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MAXImum [283]
3 years ago
15

A random sample of 45 students took an SAT preparation course prior to taking the SAT. The sample mean of their quantitative SAT

scores was 575 with a s.d. of 90, and the sample mean of their verbal SAT scores was 530 with a s.d. of 110. a) Construct 95% confidence intervals for the mean quantitative SAT and the mean verbal SAT scores of all students who take this course. b) Construct 95% confidence intervals for the standard deviations of the QSAT and VSAT scores of all students who take this course. c) What sample size would be needed to estimate the mean VSAT score with 95% confidence and with error of no more than 5 if it is assumed that the s.d. is no more than 110? a) Suppose the mean scores for all students who took the SAT at that time was 535 for the quantitative and 505 for the verbal? Do the means for students who take this course differ from the means for all students at the 5% level of significance?
Mathematics
1 answer:
Bingel [31]3 years ago
6 0

Answer:easy

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

x + 5/ 3x-11

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Ryan played in 3 basketball games. He scored 18 points in the first game and 12 points in the second game. If he scored 42 point
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The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?
Serhud [2]

Answer:

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

Step-by-step explanation:

Given - The circumference of the ellipse approximated by C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi }  =  \sqrt{\frac{a^{2} + b^{2} }{2} }

Squaring Both sides, we get

[\frac{C}{2\pi }]^{2}   =  [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2}  }   =  {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2}  }   =  {{a^{2} + b^{2} }

\frac{C^{2} }{2(\pi )^{2} }  = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} }  -  a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}  = b

∴ we get

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

8 0
3 years ago
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