Camping co. was organized to sell a single product that carries a 45-day warranty against defects. engineering estimates indicate that 8% of the units sold will prove defective and require an average repair cost of $ 65 per unit. during camping's first month of operations, total sales were 1, 000 units; by the end of the month, six defective units had been repaired. the liability for product warranties at month-end should be:
a) $1,700
b) $1.300
c) $1.500
d) $200
e) none of these

Liability for product warranty at month end is $4810

so correct option is E) None of these

Explanation:

given data

time = 45 days

defective =  8%

average repair cost = $65 per unit

total sales =  1,000 units

repaired = 6 unit defective

to find out

liability for product warranties at month end

solution

we know that First Month Sales units = 1,000

and First month estimated liability in units @8 % is = 80

and here Defective Units already repaired is 6

Additional liability in unit is = 80 - 6 = 74

and Additional liability @$65 per unit will be = 74 × 65 = 4810

So Liability for product warranty at month end is $4810

so correct option is E) None of these

start by isolating the variable on one side by adding 43w^2 to both sides.   the equation will read:

25w^2+5=0

*subtract 5 from both sides*

25w^2=-5

*divide both sides by 25*

w^2=-1/5

*take the square root of both sides*

w=√(-1/5)

the square root of -1 is i.   the square root of 5 remains √5

the answer is i/√5.

hope this

options ?

explanation: