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Arada [10]
1 year 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]1 year 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:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be X centimeters.

X follows a normal distribution with

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  • standard deviation \sigma = 2.1.

In other words,

X\sim \text{N}(11.2, 2.1^{2}),

and the probability in question is X \le 13.

Z-score table approach:

Find the z-score of this measurement:

\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143. Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that Z will be less than or equal to the z-score in question. (In case the question is asking for the probability that Z is greater than the z-score, subtract the value from table from 1.)

P(X\le 13) = P(Z \le 0.857143) \approx 0.8051.

"Technology" Approach

Depending on the manufacturer, the steps generally include:

  • Locate the cumulative probability function (cdf) for normal distributions.
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For example, on a Texas Instruments TI-84, evaluating \text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 ) gives 0.804317.

As a result,

P(X\le 13) = 0.804317.

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

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Let Y be the number of green beans out of this 150 that are shorter than 13 centimeters. Y\sim\text{B}(150,0.804317).

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

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