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Sergeu [11.5K]
2 years ago
12

Describe two types of insurance that you have or are likely to have at some point in your life. What risks are these insurance p

lans protecting you against? Why do you think these types of insurance are important? (4-8 sentences. 3.0 points)
Advanced Placement (AP)
1 answer:
Komok [63]2 years ago
8 0

Answer: Nowadays, it's very common for some people to have life insurances or auto insurances. Life insurances are given to a household when a primary wage earner dies, in order for the family to <u>have a financial help to deal with the absence of this primary wage</u>. It basically helps the household on <u>affording ordinary bills and mortgage expenses </u>for a certain period of time that might be hard for them. Auto insurances can protect you against accidents that might happen and cost you repairs or even a replacement. The insurance could also <u>cover medical expenses of the victims</u>, depending on what kind of agreement is made. Also, there's insurance against theft or natural disasters, that also help the owner financially to replace the car.

These types of insurance are important in order for you to plan your expenses better and to be prepared for unfortunate events that are unpredictable. Also, they can offer enough support for a family not to struggle to keep the household expenses if these events happen.

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Match the values associated with this data set to their correct descriptions. {6, 47, 49, 15, 43, 41, 7, 36}quartile38.5median11
cluponka [151]
Answers:
1) The first quartile (Q₁) = 11 ;  2) The median = 38.5 ; 
3) The third quartile (Q₃) = 45 ;
4) The difference of the largest value and the median = 10.5 .
_______
Explanation: 

Given this data set with 8 (eight) values:  →  {6, 47, 49, 15, 43, 41, 7, 36};
→Rewrite the values in increasing order; to help us find the median, first quartile (Q,) and third quartile (Q₃) : → {6, 7, 15, 36, 41, 43, 47, 49}.
→We want to find; or at least match; the following 4 (four) values [associated with the above data set] — 38.5, 11, 10, 45 ;

1) The first quartile (Q₁);  2) The median;  3) The third quartile (Q₃); & 
4) The difference of the largest value and the median.

Note: Let us start by finding the "median". This will help us find the correct values for the descriptions in "Numbers 2 & 4" above.
The "median" would be the middle number within a data set, when the values are placed in smallest to largest (or, largest to smallest).  However, our data set contains an EVEN number [specifically, "8" (eight)] values. In these cases , we take the 2 (two) numbers closest to the middle, and find the "mean" of those 2 (two) numbers; and that value obtained is the median.  So, in our case, the 2 (two) numbers closest to the middle are:
"36 & 41".  To get the "mean" of these 2 (two) numbers, we add them together to get the sum; and then, we divide that value by "2" (the number of values we are adding):
→  36 + 41 = 77;  → 77/2 = 38.5 ; → which is the median for our data set; and is a listed value.
→Now, examine Description "(#4): The difference of the largest value and the median"—(SEE ABOVE) ;
→ We can calculate this value.  We examine the values within our data set to find the largest value, "49".  Our calculated "median" for our dataset, "38.5".  So, to find the difference, we subtract: 49 − 38.5 = 10.5 ; which is a given value".
→Now, we have 2 (two) remaining values, "11" & "45"; with only 2 (two) remaining "descriptions" to match;
 →So basically we know that "11" would have to be the "first quartile (Q₁)";  & that "45" would have to be the "third quartile (Q₃)".
→Nonetheless, let us do the calculations anyway.
→Let us start with the "first quartile";  The "first quartile", also denoted as Q₁, is the median of the LOWER half of the data set (not including the median value)—which means that about 25% of the numbers in the data set lie below Q₁; & that about 75% lie above Q₁.). 
→Given our data set:   {6, 7, 15, 36, 41, 43, 47, 49};
We have a total of 8 (eight) values; an even number of values. 
The values in the LOWEST range would be:  6, 7, 15, 36.
The values in the highest range would be:  41, 43, 47, 49.
Our calculated median is: 38.5 .  →To find Q₁, we find the median of the numbers in the lower range. Since the last number of the first 4 (four) numbers in the lower range is "36"; and since "36" is LESS THAN the [calculated] median of the data set, "38.5" ; we shall include "36" as one of the numbers in the "lower range" when finding the "median" to calculate Q₁
→ So given the lower range of numbers in our data set:  6, 7, 15, 36 ;
We don't have a given "median", since we have an EVEN NUMBER of values.  In this case, we calculate the MEDIAN of these 4 (four) values, by finding the "mean" of the 2 (two) numbers closest to the middle, which are "7 & 15".  To find the mean of "7 & 15" ; we add them together to get a sum; 
then we divide that sum by "2" (i.e. the number of values added up);
   → 7 + 15 = 22 ;  → 22 ÷ 2 = 11 ;  ↔ Q₁ = 11.
Now, let us calculate the third quartile; also known as "Q₃".
    Q₃ is  the median of the last half of the higher values in the set, not including the median itself.  As explained above, we have a calculated median for our data set, of 38.5; since our data set contains an EVEN number of values.  We now take the median of our higher set of values (which is Q₃). Since our higher set of values are an even number of values; we calculate the median of these 4 (four) values by taking the mean of the 2 (two) numbers closest to the center of the these 4 (four) values.  This value is Q₃.  →Given our higher set of values:  41, 43, 47, 49 ;  → We calculate the "median" of these 4 (four) numbers; by taking the mean of the 2 (two) numbers in the middle; "43 & 47".
 → Method 1): List the integers from "43 to 47" ;  → 43, 44, 45, 46, 47;
→ Since this is an ODD number of integers in sequential order;
→ "45" is not only the "median"; but also the "mean" of (43 & 47); 
thus, 45 = Q₃; 
→ Method 2):  Our higher set of values:  41, 43, 47, 49 ;
→ We calculate the "median" of these 4 (four) numbers; by taking the
"mean" of the 2 (two) numbers in the middle; "43 & 47";  We don't have a given "median", since we have an EVEN NUMBER of values.  In this case, we calculate the MEDIAN of these 4 (four) values, by finding the mean of the 2 (two) numbers closest to the middle, which are "43 & 47."  To find the mean of "43 & 47"; we add them together to get a sum; then we divide that sum by "2" (i.e. the number of values added);
→ 43 + 47 = 90 ;  → 90 ÷ 2 = 45 ;  → 45 = Q₃ .
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navik [9.2K]

Answer:

The status of women in the Protestant Reformation was the role of wife and mother, just as the men's role was that of husband, father, or son.

Explanation:

The status of women in the Protestant Reformation was the role of wife and mother, just as the men's role was that of husband, father, or son.

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