When constructing an inscribed equilateral triangle, how many arcs will be drawn on the circle?
2 answers:
It can be done with 2 arcs. In the attached, I started with diameter AD, then drew arcs at B and C of the same radius as the circle from a center at A. Triangle BCD is the inscribed equilateral triangle.
Perhaps a more usual method would be to start at D, draw an arc at E, then from there at B. Resetting the compass to have radius, DB, a third arc can be drawn at C using either B or D as the center. This method would give you
.. selection A: 3 arcs
If you didn't want to reset the compass, you could draw two more arcs, say from B to A, then from A to C. This method would give you
.. selection B: 4 arcs.
Perhaps a more usual method would be to start at D, draw an arc at E, then from there at B. Resetting the compass to have radius, DB, a third arc can be drawn at C using either B or D as the center. This method would give you
.. selection A: 3 arcs
If you didn't want to reset the compass, you could draw two more arcs, say from B to A, then from A to C. This method would give you
.. selection B: 4 arcs.
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Step-by-step explanation:
Answer:
Step-by-step explanation:
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is a negative parabola with vertex at the original.
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Answer and Step-by-step explanation:
A) For the model to be a probability distribution, it has to follow two conditions:
1) The probability of each value of the discrete random variable is between, and included, 0 and 1:
2) The sum of all probabilities is 1;
In the table shown, the probabilities are from 0 to 0.3 - between 0 and 1;
Adding the probabilities: 0 + 0.3 + 0.1 + 0.3 + 0.3 = 1
Therefore, this model is a valid probability distribution model.
B) They are discrete because each value correspond to a finite number of possible values.
C) Expected value is calculated by
E(X) = 2.6
D) P(X=3) = P(3), which means probability of 3:
P(X=3) = 0.3
E) P(X<4): probabilities of values of X that are less than 4:
P(X<4) = P(0) + P(1) + P(2) + P(3)
P(X<4) = 0 + 0.3 + 0.1 + 0.3
P(X<4) = 0.7
F) P(X>0) = P(1) + P(2) + P(3) + P(4)
P(X>0) = 0.3 + 0.1 + 0.3 + 0.3
P(X>0) = 1
G) P(X=5) = P(5)
There is no probability of P(5) because the model doesn't "cover" that number.
H) P(X=0) = P(0)
P(X=0) = 0
I) The total area of any density curve is 1 because it represents all the possible values a variable can assume.