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kondor19780726 [428]
3 years ago
15

Find the exact value of sin A in simplest radical form.​

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
5 0

Using the law of sines, we have

\dfrac{AB}{\sin(C)}=\dfrac{BC}{\sin(A)}

And we know that

AB=\sqrt{85},\quad BC=6,\quad A=90

So, the equation becomes

\sin(A)=\dfrac{BC\sin(C)}{AB}=\dfrac{6\sin(90)}{\sqrt{85}}=\dfrac{6}{\sqrt{85}}

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Ira Lisetskai [31]
Just do 220 times 0.30
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Plz help me with this
xxTIMURxx [149]

It subtracts 2 then adds 5

5 0
3 years ago
Please help I don’t understand!
Olenka [21]

Answer:

16.7552

Step-by-step explanation:

The Area of a sector of a circle is θ/360 pi r²

so,

30/360 x 3.142 x 8 x 8

= 16.7551 to four decimal places is

8 0
2 years ago
Wes bought a conference table for $960. what is it worth after depreciating at a rate of 12% per year for 4 years?
murzikaleks [220]

Answer:

$594.03

Step-by-step explanation:

Using the exponential function;

A = Pe^-rt

Principal = $960

Rate r = 12% = 0.12

Time t = 4years

Substitute

A = 960e^-(0.12*)*4

A = 960e^-0.48

A = 960(0.61878)

A = 594.03

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4 0
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]

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

7 0
3 years ago
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