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saw5 [17]
3 years ago
11

A statement of an employee's biweekly earnings is given below. Earnings Deductions Week Ended Regular FED. SOC. MED STATE WITH.

WITH. CARE. WITH. NET PAY 9/10 $_____ $98.00 $52.88 $12.37 $22.54 $667.17 What is the employee's gross pay? a. $818.05 b. $830.42 c. $840.59 d. $852.96
Mathematics
2 answers:
DiKsa [7]3 years ago
8 0

Answer:  d. $852.96


Step-by-step explanation:

Given: The net pay = $667.17

Since net pay is the amount of money your employees take home after all deductions have been taken out.

We know that gross pay is the amount of money that employees receive before any taxes and deductions are taken out.

Thus to find the gross pay, we need to add all of the deductions to the net pay as:

Gross pay=667.17+98+52.88+12.37+22.54 =\$852.96

Hence, D is the right option.

Andrews [41]3 years ago
5 0

Answer:

Option D. $852.96.

Step-by-step explanation:

Employee's biweekly Net pay = $667.17

Net pay is the pay after all deductions.

So we add all deductions to his net pay to find out his gross pay

His total withholding = 98.00 + 52.88 + 12.37 + 22.54 = $185.79

Gross pay = Total withholding + Net pay

Gross pay = 185.79 + 667.17 = $852.96

Employee's gross pay would be $852.96.

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A car holds 4 people

For 34 people :

34 ÷ 4 = 8 1/2

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Answer:  39) 1              40) 2

                41) 1              42) 0

<u>Step-by-step explanation:</u>

39)     ∠A = ?        ∠B = ?       ∠C = 129°

            a = ?          b = 15         c = 45

Use Law of Sines to find ∠B:

\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36°        ∠B = 15°       ∠C = 129°       is the only valid solution.

40)      ∠A = 16°        ∠B = ?       ∠C = ?

             a = 15           b = ?         c = 19

Use Law of Sines to find ∠C:

\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1:  ∠A = 16°        ∠B = 144°       ∠C = 20°    

Solution 2:  ∠A = 16°        ∠B = 4°       ∠C = 160°    

41)       ∠A = ?        ∠B = 75°       ∠C = ?

             a = 7           b = 30         c = ?

Use Law of Sines to find ∠A:

\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

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42)      ∠A = ?         ∠B = 119°       ∠C = ?

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Use Law of Sines to find ∠A:

\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o

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If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

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