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kozerog [31]
3 years ago
6

There has been a trend toward less driving in the last few years, especially by young people. From 2001 to 2009 the annual vehic

le miles traveled by people from 16 to 34 years of age decreased from 10,300 to 7900 miles per person (U.S. PIRG and Education Fund website, April 6, 2012). Assume the standard deviation was 2000 miles in 2009. Suppose you would like to conduct a survey to develop a 95% confidence interval estimate of the annual vehicle-miles per person for people 16 to 34 years of age at the current time. A margin of error of 100 miles is desired. How large a sample should be used for the current survey?
Mathematics
1 answer:
rodikova [14]3 years ago
3 0

Answer: 1537.

Step-by-step explanation:

Formula for sample size :-

n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2

Given : \sigma=2000

Margin of error : E= 100

Critical value of 95% confidence : z_{\alpha/2}=1.96

Now, the required sample size will be :-

n=(\dfrac{1.96\cdot 2000}{100})^2\\\\=1536.64\approx1537

Hence, the minimum sample size required = 1537.

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-2\left(3x+2\right)\ge \:-6x-4

\mathrm{Expand\:}-2\left(3x+2\right),\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac,\\a=-2,\:b=3x,\:c=2,\\-2\cdot \:3x+\left(-2\right)\cdot \:2,\\\mathrm{Apply\:minus-plus\:rules},\\+\left(-a\right)=-a,\\-2\cdot \:3x-2\cdot \:2,\\\mathrm{Simplify}\:-2\cdot \:3x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6,\\-6x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4,\\-6x-4,\\-6x-4\ge \:-6x-4

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\mathrm{Therefore,\:the\:final\:solution\:is},\\True\quad \forall \:x\in \mathbb{R}

Or\:No\:Solution\:x\geq 0


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