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3 years ago

A computer valued at $6500 depreciates at the rate of 14.3% per year. Find the value of the computer after 3 years

Mathematics

1 answer:

Norma-Jean [14]3 years ago

6 0

After 3 years the computer will be valued at 3711.5. I'm sure you can round it up to 3712.

Hope this helps :)

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HELP PLEASE 5/12-(x-3)/6≤(x-2)/3 Solve with a step by step solution

zlopas [31]

Answer:x>3 1/6

Step-by-step explanation:

5/12-(x-3)/6 <(x-2)/3

5/12 *12-(x-3)/6*12<(x-2)/6*12

5-2(x-3)<4(x-2)

5-2x+6<4x-8

-2x+11<4x-8

-2x<4x-19

-6x<-19

-6x(-1)<-19(-1)

6x>19

x>19/6

x>3 1/6

just change all the inequality signs to a x is greater or equal to 3 1/6 and on the rest change to and < and equal sign only put > or equal sign on the last three lines

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2 years ago

In a binomial experiment, the outcome of each trial will depend on the ones before it.

ra1l [238]

The answer would be false

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3 years ago

What is the solution to y=3/7x -1?

DedPeter [7]

Every point on the line 3/7x -1 is a solution

Unless there is another line provided...

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2 years ago

Factor out the greatest common factor from 6y^2 + 27y using the Distributive Property.

gtnhenbr [62]

Answer:

3y(2y+9)

Step-by-step explanation:

you have to divide 6 and 2 then you have to use a factor of 27 to get 9

7 0

3 years ago

Read 2 more answers

A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order

Mazyrski [523]

Answer:

Feed 1, Soybean meal required = $\dfrac{2000}{3} \text{ lbs.}$

Feed 2, Corn meal required = $\dfrac{4000}{3} \text{ lbs.}$

Step-by-step explanation:

Total feed is 1 ton i.e. 2000 lbs.

Let x be the amount of Feed 1 required.

Feed 1 has $45\%$ of protein.

$\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)$

Then, amount of Feed 2 required = $(2000 - x) \text{ lbs.}$

Feed 2 has $15\%$ of protein.

$\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)$

As per question, total protein required is $25\%$ of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

$\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}$

Feed 1 required = $\dfrac{2000}{3}\text{ lbs .}$

Feed 2 required = $2000 - \dfrac{2000}{3}\\$

$\Rightarrow \dfrac{4000}{3}\text{ lbs.}$

7 0

3 years ago