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mylen [45]
3 years ago
6

please help me answer this please! I need it fast I need all three of these answers and please show your work! Thank you very mu

ch!

Mathematics
1 answer:
mars1129 [50]3 years ago
5 0
15. In a similar figure the sides are proportional. So you have to say which ratio of sides is also proportional or which sides are equal to the fraction 44/35.2

it would be c because :
44/35.2 = 132/105.6

16. same thing with similar triangles. if the length of sides are similar then so is perimeter. So so we find the ratio of the side lengths. We we know sq is 6 and qt is 24. So st is 30. So so the line st and sq are matching and similar so we compare their lengths and get a ratio
sq * x = st
6 * x = 30
x = 5

So so the perimeter is 5 times more than small triangle so
5 * 20 = 100

17.similar side is 0.5 and 3.6 so ratio is 3.6/0.5. we know she is 1.8 talk so
1.8 * 3.6/0.5 = 12.96
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Which Value Is The Eighth Term In The Sequence? <br><br> A. -3125<br> B. -125<br> C. -625
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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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