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Aleksandr [31]
3 years ago
6

The TI-83/84 Plus calculator can be used to generate random data from a normally distributed population. The command randNorm(10

0,15,50) generates 50 values from a normally distributed population with and . One such generated sample of 50 values has a mean of 98.4 and a SD of 16.3. Assume that is known to be 15. Estimate the mean value from the random number generator using 99% confidence
Mathematics
1 answer:
postnew [5]3 years ago
3 0

Answer:

99% confidence interval for the mean value from the random number generator is a lower limit of 85.87 and an upper limit of 110.93.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 98.4

sd = 16.3

n = 15

degree of freedom (df) = n - 1 = 15 - 1 = 14

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

t-value corresponding to 14 df and 1% significance level is 2.977.

E = t×sd/√n = 2.977×16.3/√15 = 12.53

Lower limit = mean - E = 98.4 - 12.53 = 85.87

Upper limit = mean + E = 98.4 + 12.53 = 110.93

99% confidence interval for the mean is between 85.87 and 110.93

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pychu [463]

Answer:

y = - 4

Step-by-step explanation:

x = - 1 is the equation of a vertical line parallel to the y- axis

The equation of a line perpendicular to it will therefore be a horizontal line parallel to the x- axis with equation

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where c is the value of the y- coordinates the line passes through.

The line passes through (8, - 4 ) with y- coordinate - 4 , then

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vladimir2022 [97]

Answer:

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Step-by-step explanation:

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3 years ago
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A chef makes a fruit salad using pears and mangos. Pears and mangos each cost $2.25 per fruit. The chef buys x pears and y mango
GrogVix [38]

Answer:

P=2.25x-----------1

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Step-by-step explanation:

Step one:

given data

we are told that the cost of pears= $2.25 each

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Step two:

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