The solution is the set of numbers that can be written in place of
all the variables and make all of the equations true.
If you're asking for the actual numbers, there's nobody in the world
who could find those without seeing the system of equations.
2 - 2 * 0 ≤ 1
<span>2 ≤ 1 </span>
<span>4 - 2 * (-2) ≤ 1 </span>
<span>8 ≤ 1 </span>
<span>4 - 2 * (1) ≤ 1 </span>
<span>2 ≤ 1 </span>
<span>0 - 2 * 5 ≤ 1 </span>
<span>-10 ≤ 1 </span>
<span>(TRUE</span>
Answer:
200/3n
Step-by-step explanation:
It is a strangely worded question but you are dividing the paper in 3's n times so 3n will be part of the answer and you are dividing 200. So, the answer is 200/3n.
Answer:
there will be 408$ at the end of 18 months
Step-by-step explanation:
2% of 300 is 6 and 6 x 18 = 108. 108 + 300 = 408
Answer:
![E[R]](https://tex.z-dn.net/?f=E%5BR%5D)
= 99 Ω
= 2.3094 Ω
P(98<R<102) = 0.5696
Step-by-step explanation:
The mean resistance is the average of edge values of interval.
Hence,
The mean resistance, ![E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}](https://tex.z-dn.net/?f=E%5BR%5D%20%3D%20%5Cfrac%7Ba%2Bb%7D%7B2%7D%20%20%3D%20%5Cfrac%7B95%2B103%7D%7B2%7D%20%3D%20%5Cfrac%7B198%7D%7B2%7D)
= 99 Ω
To find the standard deviation of resistance, we need to find variance first.
Hence,
The standard deviation of resistance, 
= 2.3094 Ω
To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.
From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696
