The number that is 150 more than the product of 5 and 4892 is 24700
Answer:
can you send it clearly it is blur see
I believe the answer is 8.2
Answer:
The margin of error at 90% confidence is 0.5133 gram.
Step-by-step explanation:
Margin of error (E) = (critical value × sample standard deviation) ÷ sqrt(n)
sample standard deviation = 1.5 grams
confidence level = 90% = 0.9
significance level = 1 - C = 1 - 0.9 = 0.1 = 10%
n (sample size) = 25
degree of freedom = n - 1 = 25 - 1 = 24
critical value (t) corresponding to 24 degrees of freedom and 10% significance level is 1.711
E = (1.711×1.5) ÷ sqrt(25) = 2.5665 ÷ 5 = 0.5133 gram
![\bf ln\left( \sqrt[7]{e^4} \right)\implies log_e\left( e^{\frac{4}{7}} \right)\\\\ -----------------------------\\\\ log_{{ a}}{{ a}}^x\implies x=x\impliedby \textit{let's use this cancellation rule}\\\\ -----------------------------\\\\ log_e\left( e^{\frac{4}{7}} \right)\iff \cfrac{4}{7}](https://tex.z-dn.net/?f=%5Cbf%20ln%5Cleft%28%20%5Csqrt%5B7%5D%7Be%5E4%7D%20%5Cright%29%5Cimplies%20log_e%5Cleft%28%20e%5E%7B%5Cfrac%7B4%7D%7B7%7D%7D%20%5Cright%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_%7B%7B%20%20a%7D%7D%7B%7B%20%20a%7D%7D%5Ex%5Cimplies%20x%3Dx%5Cimpliedby%20%5Ctextit%7Blet%27s%20use%20this%20cancellation%20rule%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_e%5Cleft%28%20e%5E%7B%5Cfrac%7B4%7D%7B7%7D%7D%20%5Cright%29%5Ciff%20%5Ccfrac%7B4%7D%7B7%7D)