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iogann1982 [59]
3 years ago
6

Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did

she earn? Use the formula I=Prt to find your answer, where I is interest, P is principal, r is rate and t is time. Enter your solution in decimal form rounded to the nearest hundredth. For example, if your solution is 12%, you would enter 0.12.
Mathematics
1 answer:
shutvik [7]3 years ago
5 0
Easy 450=3000*3*r R=450/9000 R=0.05*100 R=5%
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castortr0y [4]

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96

Step-by-step explanation:

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Andrei [34K]

Answer:

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Step-by-step explanation:

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8 0
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Read 2 more answers
Financial Algebra 1. Unit 1: Simple and Compound Interest Quiz A..
Nikolay [14]

Answer:

$56,558.1

Step-by-step explanation:

This is a question on compound interest.

The formula to calculate the Total Amount based on compound interest is given as:

A = P( 1 + r/n) ^nt

A = Total or Final amount in the account after t years

P = Principal/ Initial amount invested=$35,000

r = Interest rate = 12%

n = compounding Frequency = daily = using 30 days in a month = 30 × 12 = 360 days

t = time in years = 4

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A = $56558.08

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8 0
3 years ago
Add and make it lowest term:3/4+9/10
stealth61 [152]

Answer:

1 13/20

Step-by-step explanation:

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4 0
2 years ago
Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4
lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

                       f(x) =  { x^2+3x+4    , x < 3

                       f(x) =  { x^2+3x+4     , x≥3

                        Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:      

                       f(x) = x^2+3x+4

- The area "A1" bounded by f(x) is given as:

                      A1 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { -3 , 3 ] = [ a , b ]:

                     A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

                     f(x) = 4x+10

- The area "A2" bounded by f(x) is given as:

                      A2 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { 3 , 4 ] = [ a , b ]:

                     A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

                     A = A1 + A2

                     A = 42 + 26

                     A = 68 unit^2

8 0
2 years ago
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