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Novosadov [1.4K]
3 years ago
14

​$6300 is​ invested, part of it at 11​% and part of it at 6​%. For a certain​ year, the total yield is ​$533.00. How much was in

vested at each​ rate?
$___ was invested at 11%
Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0

Answer:

Invested at 11% : 3100

Invested at 6% : 3200

Step-by-step explanation:

let x= amount invested at 11%

let y= amount invested at 6%

we can write

Equation 1: x+y=6300

Equation2: .11x+.06y=533

Divide equation 2 by .06 to get

1.8333333x+y=8883.33333333

Subtract equation 2 from equation 1

.8333333x= 2583.33333333

x= 3100

Plug in x= 3100 into equation 1

3100+y= 6300

y= 3200

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Least to greatest -2, -3/4, -0.45, 3%, 0.36
Crank

Answer:

\large\boxed{-2,\ -\dfrac{3}{4},\ -0.45,\ 3\%,\ 0.36}

Step-by-step explanation:

We convert some numbers to decimal fractions:

-2 = -2.00

-3/4 = -0.75

-0.45

3% = 0.03

0.36

We order

-2, -0.75, -0.45, 0.03, 0.36

\dfrac{3}{4}=\dfrac{3\cdot25}{4\cdot25}=\frac{75}{100}=0.75

other method

\dfrac{3}{4}=3:4=0.75        (use a calculator)

p\%=\dfrac{p}{100}\to 3\%=\dfrac{3}{100}=0.03

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