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dem82 [27]
3 years ago
8

Help me Guess just help please

Mathematics
2 answers:
Dmitrij [34]3 years ago
6 0
Answer:

35

Step-by-step explanation:

√441 + 784 =

Add 441 and 784

√1225

Simplify

35
KengaRu [80]3 years ago
3 0
Answer: 35sqrt(5)

Explanation:

Sqrt(441 + 784)
= sqrt(1225)
= sqrt(5^3 x 7^2)
= 5 x 7 x sqrt(5)
= 35sqrt(5)

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  -2x  -2x

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4 0
4 years ago
May someone help me with this?
kirill115 [55]
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Round 89972 to 3 significant figure
mars1129 [50]

Answer:

90000

Step-by-step explanation:

Would you mind giving me brainliest?

8 0
4 years ago
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

 The mean is mathematically evaluated as

      \= 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}

8 0
3 years ago
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