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OverLord2011 [107]
4 years ago
14

Please help answer this anyone

Mathematics
1 answer:
satela [25.4K]4 years ago
5 0

Answer:

180  dollars

Step-by-step explanation:

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<u>Answer with explanation:</u>

As per given , we have

A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72.

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E=t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

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t_{0.05/2, 49}=t_{0.025, 49}= 2.010

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=0.204664986747\approx0.20

95% confidence interval : (\overline{x}-E,\ \overline{x}+E)

i.e.\ (3.1-0.20, 3.1+0.20)=(2.90,\ 3.30)

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Class A had 33 students and class B had 21 students. A number of students were moved from class A to class B so the classes now
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Since both the length and the width are tripled, the area will be nine times larger.  To counter this, divide by nine.  108/9=12  Divide this by 2 to get one length and one width.  You get 6.  1+5=6 and 1 is 4 smaller than 5, so the width of this rectangle is 1 cm and the length is 5 cm.  
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4 years ago
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