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ivanzaharov [21]
3 years ago
14

In the first year of a 10,000 investment the interest rate was 6%. All earned interest remained invested for the second year, an

d the interest rate increased to 7%. What was the percent increase of the entire investment after two years?
Mathematics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

13.4%

Step-by-step explanation:

First year:

$10,000*6% = $600

New balance = $10,600

Second Year:

$10,600*7% = $742

$10,600+ $742 = $11,342

Total Return:

Final Balance - Initial balance

$11,342 - $10,000 = $1,342

$10,000*x ÷ $1,342

x = $1,342/$10,000

x = 0.1342

0.134 = 13.4%

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Which number sentence is true? A. –2.0 + (–0.5) = 2.5 B. –4.5 + 6.5 = –2.0 C. 3.0 + (–0.5) = –2.5 D. 5.5 + (–2.5) = 3.0
Soloha48 [4]
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The correct number sentence is D. 5.5 + (-2.5) = 3.0

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3 years ago
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Answer:

2.4 SAUSAGES

Step-by-step explanation:

18 x 2 = 36

36/ 15 = 2.4

6 0
2 years ago
Which of the following best describes the equation below?
DanielleElmas [232]

Answer:

B

Step-by-step explanation:

3x+33+6x=9x-33

9x+33=9x-33

33=-33

no solution

7 0
3 years ago
Name the three different types of proofs you saw in this lesson
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8 0
3 years ago
Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.
sergejj [24]

Answer:

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

Step-by-step explanation:

Given that;

fx(x) = { 1/5 ; 50 < x < 100

              0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

                          1 ;   x > 100

a)

n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

Therefore P ( 75 < (x) < 90) =  ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx

= 3/50³ [ -2 (100 - x ]₇₅⁹⁰

= (3 ( -20 + 50)) / 50₃

= 9 / 12500 = 0.00072

b)

f(k) (x) = nf(x)  ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k

Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)

= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

3 0
3 years ago
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