
A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134
M = 10: 1 – 0.9973^10 = 0.0267
M = 20: 1 – 0.9973^20 = 0.0526
M = 30: 1 – 0.9973^30 = 0.0779
M = 50: 1 – 0.9973^50 = 0.126
18)
Ans=
Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values
