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vladimir2022 [97]
3 years ago
15

Geometry math question

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0

m<AFB + m<BFC + m<CFD = m<AFD

m<AFB = m<CFD = 31; m<AFD = 90

31 + m<BFC + 31 = 90

m<BFC + 62 = 90

m<BFC = 28

m<CFE = m<CFD + m<DFE

m<DFE = m<BFC = 28

m<CFE = 31 + 28

m<CFE = 59

Answer: C. 59

enot [183]3 years ago
7 0
28 because if angle afb is 31 and angle afc is equal to angle afb then angle bfc is 28 because 31+31 is 62 and 90-62=28. Angle cafe is equal to angle bfc so therefore it’s 28
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A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a
enyata [817]

Complete Question

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a sample of 18 adults.

a-1. How many of the 18 would we expect to indicate that the Treasury should continue making pennies

a-2) What is the standard deviation?

a-3) What is the likelihood that exactly 3 adults would indicate the Treasury should continue making pennies?

Answer:

a-1 \= x = 11 .52

a-2  \sigma  = 2.036

a-3 P(3) =  4.7*10^{-5}

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  18

  The proportion of adult that  felt the U.S. Treasury should continue making pennies is p =  0.64

 The proportion of adult that  feel otherwise is

q = 1- p = 1-0.64 = 0.36

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      \= x =  n * p

substituting values

  \= x = 18 * 0.64

   \= x = 11 .52

The standard deviation is mathematically represented as

       \sigma  =  \sqrt{ npq}

substituting values

      \sigma  =  \sqrt{18 * 0.64 * 0.36}

      \sigma  = 2.036

The likelihood that 3 adult would indicate  the Treasury should continue making pennies is mathematically evaluated as

    P(3) = \left  n} \atop  \right. C_3  (p)^{3} * (q)^{n-3}

Now

    \left  n} \atop  \right. C_3 =  \frac{n! }{[n-3] ! 3!}

substituting values

    \left  n} \atop  \right. C_3 =  \frac{18! }{[15] ! 3!}

    \left  n} \atop  \right. C_3 =  \frac{18 * 17 * 16 * 15! }{[15] ! (3 *2 *1 )}

   \left  n} \atop  \right. C_3 =  \frac{18 * 17 * 16 }{ (3 *2 *1 )}

   \left  n} \atop  \right. C_3 =  816    

So

    P(3) =  816 * (0.64 )^3 * (0.36 )^{18-3}

    P(3) =  4.7*10^{-5}

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