Answer:
4.
- this shows that the height of 13 year old females in one class is less variable than the other class.
- this means that the height of students in class with range 6 may vary from person to person while that in class with range 3 maybe clustered around one another.
- we're expected to see students of almost same height in class ,with range 3, more than the students in class with range 6.
5.
arranging the values in ascending order
2, 2, 4, 4, 4, 5, 6, 6, 6, 8, 8, 9
median = 1/ 2 [ n/2th term +{ (n/2)+ 1 }th term ]
for n is even
= 1/ 2 [ 6th + 7th term]
= 1/ 2 [5 + 6]
= 5.5
mode = value with highest frequency or the value that appears most frequently In a given set of data.
mode =4, 6
because both 4and 6 appear 3 times in the given set of data which is the highest frequency.
1. Downward
2. (-4, -3)
3. No x-intercept, y-intercept is at (0, -7)
4. x = -4
The correct answer would be 23% because 28+34+40+22=124, and if you do the butterfly method you would multiply 28 by 100 and you get 2800 then divide that by 124 and you get 22.5% but since you have to round to the nearest percent it's 23%
Answer:
1/5[8x³ –7x² –21x + 18]
Step-by-step explanation:
Let the polynomial be:
ax³ + bx² + cx + d
Where:
a => is the coefficient of x³
b => is the coefficient of x²
c => is the coefficient of x
d => is the constant term.
From the question given above,
Coefficient of x² (b) = a – 3 ..... (1)
Coefficient of x (c) = 3b ...... (2)
Constant term (d) = 2 + a ..... (3)
Sum of coefficient (a + b + c) = –4
a + b + c = –4
b = a – 3
c = 3b = 3(a – 3)
a + b + c = –4
a + (a – 3) + 3(a – 3) = –4
a + a – 3 + 3a – 9 = –4
Collect like terms
a + a + 3a = –4 + 3 + 9
5a = 8
Divide both side by 5
a = 8/5
Substitute the value of a into equation (1)
b = a – 3
a = 8/5
b = 8/5 – 3
b = (8 – 15)/5
b = –7/5
Substitute the value of b into equation (2)
c = 3b
b = –7/5
c = 3(–7/5)
c = –21/5
Substitute the value of a into equation (3)
d = 2 + a
a = 8/5
d = 2 + 8/5
d = (10 + 8)/5
d = 18/5
SUMMARY:
a = 8/5
b = –7/5
c = –21/5
d = 18/5
Thus, the polynomial:
ax³ + bx² + cx + d
8/5x³ + (–7/5)x² + (–21/5)x + 18/5
8/5x³ –7/5x² –21/5x + 18/5
1/5[8x³ –7x² –21x + 18]