Answer:
C: 1.25
Step-by-step explanation:
because 90/72 = 1.25
Do you have a calculator? you can solve it by substituting x.
y=16x^2
0: y = 16(0)^2 = 16(0) = 0
(x = 0 , y = 0)
0.5: y = 16(0.5)^2 = 16(0.25) = 4
(x = 0.5 , y = 4)
1: y = 16(1)^2 = 16(1) = 16
(x = 1 , y = 16)
1.5: y = 16(1.5)^2 = 16(2.25) = 36
(x = 1.5 , y = 36)
2: y = 16(2)^2 = 16(4) = 64
(x = 2 , y = 64)
2.5: y = 16(2.5)^2 = 16(6.25) = 100
(x = 2.5 , y = 100)
3 : y = 16(3)^2 = 16(9) = 144
(x = 3 , y = 144)
4: y = 16(4)^2 = 16(16) = 256
(x = 4 , y = 256)
if you multiply a negative number by itself, it will become positive. So, -4, -3, -2.5, -2, -1.5, -1, -0.5 will be the same as the positive 4, 3, 2.5, 2, 1.5, 1, 0.5.
I'm not sure about the pattern, but if you graph it, it'll be symmetrical across the y-axis.
The answer for this will be letter d. C = 22π and A = 121π. This is computed using the formula of C= Dπ which is C = (22)π. On the other hand, the area of this circle is computed by using the formula A = <span>πr^2. This is computed as follows:
A = </span><span>πr^2
A = </span><span>π (22/2)^2
A = </span><span>π (11)^2
</span>A = 121<span>π</span><span>
</span>
Answer:
Low Q1 Median Q3 High
6 9 11 12.5 14
The interquartile range = 3.5
Step-by-step explanation:
Given that:
Consider the following ordered data. 6 9 9 10 11 11 12 13 14
From the above dataset, the highest value = 14 and the lowest value = 6
The median is the middle number = 11
For Q1, i.e the median of the lower half
we have the ordered data = 6, 9, 9, 10
here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.
i.e
median = 
median = 
median = 9
Q3, i.e median of the upper half
we have the ordered data = 11 12 13 14
The same use case is applicable here.
Median = 
Median = 
Median = 12.5
Low Q1 Median Q3 High
6 9 11 12.5 14
The interquartile range = Q3 - Q1
The interquartile range = 12.5 - 9
The interquartile range = 3.5
Answer: i'm pretty sure the first option is correct. she wrote it down correctly
