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LiRa [457]
3 years ago
7

On the 1st Jan 2012 Beth invested some money into a bank account.

Mathematics
1 answer:
mixer [17]3 years ago
7 0
Account on 2014 is £17,466 when multipliled by 0.025 (interest per year) it will result to £436.65. So the money she had excluding the interest of the said year is £17,029.35 (17,466 - 436.65).
Going back to 2013, she withdraws £1000. If she has £17,029.35 in 2014, we can add £1000 to get £18, 029.35.If we take 2.5% from £18, 029.35. (18,029.35. x 0.025) the result is £450.73. Excluding the interest of year 2013, the money amounts to £17,578.62 (18,029.35 - 450.73), which is the original invested amount on her bank account from year 2012. 
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Answer:

10) It increased by appoxamately 17%.

11) 255 decreased by 20%

12) 15 increased by 66\frac{2}{3}%

13) $9,450

14) $149.8

15) $238.5

Step-by-step explanation:

10) 170 is greater than 150 by 25. 25 is 16\frac{2}{3}% of 150, so it increased by appoxamately 17%.

11) 225 is decreased by 45. 45 is 20% of 100, so 255 decreased by 20%

12) 15 is increased by 10. 10 is 66\frac{2}{3} of 15, so 15 increased by 66\frac{2}{3}%

13) 5% of 9,000 is 450. 450 + 9,000 is $9,450

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3 0
3 years ago
Which of the following are possible solutions to following system of linear inequalities:
lapo4ka [179]
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3 years ago
Chem deposited $115.00 in a savings account. Each week thereafter, he
Yanka [14]

Answer:

Part a) y=35x+115

Part b) \$1,165

Step-by-step explanation:

Part a)

Let

x -----> the number of weeks

y ----> the total amount Chem has deposited in a saving account

we know that

The equation of a line in slope intercept form is

y=mx+b

where

m is the rate or slope of the linear equation

b is the y-intercept of the linear equation or initial value (value of y when the value of x is equal to zero)

In this problem we have that

The rate or slope is equal to

m=35\frac{\$}{week} -----> amount deposited by Chem each week

The y-intercept or initial value is

b=\$115 ----> amount deposited originally in the saving account

substitute the values

y=35x+115

Part b) How much has Chem deposited 30 weeks after his initial deposit?

For x=30 weeks

substitute in the equation

y=35(30)+115

y=\$1,165

5 0
3 years ago
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