Answer:
x = 10
y = 3
Step-by-step explanation:
Given
2x - 6y = 2
x + 6y = 28
Add both equations
2x + x -6y + 6y = 2 + 28
3x = 30
Divide both sides by 3
x = 30/3
x = 10
Substitute 10 for x in either equations to get y
Using equation 2 , we have
x + 6y = 28
10 + 6y = 28
Subtract 10 from both sides
10 - 10 + 6y = 28 - 10
6y = 18
Divide both sides by by 6
y = 18/6
y = 3
Therefore
x = 10
y = 3
Answer:
12.3
Step-by-step explanation:
Step 1
We find the mean
The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100
Mean = Sum of terms/Number of terms
Number of terms = 10
Mean = 61 + 67 + 81 + 83 + 87 + 88 + 89 + 90 + 98 + 100/10
Mean = 844/10
Mean = 84.4
Step 2
Standard deviation
The formula for sample standard deviation =
√(x - Mean)²/n - 1
= √[(61 - 84.4)² + (67 - 84.4)² + (81 - 84.4)² + (83 - 84.4)² + (87 - 84.4)² + (88 - 84.4)² + (89 - 84.4)² + (90 - 84.4)² + (98 - 84.4)² + (100 - 84.4)²]/10 - 1
=√ 547.56 + 302.76 + 11.56 + 1.96 + 6.76 + 12.96 + 21.16 + 31.36 + 184.96 + 243.36/10 - 1
= √1364.4/9
= √151.6
= 12.31259518
Approximately to the nearest tenth = 12.3
The standard deviation = 12.3
D = 2*3 -10 = -4
The n-th term of the sequence is
.. t[n] = 3 +d(n -1) = 7 -4n
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.
