If two fair dice (with faces numbered 1,2,3,4,5,6) are tossed together, what is the probability that the total score will be a p
1 answer:
Answer:
Step-by-step explanation:
There are non-distinct sums that can be achieved when rolling two fair sided dice.
The smallest of these sums is and the largest of these sums is
. Within this range, there exists only one perfect cube,
.
Count how many ways we can achieve a sum of 8 with two dice:
Thus the probability the total score (sum) will be a perfect cube when rolling two fair six-sided dice is equal to
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1. Unequal Chords in Equal Circles Theorem. In equal circles (or in the same circle) if two chords are unequal, the greater chord subtends the greater minor arc. (See Figure 1)
1.1 Describe it in detail. According to Figure 1 two chords are unequal, so we assume that one of them is the greater, say, the chord in red, so it is true that this one subtends the grater minor arc, that is, ∠1 > ∠2.
1.2 Proof for the theorem.
a. Draw radii OA, OB, OC, OD
b. In ΔOAB and OCD, OA = OC, OB = OD
c. So we assume that chord AB > chord CD. Being chords AB (red) and CD (blue)
d. If the statement c is true, then necessarily ∠1 > ∠2. So the theorem has been proved.
1.3 Does the theorem that you proved relate to theorems that you previously studied?
Yes, this theorem is related to Angle at the Center theorem.
1.4 How?
We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points. But there's an exception for this theorem, that is, let's name P the vertex of the angle subtended by those points, the exception says that that angle is:
But if P is either A or B then the triangle, say, ΔOPB will be the same triangle ΔAPB shown in figure 1. The same happens as if it were taken points D and C.
1.5 Were you able to apply some theorems that you proved earlier to your newfound theorem?
Yes, applying the formula of the exception we can get angles ∠1 and ∠2. So we set the point P either at A or B and the angle ∠1 is obtained. Next, we set the point P either at C or D and the angle ∠2 is obtained.
1.6 How might your theorem be used in a real-world application?
The Ferris Wheel was invented by George Washington Gale Ferris, Jr (See Figure 2). So the use of this theorem is shown in Figure 3. The greater chord (AB) subtends the greater minor arc ∠1. So the semi-arc between A and B has more cabins than the semi-arc between C and D.
2. Circle bisected by its diameter theorem. Every diameter of a circle bisects the circumference and the circle (See Figure 4)
2.1 Describe it in detail. Given circle AMBN with center O, and AB, any diameter. We want to prove that:a. that AB bisects circumference AMBNb. that AB bisects circle AMBN.
2.2 Proof for the theorem.
a. Turn figure AMB on AB as an axis until it falls upon the plane of ANB.
b. Arc AMB will coincide with arc ANB.
c. Therefore, it is true that arc AMB = arc ANB, that is, AB bisects circumference AMBN.
d. Also, figure AMB will coincide with figure ANB. Consequently, figure AMB = figure ANB, that is AB bisects circle AMBN.
2.3 Does the theorem that you proved relate to theorems that you previously studied?
Yes, this theorem is related to Angles in a Semicircle.
2.4 How?
If we take a point, say P1, which is in the semi-arc AMB and we trace a line from A to M and next from M to B, then ∠AMB = 90°. On the other hand, if we get a point P2 which is in the semi-arc ANB and we trace a line from A to N and next from N to B, then ∠ANB = 90°.
2.5 Were you able to apply some theorems that you proved earlier to your newfound theorem?
Even though this theorem can be related to Angles in a Semicircle. It wasn't necessary to use it for this purpose.
2.6 How might your theorem be used in a real-world application?
Suppose you want to cut the cake shown in Figure 5 in two halves. Then, if A is a point of an extreme of the cake and B is another extreme point, it is true (must be like this) the diameter of the circular cake bisects the circumference and the circle (referred to the cake)
1.1 Describe it in detail. According to Figure 1 two chords are unequal, so we assume that one of them is the greater, say, the chord in red, so it is true that this one subtends the grater minor arc, that is, ∠1 > ∠2.
1.2 Proof for the theorem.
a. Draw radii OA, OB, OC, OD
b. In ΔOAB and OCD, OA = OC, OB = OD
c. So we assume that chord AB > chord CD. Being chords AB (red) and CD (blue)
d. If the statement c is true, then necessarily ∠1 > ∠2. So the theorem has been proved.
1.3 Does the theorem that you proved relate to theorems that you previously studied?
Yes, this theorem is related to Angle at the Center theorem.
1.4 How?
We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points. But there's an exception for this theorem, that is, let's name P the vertex of the angle subtended by those points, the exception says that that angle is:
But if P is either A or B then the triangle, say, ΔOPB will be the same triangle ΔAPB shown in figure 1. The same happens as if it were taken points D and C.
1.5 Were you able to apply some theorems that you proved earlier to your newfound theorem?
Yes, applying the formula of the exception we can get angles ∠1 and ∠2. So we set the point P either at A or B and the angle ∠1 is obtained. Next, we set the point P either at C or D and the angle ∠2 is obtained.
1.6 How might your theorem be used in a real-world application?
The Ferris Wheel was invented by George Washington Gale Ferris, Jr (See Figure 2). So the use of this theorem is shown in Figure 3. The greater chord (AB) subtends the greater minor arc ∠1. So the semi-arc between A and B has more cabins than the semi-arc between C and D.
2. Circle bisected by its diameter theorem. Every diameter of a circle bisects the circumference and the circle (See Figure 4)
2.1 Describe it in detail. Given circle AMBN with center O, and AB, any diameter. We want to prove that:a. that AB bisects circumference AMBNb. that AB bisects circle AMBN.
2.2 Proof for the theorem.
a. Turn figure AMB on AB as an axis until it falls upon the plane of ANB.
b. Arc AMB will coincide with arc ANB.
c. Therefore, it is true that arc AMB = arc ANB, that is, AB bisects circumference AMBN.
d. Also, figure AMB will coincide with figure ANB. Consequently, figure AMB = figure ANB, that is AB bisects circle AMBN.
2.3 Does the theorem that you proved relate to theorems that you previously studied?
Yes, this theorem is related to Angles in a Semicircle.
2.4 How?
If we take a point, say P1, which is in the semi-arc AMB and we trace a line from A to M and next from M to B, then ∠AMB = 90°. On the other hand, if we get a point P2 which is in the semi-arc ANB and we trace a line from A to N and next from N to B, then ∠ANB = 90°.
2.5 Were you able to apply some theorems that you proved earlier to your newfound theorem?
Even though this theorem can be related to Angles in a Semicircle. It wasn't necessary to use it for this purpose.
2.6 How might your theorem be used in a real-world application?
Suppose you want to cut the cake shown in Figure 5 in two halves. Then, if A is a point of an extreme of the cake and B is another extreme point, it is true (must be like this) the diameter of the circular cake bisects the circumference and the circle (referred to the cake)