Oliver would have a 1/1000 chance of getting $500, a 2/1000 chance of earning $300, and a 3/1000 in earning $1000. He would have a 994/1000 chance of making no money back
Mean = 29
Q1 = 19
Q2 = 26
Q3 = 38
IQR = Q3 - Q1 = 38 - 19 = 19
Answer:
The statement If ∠A ≅ ∠C not prove that Δ ABD ≅ Δ CBD by SAS ⇒ C
Step-by-step explanation:
* Lets revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and
including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ
≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles
and one side in the 2ndΔ
- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse
leg of the 2nd right angle Δ
* Lets solve the problem
- In the 2 triangles ABD , CBD
∵ AB = CB
∵ BD is a common side in the two triangles
- If AD = CD
∴ Δ ABD ≅ Δ CBD ⇒ SSS
- If BD bisects ∠ABC
∴ m∠ABD = m∠CBD
∴ Δ ABD ≅ Δ CBD ⇒ SAS
- If ∠A = ∠C
∴ Δ ABD not congruent to Δ CBD by SAS because ∠A and ∠C
not included between the congruent sides
* The statement If ∠A ≅ ∠C not prove that Δ ABD ≅ Δ CBD by SAS
<span>20, 21, 22, 24, 25, 26, 27, 28, 30. hope that helped</span>
The high-jumper's centre of mass is about two-thirds of the way up his body when he is standing or running in towards the take off point. He needs to increase his launch speed to the highest possible by building up his strength and speed, and then use his energy and gymnastic skill to raise his centre of gravity by H, which is the maximum that the formula U2=2gH will allow. Of course there is a bit more to it in practice! When a high jumper runs in to launch himself upwards he will only be able to transfer a small fraction of his best possible horizontal sprinting speed into his upward launch speed. He has only a small space for his approach run and must turn around in order to take off with his back facing the bar. The pole vaulter is able to do much better. He has a long straight run down the runway and, despite carrying a long pole, the world's best vaulters can achieve speeds of close to 10 metres per second at launch. The elastic fibre glass pole enables them to turn the energy of their horizontal motion 12MU2 into vertical motion much more efficiently than the high jumper. Vaulters launch themselves vertically upwards and perform all the impressive gymnastics necessary to curl themselves in an inverted U-shape over the bar,sending their centre of gravity as far below it as possible.
Pole vaulter
Let's see if we can get a rough estimate of how well we might expect them to do. Suppose they manage to transfer all their horizontal running kinetic energy of 12MU2 into vertical potential energy of MgH then they will raise their centre of mass a height of:
H=U22g
If the Olympic champion can reach 9 ms−1 launch speed then since the acceleration due to gravity is g=10 ms−2 we expect him to be able to raise his centre of gravity height of H=4 metres. If he started with his centre of gravity about 1.5 metres above the ground and made it pass 0.5 metres below the bar then he would be expected to clear a bar height of 1.5+4+0.5=6 metres. In fact, the American champion Tim Mack won the Athens Olympic Gold medal with a vault of 5.95 metres (or 19′614" in feet and inches) and had three very close failures at 6 metres, knowing he had already won the Gold Medal, so our very simple estimates turn out to be surprisingly accurate.
John D. Barrow is Professor of Mathematical Sciences and Director of the Millennium Mathematics Project at Cambridge University.
