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4 years ago

Wendy, a 32 year old female, bought a $130,000, 10-year life insurance policy. what is wendys annual premium? Use the table

Mathematics

2 answers:

Nataly_w [17]4 years ago

5 0

Answer:

Option A- $422.5

Step-by-step explanation:

Given : Wendy, a 32 year old female, bought a $130,000, 10-year life insurance policy.

To find : What is Wendy annual premium?

Solution :

In the table given,

Annual premium is per $1000 of coverage.

Wendy life insurance policy buy is $130,000

Annual premium per $1000 coverage Wendy buy at $130

From the table, (marked in the attached table below)

32 year old female 10 year cost is $3.25

Wendy's annual premium is $3.25\times 130$

$=\frac{325}{100}\times 130$

$=422.5$

Wendy's annual premium is $422.5

Therefore, Option A is correct.

steposvetlana [31]4 years ago

4 0

It would be A, because her annual premium rate is $3.25 for every $1,000. So 3.25x130=$422.50

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<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$ </h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

$x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and y_3=8$

The area of a triangle is= $\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]$

= $|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|$

= $|-54|$ = 54 square units

The length of AC = $\sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}$

= $\sqrt{(2-12)^{2} +(1-8)^2}$

= $\sqrt{149}$ units

Let the perpendicular distance from B to AC be = x

According To Problem

$\frac{1}{2} \times x \times \sqrt{149} = 54$

⇔ $x =\frac{108}{\sqrt{149} }$ units

Therefore the perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$

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