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Salsk061 [2.6K]
3 years ago
14

the area of a circle is 64π ft2. what is the circumference in feet? Express your answer in terms of π

Mathematics
1 answer:
hammer [34]3 years ago
6 0

Answer:

12π ft

Step-by-step explanation:

~Refer to the attachment for the solution. :)

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IRINA_888 [86]
\tan 61=\dfrac{y}{10}\\
y=10\tan 61\\\\
\cos 55=\dfrac{z}{10 \tan 61}\\
z=10\tan 61\cos 55\\\\
\sin x=\dfrac{10\tan 61\cos 55}{12}\\
\sin x\approx\dfrac{10\cdot1.8\cdot0.57}{12}\\
\sin x\approx0.855\\
x\approx58.76^{\circ}
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Answer: <

6.12 x 10^4 < 612,000

Step-by-step explanation:

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The answer to the question is C
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Your dog is digging a huge hole in the backyard. On average he removes 32 cubic feet of
AleksAgata [21]

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The answer will be 96

Step-by-step explanation:

32×3=96

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A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty
Sophie [7]

Answer:

The  interval is  0.7187  < p < 2.421

Step-by-step explanation:

From the question we are told that

      The  sample size is  n  = 100

       The  population  proportion is p  =  0.81

       The  confidence level is  C =  98%

The level of significance is mathematically evaluated as

     \alpha  =  100 -98

    \alpha  =  2%%

    \alpha  =  0.02

Here this level of significance represented the left and the right tail

The degree of  freedom is evaluated as

     df =  n-1

substituting value  

    df =  100 - 1

     df = 99

Since we require the critical value of one tail in order to evaluate the  98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The  critical value of  \frac{\alpha}{2} and the evaluated degree of freedom is  

      t_{df , \alpha } =  t_{99 , \frac{0.02}{2}  }  = 2.33

this is obtained from the critical value table  

The standard error is mathematically evaluated as

             SE =  \sqrt{\frac{p(1-p )}{n} }  

substituting value  

           SE =  \sqrt{\frac{0.81(1-0.81 )}{100} }  

           SE = 0.0392  

The 98%  confidence interval is evaluated as

      p  - t_{df ,  \frac{\alpha }{2} } *  SE  < p <  p  + t_{df ,  \frac{\alpha }{2} }

substituting value  

     0.81  - 2.33  *  0.0392  < p <  0.81  +2.33 *  0.0392

      0.7187  < p < 2.421

     

4 0
3 years ago
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