Answer: The sum that will be the upper limit of this population is 1280.
Step-by-step explanation:
Since we have given that
Initial population a₁ = 960
Common ratio = ![\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D)
So, We have to write the sum in sigma notation:
![\sum (ar^{n-1})\\\\=\sum 960(\frac{1}{4})^{n-1}\\\\](https://tex.z-dn.net/?f=%5Csum%20%28ar%5E%7Bn-1%7D%29%5C%5C%5C%5C%3D%5Csum%20960%28%5Cfrac%7B1%7D%7B4%7D%29%5E%7Bn-1%7D%5C%5C%5C%5C)
Since ![r=\frac{1}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B4%7D%3C1)
so, the sum is convergent, then,
![\sum 960(\frac{1}{4})^{n-1}=\frac{a}{1-r}=\frac{960}{1-\frac{1}{4}}=\frac{960}{\frac{3}{4}}=\frac{960\times 4}{3}=320\times 4=1280](https://tex.z-dn.net/?f=%5Csum%20960%28%5Cfrac%7B1%7D%7B4%7D%29%5E%7Bn-1%7D%3D%5Cfrac%7Ba%7D%7B1-r%7D%3D%5Cfrac%7B960%7D%7B1-%5Cfrac%7B1%7D%7B4%7D%7D%3D%5Cfrac%7B960%7D%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Cfrac%7B960%5Ctimes%204%7D%7B3%7D%3D320%5Ctimes%204%3D1280)
Hence, the sum that will be the upper limit of this population is 1280.
Answer:
0.975
Step-by-step explanation:
Answer:
a) distance = (rate) x (time) or do they want rate = distance / time
Step-by-step explanation:
the rate is 420 km / 3 hr = 140 km/hr
<span>The first group is 5 - 9, the second group is 10 - 14 and the third group is 15 - 19. We now find the number of data within the interval 15 - 19. In the given frequency distribution, 17, 15, 18, 19, 19 and 16 are within the interval 15 - 19. Therefore, the frequency of the third group is 6.</span>
A number family hope it helps