Answer:
2% gain
Step-by-step explanation:
We assume the shopkeeper bought the appliances at the indicated prices, and that gain is computed on the basis of that cost price.
Since the base cost is the same for each appliance, the percentages can be added directly to find the percentage gain on 8000. However, the shopkeeper's total outlay was 16000, not 8000, so the final gain percentage is half of that total.
gain percent = (-4% + 8%)/2 = 2%
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If you want to see the actual numbers:
Loss on VCR = 4% × 8000 = 320.
Gain on TV = 8% × 8000 = 640.
Total gain on 16000 is -320 +640 = 320. As a percentage, that is ...
320/16000 × 100% = 2%
Step-by-step explanation:
The GCF of 28 and 80 is 4.
<u>Distribute 4 for 28 and 80, then solve:</u>
The total amount charged for each work can be expressed by the equation : 100 + 5x
Fixed charged per work = 100.00
Charge per page printed = 5.00
<u>Let the number pages printed be represented as x :</u>
Total amount charged for n pages can be related thus :
Fixed charge per work + (cost per page × number of pages)
100 + 5x
Therefore, the total amount charged can be expressed with the equation 100 + 5x
Learn more :brainly.com/question/18796573
To factor by grouping, you group up the 1st 2 terms together and the last 2 terms together and then factor those pieces.
(x^3 + x^2) + (x + 1)
x^2(x + 1) + 1(x +1) - now factor out the (x + 1)
(x + 1)(x^2 + 1) - final answer
Answer:
A. The value of the sample proportion is 0.4
B. The standard error of the sample proportion is 0.02619
C. 0.3487 ≤ p ≤ 0.4513
Step-by-step explanation:
The value of the sample proportion p' is calculated as:
Where x is the number of success in the sample or the number of students that use a laptop in class to take notes and n is the size of the sample or 350 students.
On the other hand, the standard error SE of the sample proportion is calculated as:
so, replacing the values, we get:
Finally, an approximate 95% confidence interval for the true proportion p is calculate as:
Where 1-α is equal to 95%, so 
is equal to 1.96. Then, replacing the values we get:
0.4 - 0.0513 ≤ p ≤ 0.4 + 0.0513
0.3487 ≤ p ≤ 0.4513
