5578/68=82.029411764705882352941176470588
that would be rounded to 82.02 which is the answer
Answer: E) Mean will decrease, standard deviation will decrease
Step-by-step explanation:
Initial mean = 81
Initial standard deviation = 9
New score = 78
Taking a look at the initial mean(averahe) score, which is 81 and the new score to be added to the initial scores, the initial average is greater than the new score. Hence, this will result in a decrease in the new mean score after adding the new score of 78.
Also, taking a look at the standard deviation which is 9, we can conclude that the variability in initial scores from the mean scores is high. However, the new score of 78 and the initial mean are very close, and tend to low variation. Hence, adding the new score will lead to a decrease in variability and hence a decrease in the value of standard deviation.
Answer:
y = 4 or y = 6
Step-by-step explanation:
2log4y - log4 (5y - 12) = 1/2
2log_4(y) - log_4(5y-12) = log_4(2) apply law of logarithms
log_4(y^2) + log_4(1/(5y-12)) = log_4(/2) apply law of logarithms
log_4(y^2/(5y-12)) = log_4(2) remove logarithm
y^2/(5y-12) = 2 cross multiply
y^2 = 10y-24 rearrange and factor
y^2 - 10y + 24 = 0
(y-4)(y-6) = 0
y= 4 or y=6
The markup percentage is 45.14%
Step-by-step explanation:
The given is:
- The selling price of a box of crackers is $1.75
- You mark the crackers up to $2.54
We need to find the markup percentage
The markup percentage = 
× 100%
∵ The selling price of a box of crackers is $1.75
∴ Old = 1.75
∵ You mark the crackers up to $2.54
∴ New = 2.54
- Substitute these values in the rule above
∵ The markup percentage = 
× 100%
∴ The markup percentage = 
× 100%
∴ The markup percentage = 0.4514 × 100%
∴ The markup percentage = 45.14%
The markup percentage is 45.14%
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If one labour works 200 hours per month, the amount of hot water heaters the labour can produce is = 200 × 0.25 = 50 hot water heaters
The demand is to produce 57600 hot water heaters
The number of labourers employed is 57600 ÷ 50 = 1152 labourers
