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satela [25.4K]
3 years ago
5

Tate and Anna have two children who need orthodontic treatment of braces and retainers, which insurance should they purchase?

Advanced Placement (AP)
2 answers:
madreJ [45]3 years ago
8 0
I think the insurance that they should purchase is: C. supplemental insurance

Dental insurance is not covered by Medicare, so both HMO and  Point of service is out of the option.
I think supplemental insurance will be the best option because it is the type of insurance sold by private companies that specifically cover the policies that are not covered by normal Medicare.


sammy [17]3 years ago
3 0

Answer:

c

Explanation:

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Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the
KIM [24]

Answer:

T_{2020} = 15

Explanation:

Given

T_1 = 10

T_2 = 20

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

<em>This is proved as follows;</em>

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1

<em>Assume n = 3</em>

T_{3} = \frac{T_1 + T_2}{2}

<em>Multiply both sides by 2</em>

2 * T_{3} = \frac{T_1 + T_2}{2} * 2

2T_{3} = T_1 + T_2

<em>Assume n = 4</em>

T_{4} = \frac{T_1 + T_2 + T_3}{3}

T_{4} = \frac{(T_1 + T_2) + T_3}{3}

Substitute 2T_{3} = T_1 + T_2

T_{4} = \frac{2T_3 + T_3}{3}

T_{4} = \frac{3T_3}{3}

T_{4} = T_3

Assume n = 5

T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}

T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}

Substitute 2T_{3} = T_1 + T_2 and T_{4} = T_3

T_{5} = \frac{2T_3 + T_3 +T_3}{4}

T_{5} = \frac{4T_3}{4}

T_{5} = \frac{(5-1)T_3}{5-1}

<em>Replace 5 with n</em>

T_{n} = \frac{(n-1)T_3}{n-1}

<em>(n-1) will definitely cancel out (n-1); So, we're left with</em>

T_{n} = T_3

Hence,

T_{2020} = T_3

Calculating T_3

T_{3} = \frac{10 + 20}{2}

T_{3} = \frac{30}{2}

T_{3} = 15

Recall that T_{2020} = T_3

T_{2020} = 15

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