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lisabon 2012 [21]

3 years ago

6

I need help with this problemmmm.

2 answers:

AlladinOne [14]3 years ago

3 0

Answer:

x = 1 1/6, x = 1.16

There you go! :D

stiks02 [169]3 years ago

3 0

Me multiply both side by 4 so we can remove the fraction, 12x - 5.6 = 2(3x + 0.7). Distribute 2, 12x - 5.6 = 6x + 1.4. Collect like terms, 6x = 7. Divide both sides by 6 and get x = 7/6!

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klio [65]

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Step-by-step explanation:

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3 years ago

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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

or

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

or

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

or

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

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3 years ago

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qaws [65]

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3 years ago

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