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Helga [31]
3 years ago
8

How much less would a 23 year old female pay for a $25,000 policy of 20 year life insurance (@ $2.90 per $1000) than straight li

fe (@ $15.78 per $1000)?
$662.30

$644.00

$322.00

$585.32
Mathematics
2 answers:
Yakvenalex [24]3 years ago
8 0
Hi there
number of thousands on 25000
25000/1000=25
life insurance
25×2.9
=72.5
straight life
25×15.78
=394.5
How much less
394.5−72.5
=322...answer

Hope it helps
zhenek [66]3 years ago
8 0

Answer: Third Option is correct.

Step-by-step explanation:

Since we have given that

Amount for policy = $25000

If she opt for 20 year life insurance @ $2.90 per $1000.

so, her amount of premium becomes

25000\times \frac{2.9}{1000}\\\\=\$72.50

If she opt for straight life insurance @ $15.78 per $1000,

Then, her amount of premium becomes

25000\times \frac{15.78}{1000}\\\\=\$394.50

Difference between them is given by

\$394.50-\$72.5=\$322.00

Hence, Third Option is correct.

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Artyom0805 [142]

Answer:

Range = 115$

Standard Deviation = 43.76$

Variance = 1915.142$

Option  A) The measures of variation are not very useful because when searching for a​ room, low​ prices, location, and good accommodations are more important than the amount of variation in the area.

Step-by-step explanation:

We are given the data for  prices in dollars for one night at different hotels in a certain region.

234, 160, 119, 131, 218, 207, 146, 141        

Range:

Sorted data: 119, 131, 141, 146, 160, 207, 218, 234

\text{Range} = 234-119 = 115\$

Standard Deviation:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1356}{8} = 169.5

Sum of squares of differences = 4160.25 + 90.25 + 2550.25 + 1482.25 +  2352.25 + 1406.25 + 552.25 + 812.25 = 13406

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Variance =

\sigma^2 = 1915.142\$

Measure of variance for someone searching for room:

Option  A) The measures of variation are not very useful because when searching for a​ room, low​ prices, location, and good accommodations are more important than the amount of variation in the area.

5 0
3 years ago
A certain bookstore chain has two stores, one in San Francisco and one in Los Angeles. It stocks three kinds of books: hardcover
riadik2000 [5.3K]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

\left[\begin{array}{cccc}T&H&S&P\\S&600&1300&2000\\L&400&300&400\end{array}\right] = A.

In the above matrix A, the columns refers the three type of books and the rows refers the from which stores the books are been sold.

The numbers represents the corresponding sales in the month of January.

The sale is same for the 6 months.

Hence, 6A = \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right]. This matrix 6A represents the total sales over the 6 months.

(b)

If we denote the books in stock at the starting of January by B, then

B = \left[\begin{array}{cccc}T&H&S&P\\S&1000&3000&6000\\L&1000&6000&3000\end{array}\right].

Each month, the chain restocked the stores from its warehouse by shipping 500 hardcover, 1,400 softcover, and 1,400 plastic books to San Francisco and 500 hardcover, 500 softcover, and 500 plastic books to Los Angeles.

If we represent the amount restocked books at the end of each month by another matrix C, then

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B + 5C = \left[\begin{array}{cccc}T&H&S&P\\S&1000+2500&3000+7000&6000+7000\\L&1000+2500&6000+2500&3000+2500\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right].

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