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nydimaria [60]
1 year ago
15

An employee receives a pay increase from $35,000 per year to $37,500 per year. Calculate the percent increase. Round to the near

est tenth of a percent.
Mathematics
2 answers:
arsen [322]1 year ago
8 0

Answer:

The percentage increase in salary of employee is 7.14 %  

Step-by-step explanation:

Given as :

The initial pay of employee = $35,000 per year

The increase pay of employee = $37,500 per year

Let The percentage increase in salary =  X %

So, X  = \frac{increase salary - initial salary}{initial salary}  × 100

Or, X  = \frac{$37,500 - $35,000}{$35,000}  × 100

Or,  X  = \frac{$2,500}{$35,000}  × 100

∴     X = 0.0714  × 100  = 7.14 %

Hence The percentage increase in salary of employee is 7.14 %   Answer

aleksklad [387]1 year ago
7 0

Answer:

Percentage increase = 7.14%

Step-by-step explanation:

An employee receives a pay increase from $35,000 per year to $37,500 per year. This means that the previous amount that the employee was receiving is $35,000 per year.

The amount that the employee is currently receiving is $37,500

Increase in pay =

The amount that the employee is currently receiving per year - previous amount that the employee was receiving per year. This means

Increase in pay = 37,500 - 35,000

= $2500

The percent increase = Increase in pay / previous amount that the employee was receiving per year × 100. It becomes

2500/35000 × 100

7.14%

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Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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In this problem, we have that:

\mu = 3.8, \sigma = 0.2, n = 4, s = \frac{0.2}{\sqrt{4}} = 0.1

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