Answer:not sure sorry
Step-by-step explanation: I need points
Answer:
a) Profit = 84000
b) at break even number of pens sold = 46
Step-by-step explanation:
Profit = Selling price - Cost price
Total revenue generated = 99 * 1000
Total revenue generated = 99000
Total cost on making the pen = 11 * 1000
Total cost on making the pen = 11000
Total cost including the initial cost = 11000 + 4000
Total cost including the initial cost = 15000
Profit = 99000 - 15000
Profit = 84000
Break even is when the cost are equal to Revenue thus no profit or loss
Revenue = total cost (break even)
9x = 1x + 4000
9x - x = 4000
8x = 4000
x = 500
At breakeven Revenue = 9 * 500
At breakeven Revenue = 4500
since one pen is sold at 99 therefore at break even number of pens sold = 4500/99 = 45.45( to 2 decimal place)
at break even number of pens sold = 46
Answer:
62.5 %
Step-by-step explanation:
you have to subtract 32-12, which equals 20
then divide 20 by the original amount (20/32)
20/32= .625, then you have to multiply to get the final answer...hope this helped!
The maximum number of hours that can be allocated in a day is 21
Data given;
- Sale = $3500
- Employee pay = $10.00 per hour
- Percentage allocated to employee = 6% of sales
<h3>Percentage</h3>
This is the process of finding a fraction of a value multiplied by 100.
Let's find the actual figure allocated to employee pay.
6% of $3500 is;
The percentage out of sale allocated for employee pay is $210
Now we know the actual amount allocated to employee pay and it is $210.
Divide the amount allocated to employee pay by $10:00 to find the actual hours we can allocate in a day.
The maximum number of hours we can allocate in a day is 21 hours.
Learn more on percentage here;
brainly.com/question/843074
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
