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Sliva [168]
3 years ago
8

Stanley McBride is employed by the True Blue company. The PPO annual premium is $7,058. His employer pays 70% of the total cost.

His contribution is deducted from his paycheck. What is Stanley’s weekly deduction?
Mathematics
2 answers:
Savatey [412]3 years ago
6 0

Answer:

$40.72.

Step-by-step explanation:

We have been given that The PPO annual premium is $7,058. Stanley's employer pays 70% of the total cost.

First of all, let us find the annual amount paid by Stanley for PPO. Since his employer pays 70% of the total cost, so the amount paid by Stanley will be 30% (100%-70%) of the total cost.

\text{The annual amount paid by Stanley for PPO}=\frac{30}{100}\times \$7,058

\text{The annual amount paid by Stanley for PPO}=0.30\times \$7,058

\text{The annual amount paid by Stanley for PPO}=$2117.40

To find Stanley' weekly deduction, we need to divide annual amount paid by Stanley by 52 as 1 year is equal to 52 weeks.

\text{Stanley's weekly deductions}=\frac{\$2117.40}{52}

\text{Stanley's weekly deductions}=\$40.719230769\approx \$40.72

Therefore, Stanley's weekly deduction is approximately $40.72.

Vitek1552 [10]3 years ago
4 0
First deducte his contribution
7,058−7,058×0.70
=2,117.4
Then find weekly deduction
2,117.4÷52
=40.72
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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5
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\large\underline{\sf{Solution-}}

Given that,

In <u>triangle TPQ, </u>

  • RS || PQ,

  • RS = 3 cm,

  • PQ = 6 cm,

  • ar(∆ TRS) = 15 sq. cm

As it is given that, <u>RS || PQ</u>

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)}  = \dfrac{ {PQ}^{2} }{ {RS}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{ {6}^{2} }{ {3}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{36 }{9}

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

\rm\implies \:ar( \triangle \: TPQ)  = 60 \:  {cm}^{2}

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