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Arada [10]
10 months ago
11

Lisa receives a net pay of $619.06 biweekly. She has $143withheld from her pay each pay period. What is her annual gross salary?

a. $ 762.06b. $18,289.44c. $19,813.56d. $39,627.12
Mathematics
1 answer:
Butoxors [25]10 months ago
6 0

Answer:

c. $19,813.56

Explanation:

Given:

• Lisa receives a net pay of $619.06 biweekly.

,

• $143 is withheld from her pay each pay period.

We are required to find her annual gross salary.

First, determine her gross salary for each pay period.

\begin{gathered} \text{Gross Salary}=\text{Net Pay+Deduction} \\ =619.06+143 \\ =\$762.06 \end{gathered}

Next, determine the number of payment periods.

\begin{gathered} \text{Lisa is paid biwe}ekly,\text{ that is every 2 weeks.} \\ The\text{ number of weeks in a year}=52 \\ \text{Therefore:} \\ \text{The number of payment periods}=\frac{52}{2}=26 \end{gathered}

Finally, multiply her gross salary per period by the number of periods to get her annual gross salary.

\begin{gathered} \text{Gross annual salary}=26\times762.06 \\ =\$$19,813.56$ \end{gathered}

Lisa's annual gross salary is $19,813.56.

Option C is correct.

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Answer:

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Step-by-step explanation:

Data given and notation

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We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

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Alternative hypothesis:\mu_{1} > \mu_{2}

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

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For this case the value is t_{1-\alpha/2}=1.66

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Since we have all the values we can replace in formula (1) like this:

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The best option would be:

Yes, since the test statistic value is greater than the critical value.

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