Answer:
<u>There are four different steps involved in determining extraneous solutions in any of the rational equation.</u>
Step-by-step explanation:
<u>Steps involved are:</u>
1. The very step involved is to find the "common denominator".
2. The second step includes multiplying every single thing with the "common denominator".
3. The third step includes "simplifying".
4. The last and the fourth step involves checking the answer obtained in order to assure that there's no "extraneous solution".
<u>Example of an extraneous solution in a rational equation:</u>
4÷x+3 + 7÷x²+3x = 4-x÷x
Ok so the answer is
![4 {x}^{2}](https://tex.z-dn.net/?f=4%20%7Bx%7D%5E%7B2%7D%20)
so we multiply numbers that have the same variable
![2 {x}^{3} \: \: \times {2x}^{ - 1} \\ \\](https://tex.z-dn.net/?f=2%20%7Bx%7D%5E%7B3%7D%20%5C%3A%20%5C%3A%20%5Ctimes%20%7B2x%7D%5E%7B%20-%201%7D%20%5C%5C%20%5C%5C%20)
multiply 2 x 2
![2 \times 2 = 4](https://tex.z-dn.net/?f=2%20%5Ctimes%202%20%3D%204)
then multiply the exponents
![x ^{3} \times x ^{ - 1}](https://tex.z-dn.net/?f=x%20%5E%7B3%7D%20%5Ctimes%20x%20%5E%7B%20-%201%7D%20)
multipling like this will look like addition.
![{x}^{3} \times {x}^{ - 1} = {x}^{2}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B3%7D%20%5Ctimes%20%7Bx%7D%5E%7B%20-%201%7D%20%3D%20%7Bx%7D%5E%7B2%7D%20)
so in other words you could do it like this too
![{x}^{3} + {x}^{ - 1} = {x}^{2}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B3%7D%20%2B%20%7Bx%7D%5E%7B%20-%201%7D%20%3D%20%7Bx%7D%5E%7B2%7D%20)
now add the exponent with 4
![4 + {x}^{2} = 4x ^{2}](https://tex.z-dn.net/?f=4%20%2B%20%7Bx%7D%5E%7B2%7D%20%3D%204x%20%5E%7B2%7D%20)
now for the last part
![{y}^{3} + {y}^{ - 3} = 0](https://tex.z-dn.net/?f=%20%7By%7D%5E%7B3%7D%20%2B%20%7By%7D%5E%7B%20-%203%7D%20%3D%200)
because -3 and 3 added together make 0
![0 + 4 {x}^{2} = 4 {x}^{2}](https://tex.z-dn.net/?f=0%20%2B%204%20%7Bx%7D%5E%7B2%7D%20%3D%204%20%7Bx%7D%5E%7B2%7D%20)
I hope this help(it looks complicated but it's very easy and I hope I helped you out)
Answer:
<u>Alternative hypothesis 1</u>: the mean amperage at which the fuses burn out is > 40 amperes.
<u>Alternative hypothesis 2</u>: the mean amperage at which the fuses burn out is < 40 amperes.
Step-by-step explanation:
Recall that the null hypothesis is the fact you want to refute and is in doubt.
So, in this specific case, <em>the null hypothesis would be that the mean amperage at which the fuses burn out is 40 amperes.
</em>
The alternative hypothesis are those that want to refute the null hypothesis, in this case there are 2:
<u>Alternative hypothesis 1:</u> the mean amperage at which the fuses burn out is > 40 amperes.
<u>Alternative hypothesis 2:</u> the mean amperage at which the fuses burn out is < 40 amperes.
Answer:
-17
Step-by-step explanation:
-16r=272
16r=-272
r=-17
No since the between consecutive numbers doesn't give a constant