Answer:
Step-by-step explanation:
<u>Simplify the numerator:</u>
- (2x - y)/(2x + y) + (2x + y)/(3y - 6x) + 8xy/(12x² - 3y²) =
- (2x - y)/(2x + y) - (2x + y)/3(2x - y) + 8xy/3(2x + y)(2x - y) =
- [3(2x - y)² - (2x + y)² + 8xy] / [3(2x + y)(2x - y)] =
- [12x² - 12xy + 3y² - 4x² - 4xy - y² + 8xy] / [3(2x + y)(2x - y)] =
- [8x² - 8xy + 2y²] / [3(2x + y)(2x - y)] =
- 2{2x - y)² / [3(2x + y)(2x - y)] =
- 2(2x - y) / [3(2x + y)]
<u>Simplify the denominator:</u>
- (4x²y - 2xy²) / (6x + 3y) =
- 2xy(2x - y) / [3(2x + y)]
<u>Now simplify the remainder of the expression:</u>
- 2(2x - y) / [3(2x + y)] × [3(2x + y)]/[2xy(2x - y}] =
- 1/xy
4 and 1.5
4x1.5=6
4+1.5=5.5
Answer:
100+70+8+0.2+0.05. is the answer
Answer:
D. The 95% confidence interval is (3.4102, 4.1898). We are 95% confident that the true population mean potassium level for professional wrestlers will be between 3.4102 millimoles per liter and 4.1898 millimoles per liter.
Step-by-step explanation:
Hello!
The variable if interest is X: potassium level of a professional wrestler (mmol/L)
A sample of n= 52 professional wrestlers was taken and a sample mean of X[bar]= 3.8 mmol/L and a sample standard deviation of S= 1.4 mmol/L
You have to estimate the population average potassium level using a 95% confidence interval. Assuming that the variable has a normal distribution, and since the population variance is unknown, I'll use a student t to construct the interval:
[X[bar] ±
*
]

[3.8±2.008*
]
[3.4102; 4.1898]
With a 95% confidence level you'd expect that the interval [3.4102; 4.1898]mmol/L will contain the average potassium level of professional wrestlers.
I hope this helps!