Answer:
The answer is below
Step-by-step explanation:
The question is not complete, what are the coordinates of point Q and R. But I would show how to solve this.
The location of a point O(x, y) which divides line segment AB in the ratio a:b with point A at (
) and B(
) is given by the formula:
If point Q is at () and S at () and R(x, y) divides QS in the ratio QR to RS is 3:5, The coordinates of R is:
Let us assume Q(−9,4) and S(7,−4)
First, let's find out the equivalent amount of one-sixth of the total length of 8 ft.
Length of cut = 8(1/6) = 4/3 ft
So, the remaining length would be:
Remaining length = 8 ft - 4/3 ft = 20/3 ft or that's 6 and 2/3 ft.
Since there are 12 inches in 1 ft:
2/3 ft * 12 in/ft = 8 inches
Thus, the remaining length is 6 ft and 8 inches.
In order to find the percent error, we need to first find the difference between what was expected and what is actually costed. We do this by subtracting:
So now we know that the expected amount was off by $63. To find the percent error, we need to take this $63, and divide it by the amount that was estimated. Let's do that now:
However this is in decimal form. We need to multiply by 100 in order to get it in a percent:
Now we know that
the percent error of the hospital bill estimate is 13.64%.
Answer:
C None of the above
Step-by-step explanation:
The expression
(5g+3h+4)⋅2
can be expanded using distributive property as follows:
5g⋅2 + 3h⋅2 + 4⋅2 =
= 10g + 6h + 8
option A expression
(5g+3h)⋅8
can be expanded using distributive property as follows:
5g⋅8+3h⋅8 =
= 40g + 24h
which is different from 10g + 6h + 8
Option B expression
(5g+3h)⋅6
can be expanded using distributive property as follows:
5g⋅6+3h⋅6 =
= 30g + 18h
which is different from 10g + 6h + 8
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.
