Answer:
Option 2 50 ≤ s ≤ 100
Option 5 She can make a deposit of $50
Option 6 She can make a deposit of $75
Detailed explanation:
Let
s represent the amount of money Layla puts into her savings account.
We know that
25% = 25/100 = 0.25
50% = 50/100 = 0.50
Thus
-----> 
-----> 
The compound inequality is

Check each case
case 1) 25 ≤ s ≤ 50
This statement is false
Refer to the procedure
case 2) 50 ≤ s ≤ 100
This statement is true
Refer to the procedure
case 3) s ≤ 25 or s ≥ 50
This statement is false
Because s ≤ 100 and s ≥ 50
case 4) s ≤ 50 or s ≥ 100
This statement is false
Because s ≤ 100 and s ≥ 50
case 5) She can make a deposit of $50
This statement is true
As the value of s meets the compound inequality 
case 6) She can make a deposit of $75
This statement is true
As the value of s meets the compound inequality 
I will designate the hourly rate for weekdays as x and for weekends as y. The equations are arranged as follows:
13x + 14y = $250.90
15x + 8y = $204.70
This gives us a system of equations which can be solved by multiplying the first equation by 4 and the second by -7. This leads to:
52x + 56y = $1003.60
-105x - 56y = -$1432.90
By summing these two equations, we arrive at:
-53x = -$429.30 --> 53x = $429.30 --> (dividing both sides by 53) x = 8.10. This represents her hourly wage on weekdays.
Substituting our value for x allows us to determine y. I will utilize the first equation, but either could work.
$105.30 + 14y = $250.90. To isolate y, subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40
Thus, we find that her earnings are $8.10 per hour on weekdays and $10.40 per hour on weekends. The difference shows she earns $2.30 more on weekends than on weekdays.
Response:
The problem is summarized in the following explanation segment.
Detailed explanation:
The estimate of the slots or positions lost due to simultaneous transmission attempts can be calculated as follows:
Evaluating the likelihood of transmitting gives us "p".
When considering two or more attempts, we arrive at
Fraction of slots 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)
Substituting the values yields
= ![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)
Thus, the answer appears to be correct.