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4 years ago

What are the two essential types of insurance?

Mathematics

1 answer:

Maksim231197 [3]4 years ago

4 0

Answer:

In not sure if this is the insurence you where talking about but here you go There are two major types of life insurance—term and whole life. Whole life is sometimes called permanent life insurance, and it encompasses several subcategories, including traditional whole life, universal life, variable life and variable universal life.

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Y=6x+20 substitution

Oksana_A [137]

Answer:

-10/3 or -3.33

Step-by-step explanation:

substitute 0 for y

0 = 6x + 20

-20 = 6x + 20 - 20 subtract 20 on both sides

-20 = 6x divide by 6 on each side

-20/6 = 6x/6

x = -20/6 simplify

x = -10/3

8 0

3 years ago

A right triangle has legs with lengths 20 cm and 21 cm.

Bond [772]

to find the hypotenuse length you need to use the pythegorian therum and that will give you 29cm

7 0

3 years ago

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Is 7/3 a rational number ?

boyakko [2]

Answer: Yes

Step-by-step explanation: To determine whether 7/3 is a rational number, remember all fractions either positive or negative are rational numbers.

Therefore, 7/3 is a rational number.

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3 years ago

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Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$