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kykrilka [37]
2 years ago
5

Find the value of x in the triangle shown below

Mathematics
2 answers:
konstantin123 [22]2 years ago
8 0

Answer:

x = 55

Step-by-step explanation:

Since 2 sides of the triangle are congruent, both 5, then the triangle is isosceles and the 2 base angles are congruent, both x , then

x + x + 70 = 180 ← sum of 3 angles = 180°

2x + 70 = 180 ( subtract 70 from both sides )

2x = 110 ( divide both sides by 2 )

x = 55

alexandr1967 [171]2 years ago
8 0

Answer:

55.52

Step-by-step explanation:

By sine rule:

\frac{ \sin x \degree}{5}  =  \frac{ \sin 70 \degree}{5.7} \\  \\  \sin x \degree = \frac{ 5 \: \sin 70 \degree}{5.7} \\  \\ \sin x \degree = \frac{ 5 \: \sin 70 \degree}{5.7} \\  \\ x \degree =  {sin}^{ - 1} ( \frac{ 5 \: \sin 70 \degree}{5.7}) \\  \\ x \degree = 55.516753493 \degree\\  \\ x = 55.52

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Q11

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3 years ago
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]

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CI = (98.11 , 98.49)

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Step-by-step explanation:

Data provided in the question:

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°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

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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}})

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or

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

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Lower limit of CI = 98.11

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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

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CI = (98.11 , 98.49)

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