The answer is 7.41717 x 10^4 because you move the decimal to the left 4 spaces
B not quiet sure but if anything try again : )
Given: Security Service Company:
1.4 1.8 1.6 1.7 1.5 1.5 1.7 1.6 1.5 1.6
Mean: 1.59
Standard Deviation: 0.014333
Other companies: 1.8 1.9 1.6 1.7 1.6 1.8 1.7 1.5 1.8 1.7
Mean: 1.71
Standard deviation: 0.014333
The coefficient of variation for security Service Company:
CV = (Standard Deviation/Mean) * 100.
= (0.14333/1.59) * 100
= 9.01%
The coefficient of variation for other companies:
CV = (Standard Deviation/Mean) * 100.
= (0.014333 / 1.71) * 100
= 8.38%
the limited data listed here show evidence of stealing by the security service company's employees because there is a significant difference in the variation.
1.)11x^2-3x
2.)8x+1-x^2
3.)-x^2-x+8
4.)-2y+10-x^2
Answer:
Step-by-step explanation:
![\ln \dfrac{4y^5}{x^2}\\\\=\ln(4y^5) - \ln(x^2)~~~~~~~~~~~;\left[ \log_b\left( \dfrac mn \right) = \log_b m - \llog_b n \right]\\\\=\ln 4 + \ln y^5 - 2\ln x~~~~~~~~~~~~;[\log_b m^n = n \log_b m ~\text{and}~\log_b(mn) = \log_b m + \log_b n ]\\\\=\ln 4 + 5 \ln y -2 \ln x\\\\=\ln 4 -2 \ln x +5 \ln y](https://tex.z-dn.net/?f=%5Cln%20%5Cdfrac%7B4y%5E5%7D%7Bx%5E2%7D%5C%5C%5C%5C%3D%5Cln%284y%5E5%29%20-%20%5Cln%28x%5E2%29~~~~~~~~~~~%3B%5Cleft%5B%20%5Clog_b%5Cleft%28%20%5Cdfrac%20mn%20%5Cright%29%20%20%3D%20%5Clog_b%20m%20-%20%5Cllog_b%20n%20%5Cright%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%20%5Cln%20y%5E5%20-%202%5Cln%20x~~~~~~~~~~~~%3B%5B%5Clog_b%20m%5En%20%3D%20n%20%5Clog_b%20m%20~%5Ctext%7Band%7D~%5Clog_b%28mn%29%20%3D%20%5Clog_b%20m%20%2B%20%5Clog_b%20n%20%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%205%20%5Cln%20y%20-2%20%5Cln%20x%5C%5C%5C%5C%3D%5Cln%204%20-2%20%5Cln%20x%20%2B5%20%5Cln%20y)
