Answer:
4, 6, 1
Step-by-step explanation:
We can solve this problem using a system of equations:
1) a + b + c = 11
2) 2a + 5b + 6c = 44
3) 4a - b = 10
First, we can substitute the value of b from equation #3 into equation #1:
b = 4a - 10
a + (4a - 10) + c = 11
5a - 10 + c = 11
5a + c = 21
c = 21 - 5a
Now, we can plug the values of b and c into equation #2, as b and c are represented in terms of a:
2a + 5(4a - 10) + 6(21 - 5a) = 44
2a + 20a - 50 + 126 - 30a = 44
-8a + 76 = 44
-8a = -32
a = 4
b = 4a - 10 = 4(4) - 10 = 6
c = 21 - 5a = 21 - 5(4) = 1
D would be the best option in my opinion
Given the data below <span>representing the body mass index (bmi) values for 20 females.
17.7 33.5 26.9 22.7 22.2 29.9 23.6 18.3 27.7 23.4 19.2 25.9 22.9 37.7 31.6 28.1 44.9 31.6 25.2 23.9
From the data above, we construct a frequency distribution beginning with a lower class limit of
15.0 and use a class width of 6.0 as follows:
Class iinterval: 15.0 - 21.0 21.0 - 27.0 27.0 - 33.0 33.0 - 39.0 39.0 - 45.0
Frequency: 3 9 5 2 1
From the frequency table, it can be seen that the frequencies started low, increased to a point and then decrease. It can also be seen that the highest frequency of the data is not at the center of the distribution, so the distribution is not symetric.
Therefore, the frequency
distribution does not appear to be roughly a normal distribution, because, "although the frequency start low, increase to some maximum, then decrease, the distribution is not symmetric."</span>
The answer to your problem would be answer
Answer: i think itss A???
