80 is the estimate that I say it's pretty close
Answer:
5 times as many
<em>BRAINLIEST, please!</em>
Step-by-step explanation:
I'd need more information about how many students there are at one of the schools. For example, if there are 200 students at Park Elementary, then there are 1000 at Lane High School since 200 x 5 = 1000. 1000 - 200 is 800, so there would be 800 more students at Lane High.
Answer:
Jessiah has 14 pencils
Miles has 28 pencils
Grace has 8 pencils
Step-by-step explanation:
I'm going to assume that you mean Jessaih instead of Jessia because otherwise it would be impossible to solve
Let the # of Jessaih's pencils equal j, # of Grace's pencils equal g, # of Miles' pencils equal m,
j=g+6
m=2j
So m=2(g+6) or 2g+12
Therefore, g+g+6+2g+12, or 4g+18=50, so g = 8
j=8+6, which is 14
m = 2(14), which is 28
Answer:
The overview of the given problem is outlined in the following segment on the explanation.
Step-by-step explanation:
The proportion of slots or positions that have been missed due to numerous concurrent transmission incidents can be estimated as follows:
Checking a probability of transmitting becomes "p".
After considering two or even more attempts, we get
Slot fraction wasted,
= ![[1-no \ attempt \ probability-first \ attempt \ probability-second \ attempt \ probability+...]](https://tex.z-dn.net/?f=%5B1-no%20%5C%20attempt%20%5C%20probability-first%20%5C%20attempt%20%5C%20probability-second%20%5C%20attempt%20%5C%20probability%2B...%5D)
On putting the values, we get
= ![1-no \ attempt \ probability-[N\times P\times probability \ of \ attempts]](https://tex.z-dn.net/?f=1-no%20%5C%20attempt%20%5C%20probability-%5BN%5Ctimes%20P%5Ctimes%20probability%20%5C%20of%20%5C%20attempts%5D)
= ![1-(1-P)^{N}-N[P(1-P)^{N}]](https://tex.z-dn.net/?f=1-%281-P%29%5E%7BN%7D-N%5BP%281-P%29%5E%7BN%7D%5D)
So that the above seems to be the right answer.
