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ankoles [38]
3 years ago
10

Simplify

7Bx%7D%20" id="TexFormula1" title=" \frac{ \frac{1}{x+h} + \frac{1}{x} }{x} " alt=" \frac{ \frac{1}{x+h} + \frac{1}{x} }{x} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
liubo4ka [24]3 years ago
7 0
\bf \cfrac{\frac{1}{x+h}+\frac{1}{x}}{x}\implies \cfrac{\frac{(x\cdot 1)+(1\cdot (x+h))}{x(x+h)}}{\frac{x}{1}}\implies \cfrac{\frac{x+x+h}{x(x+h)}}{\frac{x}{1}}\implies \cfrac{x+x+h}{x(x+h)}\cdot \cfrac{1}{x}
\\\\\\
\cfrac{2x+h}{x^2(x+h)}
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Answer:

$2

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Assume the general population gets an average of 7 hours of sleep per night. You randomly select 45 college students and survey
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Answer:

a) ii. This is a left-tailed test.

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d) i. reject null hypothesis

e) Option i) The data supports the claim that college students get less sleep than the general population.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 7 hours

Sample mean, \bar{x} = 6.87 hours

Sample size, n = 45

Alpha, α = 0.10

Sample standard deviation, s =  0.55 hours

First, we design the null and the alternate hypothesis

H_{0}: \mu = 7\text{ hours}\\H_A: \mu < 7\text{ hours}

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t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{6.87 - 7}{\frac{0.55}{\sqrt{45}} } =-1.59

c) Now,

t_{critical} \text{ at 0.10 level of significance, 44 degree of freedom } = -1.301

Since,                    

t_{stat} < t_{critical}

d) We fail to accept the null hypothesis and reject it.

We accept the alternate hypothesis and conclude that  mean number of hours of sleep for all college students is less than 7 hours.

e) Option i) The data supports the claim that college students get less sleep than the general population.

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