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Answer:
The height of the statue is 152 feet
Step-by-step explanation:
<u><em>The complete question is :</em></u>
The total height of the Statue of Liberty and its pedestal is 305 feet. This is 153 more than the height of the statue. Write and solve an equation to find the height h (in feet) of the statue.
Let
h ----> the height of the statue in feet
p ---> the height of the pedestal in feet
we know that
----> equation A
---> equation B
so
substitute equation A in equation B and solve for h
subtract 153 both sides
Answer:
a) The probabilities are:
Pr(1) = 1/4
Pr(2) = 1/4
Pr(3) = 1/4
Pr(4) = 1/4
b) Pr(sum of 6 or more on two throw) = 6/16 = 3/8.
c) If the die was fair, it should have six (6) numbers => (1,2,3,4,5,6) and the sample space from two throw will be = 36.
Hence, Pr(sum of 6 or more on two throw) = 26/36 = 13/18.
Step-by-step explanation:
a) Since the probability of a face is proportional to the number of dots (there are 1, 2, 3, and 4 dots on the respective sides), then the occurrence of any outcome in a single throw are equally likely. Hence, there is equal probability of occurrence.
By probability, we say
Pr(x) = number of favorable event/ total number of possible events
Therefore, the total number = 4, and each can occur at equal probability. This is why the answer are:
Pr(1) = 1/4
Pr(2) = 1/4
Pr(3) = 1/4
Pr(4) = 1/4
b) The sample space when the die is thrown twice = = 16 possible outcome.
Among the outcomes, those that will have 6 or more when sum are just six (6). They are:
(2,4), (3,3), (3,4), (4,2), (4,3), and (4,4)
Thus, Pr(sum of 6 or more on two throw) = 6/16 = 3/8.
Just for understanding:
Other outcome from the sample space are:
(1,1), (1,2), (1,3), (1,4)
(2,1), (2,2), (2,3), ___
(3,1), (3,2), ___, ___
(4,1), ___, ___, ___
If you observe, the dash are the ones we have listed above!
c) When the die is fair, this is what happen.
Just for understanding:
Other outcome from the sample space when the die is fair are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
The bold outcome are values when sum are 6 or more. And there are 26 of them.