You wanna find a common denominator so put the problem like this
3 1/3.< You wanna take 3 (denominator)
And times it
By 5 then you wanna take the 5. - 2 2/5.< (deniminator) and times it by 3.
____ Your common denominator is 15
now it will look like this
3 1/15
-2 2/15
———-
Now subtract. You will need to borrow 1 from 3 since you cant subtract 2 by 1 so the 3 will become a 2 and the 1 will be come 11 and here you can subtract so this will be your answer.
2 11/15
-2 2/15
-———-
0 9/15
Thats your official answer 9/15 hope I helped :)
I'm pretty sure the answer is C
Answer:1/3
Step-by-step explanation:
gradient equals
=
=
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Answer:
The sample size to obtain the desired margin of error is 160.
Step-by-step explanation:
The Margin of Error is given as
Rearranging this equation in terms of n gives
![n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2](https://tex.z-dn.net/?f=n%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2)
Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as
![n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n](https://tex.z-dn.net/?f=n_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM_2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%2F2%7D%5Cright%5D%5E2%5C%5Cn_2%3D%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B2%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D2%5E2%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4%5Cleft%5Bz_%7Bcrit%7D%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7BM%7D%5Cright%5D%5E2%5C%5Cn_2%3D4n)
As n is given as 40 so the new sample size is given as
So the sample size to obtain the desired margin of error is 160.
Hello there! The answer would be the first one, or No. At least one output results in two inputs.
When dealing with functions: you must remember that x does not repeat. So, lets look at the relation given, {(6, 1), (8, –3), (6, 7)}. You can see that x does repeat, so this is not a function. This eliminates second and fourth option choices. Out of the options A and C, A would be your choice since x is the input value and y is the output value, and there are two input values.
Hope his helps and have a great day!
