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max2010maxim [7]
3 years ago
6

Kyle plans to invest $8500, part at 6% simple interest and the rest at 7% simple interest. What is the most that he can invest a

t 6% and still be guaranteed at least $535 in interest per year?
Mathematics
1 answer:
Travka [436]3 years ago
3 0

Answer:

Most of the money which can be invested at 6% is $6000

Step-by-step explanation:

Total amount of money = $8500

Let most of the money which is invested at 6% = x

So, most of the money which is invested at 7% = 8500 - x

Now, amount of interest which is guaranteed per year = $535

So, according to the given condition, We have :

6% of x + 7% of (8500 - x) = 535

⇒ 0.06·x + 0.07·(8500 - x) = 535

⇒ 0.06·x + 0.07  × 8500 - 0.07·x = 535

⇒ -0.01·x = -60

⇒ x = 6000

Hence, Most of the money which can be invested at 6% is $6000

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$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest e
ch4aika [34]

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102  

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = p_1  = $A

And The money invested at 6% = p_2 = $13957 - $A

Let The interest earn at 7% = I_1

And The interest earn at 6% = I_2

I_1 -  I_2 = $833.73

Let The time period = 1 year

Now,<u> From Simple Interest method</u>

Simple Interest = \dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}

Or,  I_1 = \dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}

Or,  I_1 = \dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}

And

I_2 = \dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}

Or,  I_2 = \dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}

∵  I_1 -  I_2 = $833.73

So, \dfrac{\textrm A\times \textrm 7\times \textrm 1}{100} -  \dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100} = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = \dfrac{167115}{13}

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102   . Answer

8 0
3 years ago
Is (-1,2) a solution of the system y&gt;x and x+y=&gt;
valentina_108 [34]
(-1,2) is a solution

y>x

2>-1 is correct

in this case yes

x+y≥x

 -1+2≥-1 

1≥-1 is correct: Yes

in this case no

x+y≥y

1≥2 

is not correct. No


8 0
4 years ago
In a certain residential suburb, 60% of all households get Internet service from the local cable company, 80% get television ser
Serggg [28]

Answer:

a) 0.900

b) 0.400

Step-by-step explanation:

Let I = households getting internet

T = households getting television

Then I = 60%

and T = 80%

I\cap T = 50\% (Households that get both services)

a) Households getting at least one service, I\cup T = I + T - I\cap T= 60+80-50 =90\%=0.900

b) Those getting internet only = I - I\cap T = 60 - 50 = 10\% = 0.1

Those getting television only = T - I\cap T = 80- 50=30\%=0.3

Probability of getting exactly one service = 0.1 + 0.3 = 0.400

6 0
3 years ago
Marylou buys a package of 500 jewels to decorate 4 different pairs of jeans. She uses the same number of jewels on each pair of
patriot [66]
500 divided by 4 = 125 so 
125 jewels on each pair of jeans

4 0
4 years ago
Please help ASAP!
Allushta [10]

Answer:

108

Step-by-step explanation:

6 0
3 years ago
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