3 years ago

I NEED HELP ASAPPPPPPPPPPPP

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

7 0

The answer to your question is C

Bingel [31]3 years ago

4 0

Answer:

Step-by-step explanation:

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3 years ago

Help me please this is due in 10 mins

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A total of $6000 was invested, part of it at 3% interest and the remainder at 8%. If the total yearly interest amounted to $380,

Nastasia [14]

$ 2000 is invested at 3 % interest and $ 4000 is invested at 8 % interest

Solution:

Given that, total of $6000 was invested

Let "x" be the amount invested at 3 % interest

Then, (6000 - x) is the amount invested at 8 % interest

Given that,

The total yearly interest amounted to $380

Then, we can frame a equation as:

$x \times 3 \% + (6000 - x) \times 8 \% = 380$

Solve the above expression for "x"

$x \times \frac{3}{100} + (6000-x) \times \frac{8}{100} = 380\\\\0.03x + (6000-x) \times 0.08 = 380\\\\0.03x + 480 - 0.08x = 380\\\\-0.05x = -100\\\\\text{Divide both sides by -0.05 }\\\\x = 2000$

Thus, $ 2000 is invested at 3 % interest

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$ 4000 is invested at 8 % interest

4 0

4 years ago

so I'm pretty good at math, but i just fomt know how to do this. i mean i fo, but I don't understand what the question wants me

WINSTONCH [101]

I'd use 6 because after you multiply everything by 6 the fractions reduce to a whole number.

3 0

3 years ago

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