6. A leather bag is bought for Rs 18000 and sold with 15% loss. Find the selling price of the bag.
2 answers:
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Answer:
Chandra's Average Monthly Expenses are;
1) For Both Jan and Feb are $269
2) For the remaining 10 months each are $244
Step-by-step explanation:
From Chandra's matrix all monthly expenses are all same that is,
Cell phone $71, Rent $1,025, Gym $75, Internet $25, Auto insurance $425, Gas $ 120, Food $145. which are all the expenses carried out every month for 12 months.
That means Chandra carries out a total of 7 expenses for the month of January and February while she carried a total of 6 expenses for the remaining 10 month which was noted from the matrix that Auto insurance was carried out only in the month of January and February.
Therefore, you start by adding up each month total expenses, which are ;
January = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886
February = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886
March = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
April = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
May = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
June = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
July= $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
August = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
September = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
October = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
November = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
December = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461
Therefore Chandra's Average Monthly Expenses are:
1) January & February = =
= 269.4286 to the nearest cent
≅ $269
2) For the remaining 10 month are = =
= 243.5 to the nearest cent
≅ $244
Not on your list, but an easy way is
.. a) swap coefficients of x and y, negating one. (Now you have 2x -3y.)
.. b) set any constant term to zero (now you have 2x -3y = 0)
.. c) translate the line to the point (5, 2) by substituting x ⇒ x-5, y⇒ y-2
2(x -5) -3(y -2) = 0
The way you've been taught, selection C is the proper choice.
.. a) swap coefficients of x and y, negating one. (Now you have 2x -3y.)
.. b) set any constant term to zero (now you have 2x -3y = 0)
.. c) translate the line to the point (5, 2) by substituting x ⇒ x-5, y⇒ y-2
2(x -5) -3(y -2) = 0
The way you've been taught, selection C is the proper choice.