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svlad2 [7]
3 years ago
14

A 30-year-old female purchased a 20-Year Endowment insurance policy at the age of 23. The face value of the policy was $78,250.

She asks her insurance agent for the extended term nonforfeiture option. Given the table below and the permanent insurance amount of $37.80 per thousand, determine the annual premium and the Extended Term.
A 5-column table with 4 rows titled 20-year endowment options. Column 1 is labeled End of year with entries 7, 10, 15, 20. Column 2 is labeled option 1 cash value (dollars) with entries 226, 364, 687, 1000. Column 3 is labeled Option 2 Reduced paid-up insurance (dollars) with entries 421, 562, 834, 1000. Column 4 is labeled option 3 extended term years with entries 26, 31, 37, life. Column 5 is labeled option 3 extended term days with entries 10, 182, 50, life.
a.
Annual Premium of $2,532.10; 31 years and 182 days
c.
Annual Premium of $3,210.23; 37 years and 50 days
b.
Annual Premium of $2,957.85; 26 years and 10 days
d.
Annual Premium of $3,298.54; Life
Mathematics
2 answers:
FrozenT [24]3 years ago
8 0

Answer:

B

Step-by-step explanation:

Annual Premium of $2,957.85; 26 years and 10 days

Igoryamba3 years ago
8 0

Answer:

B). Annual premium of $2957.85; 26 years and 10 days

Step-by-step explanation:

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4 0
3 years ago
The center of the inscribed circle of Triangle ABC is point , and the center of the circumscribed circle of Triagnle ABC is poin
quester [9]

Answer:

Part A: The center of The inscribed circle is the point S

Part B: The center of The circumscribed circle is the point P

Step-by-step explanation:

Part A: Find the center of The inscribed circle of ΔABC

The inscribed circle will touch each of the three sides of the triangle in exactly one point,

The center of the inscribed circle is the point of intersection between the angle bisectors of the triangle.

<u>So, </u>According to the previous definition:

A₁A₂ and B₁B₂ are the angle bisectors of ∠A and ∠B

So, the inter section between them is the center of <u>The inscribed circle</u>

<u>So, the center of The inscribed circle is the point S</u>

=====================================================

Part B: Find the center of The circumscribed circle of ΔABC

The circumscribed circle is the circle that passes through all three vertices of the triangle.

The center of the circumscribed circle is the point of intersection between the perpendicular bisectors of the sides.

<u>So,</u> According to the previous definition:

P₁P₂ and Q₁Q₂ are the perpendicular bisectors of AB and BC

So, the inter section between them is the center of <u>The circumscribed circle</u>

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5 0
3 years ago
Probability. <br><br> CORRECT answer gets brainliest.
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The probability of spinning on yellow would be 1/3 because this is broken into 3 pieces and that would be the probability of something that has been broken into 3 pieces, and 1/3 in decimal form would be 0.33.

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Therefore your answers are,

Blue=0.5

Yellow=0.33

3 0
3 years ago
Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

5 0
3 years ago
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