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Marilyn's finance charge at the end of the first month will be
$991.38 × 0.199/12 = $16.44
The balance subject to the next month's finance charge will be
$991.38 +16.44 -410.00 = $597.82
The finance charge at the end of the second month will be
$597.82 × 0.199/12 = $9.91
The balance remaining after the second payment will be
$597.82 +9.91 -410.00 = $197.73
The finance charge applied at the end of the third month is
$197.73 × .199/12 = $3.28
so Marilyn can make one final payment of
$197.73 +3.28 = $201.01
to pay off the balance.
In all, Marilyn has paid 2×$410.00 +201.01 =
$1021.01 . . . . . . . . corresponds to the first choice
_____
In real life, Marilyn's credit card may not accrue any finance charge until after the first statement on which the charge appears. Thus the total cost of the purchase may be only $1004.02. The attached spreadsheet shows the beginning balance and the finance charges for each month for the two different scenarios.
$991.38 × 0.199/12 = $16.44
The balance subject to the next month's finance charge will be
$991.38 +16.44 -410.00 = $597.82
The finance charge at the end of the second month will be
$597.82 × 0.199/12 = $9.91
The balance remaining after the second payment will be
$597.82 +9.91 -410.00 = $197.73
The finance charge applied at the end of the third month is
$197.73 × .199/12 = $3.28
so Marilyn can make one final payment of
$197.73 +3.28 = $201.01
to pay off the balance.
In all, Marilyn has paid 2×$410.00 +201.01 =
$1021.01 . . . . . . . . corresponds to the first choice
_____
In real life, Marilyn's credit card may not accrue any finance charge until after the first statement on which the charge appears. Thus the total cost of the purchase may be only $1004.02. The attached spreadsheet shows the beginning balance and the finance charges for each month for the two different scenarios.
Answer:
y = ix + 6i + 4
Step-by-step explanation:
Since the line has undefined slope or indefinite slope we connote that the slope is (i)
Slope(m) of straight line = change in y ÷ change in x
Taking another point (x,y) on the line:
i =
i =
Cross multiplying gives;
y - 4 = ix + 6i
y = ix + 6i + 4