The rubber ball, in this case, can be considered a circle with a circumference of 13 inches. So, we can find the diameter of the rubber ball using the formula
![C=\pi d](https://tex.z-dn.net/?f=C%3D%5Cpi%20d)
or
![d=\frac{C}{\pi }](https://tex.z-dn.net/?f=d%3D%5Cfrac%7BC%7D%7B%5Cpi%20%7D)
. That is
![d=\frac{C}{\pi }=\frac{13\:inches}{\pi }=\frac{13}{\pi }=4.14\:inches](https://tex.z-dn.net/?f=d%3D%5Cfrac%7BC%7D%7B%5Cpi%20%7D%3D%5Cfrac%7B13%5C%3Ainches%7D%7B%5Cpi%20%7D%3D%5Cfrac%7B13%7D%7B%5Cpi%20%7D%3D4.14%5C%3Ainches)
Therefore, the diameter is greater than 4 inches and less than 5 inches. That means the smallest box that the ball will fit into with the top on is the
5 inches box.
Answer:
......................................................
Step-by-step explanation:
Answer:
![\ln(1\cdot2^{2} \cdot3^{3}\cdot4^{4} ...\cdot10^{10} )](https://tex.z-dn.net/?f=%5Cln%281%5Ccdot2%5E%7B2%7D%20%5Ccdot3%5E%7B3%7D%5Ccdot4%5E%7B4%7D%20%20%20...%5Ccdot10%5E%7B10%7D%20%20%20%29)
Step-by-step explanation:
By logarithm rules
![\ln1+2\ln2+3\ln3+....+10\ln10\\\ln1+\ln2^2+\ln3^3+....\ln10^{10}\\\ln(1\cdot2^{2} \cdot3^{3}\cdot4^{4} ...\cdot10^{10} )](https://tex.z-dn.net/?f=%5Cln1%2B2%5Cln2%2B3%5Cln3%2B....%2B10%5Cln10%5C%5C%5Cln1%2B%5Cln2%5E2%2B%5Cln3%5E3%2B....%5Cln10%5E%7B10%7D%5C%5C%5Cln%281%5Ccdot2%5E%7B2%7D%20%5Ccdot3%5E%7B3%7D%5Ccdot4%5E%7B4%7D%20%20%20...%5Ccdot10%5E%7B10%7D%20%20%20%29)
Answer:
The answer is "NOT"
Step-by-step explanation:
- To demonstrate a similar statement we want to have yet another side or some other angle, however, the third section is not proportional.
- Its sides were equivalent in proportions (yeah sure they are) + corner by the by is not given.
- There is no triangle angle provided so, that we can not enforce the SAS argument.
- It is not evidence of similarity between ABC ≈DEF.
Sample space = 17 + 7 + 4 = 28
P(pink) = 7/28
P(orange) = 17/28
P(pink or orange) = P(pink) + P(orange)
P(pink or orange) = 7/28 + 17/28
P(pink or orange) = 24/28
P(pink or orange) = 6/7 or 0.857143 or 85.7%