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.
Answer:
43,300
Step-by-step explanation:
Starting with 2 since its in the hundredths and going over one place to the tenths, we see 7 <em>is </em> greater than 5 so we make the 2 a 3 and replace everything after with 0's
43,300!!
5.9 × 10^24 = 5,900,000,000,000,000,000,000,000 minus
13,000,000,000,000,000,000,000 equals
5,887,000,000,000,000,000,000,000 kg
which is the amount greater that Earth's mass is. you could write this problem as
5.9 × 10^24 - 13 × 10^21 = 5.887 × 10^21
Answer:
segments LJ and LI are congruent
Step-by-step explanation:
look for the little lines (tick marks)
similar marks mean congruent to each other
