Answer:
A.
Class interval frequency
12-14 6
15-17 7
18-20 9
21-23 9
24-26 8
Total 39
B.
Class interval frequency Percent frequency
12-14 6 0.15
15-17 7 0.18
18-20 9 0.23
21-23 9 0.23
24-26 8 0.21
Total 39 1
Step-by-step explanation:
a)
There are total 39 observation in the data. Note that sum of frequencies should be equal to the number of data values i.e. sum f=n=39.
Arranging the data values in the frequency distribution. The frequency depicts that number of times data value occurs in the data
Class interval frequency
12-14 6
15-17 7
18-20 9
21-23 9
24-26 8
Total 39
The class interval were given and frequency for each class interval depicts the number of data values falls in this interval.
b)
The class interval are given and we already found the frequencies. Now we have to find percent frequencies for each class which can be calculated by dividing the frequency of respective class to the sum of frequencies. Also, sum of percent frequencies should be equal to 1.
Class interval frequency Percent frequency
12-14 6 6/39=0.15
15-17 7 7/39=0.18
18-20 9 9/39=0.23
21-23 9 9/39=0.23
24-26 8 8/39= 0.21
Total 39 1
Answer:
Age of Lee = y years
Age of Toby = (y - 8) years
Sum of their ages < 41
or, y +y - 8 < 41
or, 2y - 8 < 41
Adding '8' on both sides,
or, 2y - 8 + 8 < 41
or, 2y < 41
It was chord less sorry if wrong
Answer:
90
Step-by-step explanation:
Evaluate x^2 + 2 x^2 + 3 x^2 + 4 x^2 where x = 3:
x^2 + 2 x^2 + 3 x^2 + 4 x^2 = 3^2 + 2×3^2 + 3×3^2 + 4×3^2
3^2 = 9:
3^2 + 2×9 + 3×3^2 + 4×3^2
3^2 = 9:
3^2 + 2×9 + 3×9 + 4×3^2
3^2 = 9:
3^2 + 2×9 + 3×9 + 4×9
3^2 = 9:
9 + 2×9 + 3×9 + 4×9
2×9 = 18:
9 + 18 + 3×9 + 4×9
3×9 = 27:
9 + 18 + 27 + 4×9
4×9 = 36:
9 + 18 + 27 + 36
| 3 |
| 3 | 6
| 2 | 7
| 1 | 8
+ | | 9
| 9 | 0:
Answer: 90
