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trasher [3.6K]
3 years ago
14

A miner descended 1200 feet into a mine. While in the mine, he then rose 900 feet and then descended 350 feet. What was the mine

r's net gain or loss in altitude? A. –1200 ft B. 650 ft C. –650 ft D. 50 ft
Mathematics
1 answer:
solniwko [45]3 years ago
6 0
We can just add all these numbers together because they are just addition and subtraction.

We assume the miner starts at 0 (the surface).

0 - 1200 (down) + 900 (up) - 350 (down) = -650 (C)
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Step-by-step explanation:

Hello!

The objective of the research is to compare the newly designed drug to reduce blood pressure with the standard drug to test if the new one is more effective.

Two randomly selected groups of subjects where determined, one took the standard drug (1- Control) and the second one took the new drug (2-New)

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The parameter of study is the difference between the two population means (no order is specified, I'll use New-Standard) μ₂ - μ₁

Assuming both variables have a normal distribution, there are two options to estimate the difference between the two means using a 95% CI.

1) The population variances are unknown and equal:

[(X[bar]₂-X[bar]₁)±t_{n_1+n_2-2;1-\alpha /2}*(Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  })]

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Sa=\sqrt{\frac{(n_1-1)*S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{22*7.15^2+20*8.01^2}{42} }= 7.57

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2) The population variables are unknown and different:

Welche's approximation:

[(X[bar]₂-X[bar]₁)±t_{Dfw;1-\alpha /2}*( \sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} })]

Df_{w}= \frac{(\frac{S_1^2}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S_1^2}{n_1} )^2}{n_1-1}+ \frac{(\frac{S_2^2}{n_2} )^2}{n_2-1}  } =  \frac{(\frac{7.15^2}{23} +\frac{8.01^2_2}{21} )^2}{\frac{(\frac{7.15^2}{21} )^2}{20}+ \frac{(\frac{8.01^2}{23} )^2}{22}  } = 42.85= 42

t_{Df_w;1-\alpha /2}= t_{42; 0.975}=  2.018

[(23.48-18.52)±2.018\sqrt{\frac{7.15^2}{23} +\frac{8.01^2}{21} }]

[0.324; 9.596]

I hope this helps!

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