Answer:
I would say A
Explanation:
Cant be D because cheaper labor would attract more companies
While lower company taxes may seem nice, it does not make up for high production costs
And infrastructure would be better, but that generally goes with higher taxes and wages.
A Perfect example is Bangladesh and other countries like it vs Europe and North America
Bangladesh has bad infrastructure, less wages, etc. but it has lower production costs because there are fewer regulations and protection plans. Therefore, more production of shoes or clothing there rather than Western civilization.
I’m going to say achievement . Why? As you try your hardest & tend to your needs for prestige and leadership opportunities, as well as opportunity for advancement , you’re moving up & your ultimate goal is to achieve that .
Answer:
Hello. You did not put the answer options, but the role of moss in the transition to stage III is the creation of protonemes where the gametes that will act in stage III will be formed.
Explanation:
Stage III consists of the fertilization of the female gamete of the moss with the male gamete of another moss, thus creating a zygote that will give rise to other mosses in the future.
Moss is very important for this phase, because it will be responsible for creating a structure called protonemas. In the protonemas are the gametophytes that will have a structure called gametangium that will allow the formation of gametes.
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 .
