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Approximate f(0.05) use the following data and the Newton forward-difference formula x 0.0 0.2 0.4 0.6 0.8 f(x) 1.00000 1.22140 1.49182 1.82212 2.22554 b. use the Newton backward-difference formula to approximate f(0.65) c.Use stirlings formula to approximate f(0.43).

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Approximate f(0.05) use the following data and the Newton
forward-difference formula

x 0.0 0.2 0.4 0.6 0.8

f(x) 1.00000 1.22140 1.49182 1.82212 2.22554

b. use the Newton backward-difference formula to approximate f(0.65)

c.Use stirlings formula to approximate f(0.43).

Joshua graduated from college and began working in the family restaurant business. At the end of the third month of the first year he began putting R7440,00 per quarter in an individual retirement account and contributed to it for a total of ten years. The account earned interest at 11% per annum, compounded quarterly. The amount that was available to him after the ten years is

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Joshua graduated from college and began working in the family restaurant business. At the end of the third month of the first year he began putting R7440,00 per quarter in an individual retirement account and contributed to it for a total of ten years. The account earned interest at 11% per annum, compounded quarterly. The amount that was available to him after the ten years is

If the Npv of a shop is R195000 and the profitability index is 1,24375, the initial investment in the shop is

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If the Npv of a shop is R195000 and the profitability index is 1,24375, the initial investment in the shop is

If you deposit $10,000 in a bank account that pays 10% interest annually, how much will be in your account after 5 years?

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If you deposit $10,000 in a bank account that pays 10% interest
annually, how much will be in your account after 5 years?

Mrs smith invests R25000,00 in an account earning 7,5% interest per year,compounded weekly. After a number of years, she receives double the amount she invested. Determine the period under consideration. Round your answer to the nearest year

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Mrs smith invests R25000,00 in an account earning 7,5% interest per year,compounded weekly. After a number of years, she receives double the amount she invested. Determine the period under consideration. Round your answer to the nearest year

How long will it take an investment of R8 000 to mature to R15 000 at a simple interest rate of 11% per annum?

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How long will it take an investment of R8 000 to mature to R15 000 at a simple interest rate of 11% per annum?

An estate agent suspects that there is a linear relationship between the number of houses sold and the monthly loan payments. She analyses the following data over the past six months. Number of houses Monthly loan sold payments (in R1 000’s) x y 160 3,7 250 5,6 800 7,5 450 11,3 120 18,9 50 28,4 The regression line equation is [1] y = 480,89x − 13,99. [2] y = −0,016x + 17,45. [3] y = 17,45x − 0,016. [4] y = −13,99x + 480,89. [5] none of the above.

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An estate agent suspects that there is a linear relationship between the number of houses sold and the
monthly loan payments. She analyses the following data over the past six months.
Number of houses Monthly loan
sold payments (in R1 000’s)
x y
160 3,7
250 5,6
800 7,5
450 11,3
120 18,9
50 28,4
The regression line equation is
[1] y = 480,89x − 13,99.
[2] y = −0,016x + 17,45.
[3] y = 17,45x − 0,016.
[4] y = −13,99x + 480,89.
[5] none of the above.

The following figures show the profit of a greengrocer for the past five years: R360 000, R550 000, R200 000, R80 000 and R700 000. The arithmetic mean of the data is [1] R225 424. [2] R252 032. [3] R1 890 000. [4] R378 000. [5] none of the above.

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The following figures show the profit of a greengrocer for the past five years: R360 000, R550 000, R200 000,
R80 000 and R700 000. The arithmetic mean of the data is
[1] R225 424.
[2] R252 032.
[3] R1 890 000.
[4] R378 000.
[5] none of the above.

The next coupon date that follows the settlement date of a bond is 28 October 2021. The half-yearly coupon rate is 7,375%. The accrued interest equals R5,49589%. If this is a cum interest case, the settlement date for this bond is [1] 11 September 2021. [2] 14 June 2021. [3] 30 July 2021. [4] 29 August 2021. [5] none of the above.

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The next coupon date that follows the settlement date of a bond is 28 October 2021. The half-yearly coupon
rate is 7,375%. The accrued interest equals R5,49589%. If this is a cum interest case, the settlement date
for this bond is
[1] 11 September 2021.
[2] 14 June 2021.
[3] 30 July 2021.
[4] 29 August 2021.
[5] none of the above.

The equation for the present value of Bond 123 on 2021-06-17 is given by 107,55174 = da n z + 100 1 + 0,135 2 −29 . The yearly coupon rate is [1] 6,75%. [2] 7,35%. [3] 14,70%. [4] 8,55%. [5] none of the above.

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The equation for the present value of Bond 123 on 2021-06-17 is given by
107,55174 = da n z + 100
1 +
0,135
2
−29
.
The yearly coupon rate is
[1] 6,75%.
[2] 7,35%.
[3] 14,70%.
[4] 8,55%.
[5] none of the above.