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Blababa [14]
3 years ago
14

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]3 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]3 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>                          Question # 3</h2>

Answer:

x = 13.094

Step-by-Step Explanation

Considering the exponential equation

e^{x-9}-6=54

Solving the exponentiation equation

e^{x-9}-6=54

\mathrm{Add\:}6\mathrm{\:to\:both\:sides}

e^{x-9}-6+6=54+6

e^{x-9}=60

\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)

\ln \left(e^{x-9}\right)=\ln \left(60\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

\ln \left(e^{x-9}\right)=\left(x-9\right)\ln \left(e\right)

\left(x-9\right)\ln \left(e\right)=\ln \left(60\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1

\ln \left(e\right)=1

x-9=\ln \left(60\right)

x=\ln \left(60\right)+9

x=13.094          ∵ \ln \left(60\right)=4.094

Therefore, x = 13.094

<h2>                       Question # 4</h2>

Answer:

x=0.386

Step-by-Step Explanation

Considering the exponential equation

3e^{9x}-6=91

Solving the exponentiation equation

3e^{9x}-6=91

\mathrm{Add\:}6\mathrm{\:to\:both\:sides}

3e^{9x}=97

\mathrm{Divide\:both\:sides\:by\:}3

\frac{3e^{9x}}{3}=\frac{97}{3}

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\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)

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\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1

\ln \left(e\right)=1

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x=0.386    

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