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icang [17]
3 years ago
9

Nancy had $2,848.52 in a savings account with simple interest. She had opened the account with $2,006 exactly 3 years earlier. W

hat was the interest rate?
Mathematics
1 answer:
Elanso [62]3 years ago
7 0

Answer:

  14%

Step-by-step explanation:

A = P(1 +rt) . . . . . . . . . . . . . . . . . . formula for the balance for simple interest

2848.52 = 2006(1 +3r) . . . . . . . filling in the given numbers

1.42 = 1 +3r . . . . . . . . . . . . . . . . . .divide by 2006

0.42 = 3r . . . . . . . . . . subtract 1

0.14 = r = 14%

Nancy's account paid interest at 14% per year.

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The product of a number and 16 is equal to the sum of the number and 14​
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Step-by-step explanation:

let number be x

16x = 14 + x

16x - x = 14

15x = 14

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On a standardized test with a normal distribution the mean score was 67.2. The standard deviation was 4.6. What percent of the d
Reika [66]

Answer:

P ( -1 < Z < 1 ) = 68%

Step-by-step explanation:

Given:-

- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :

                         u = 67.2

                         s.d = 4.6

- The random variable (X) that denotes standardized test scores following normal distribution:

                         X~ N ( 67.2 , 4.6^2 )

Find:-

What percent of the data fell between 62.6 and 71.8?

Solution:-

- We will first compute the Z-value for the given points 62.6 and 71.8:

                          P ( 62.6 < X < 71.8 )

                          P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )

                          P ( -1 < Z < 1 )

- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:

                          P ( -1 < Z < 1 ) = 68%

                          P ( -2 < Z < 2 ) = 95%

                          P ( -3 < Z < 3 ) = 99.7%

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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