Answer:
16=16^1
1/4=(1/2)^2
25= 5^2
49=7^2
16=4^2
Step-by-step explanation:
A graph that uses bars of various heights to represent the frequencies is a <u>Histogram</u>
A histogram is an approximate representation of the distribution of numerical data. The term was first introduced by Karl Pearson. To construct a histogram, the first step is to "bin" (or "bucket") the range of values—that is, divide the entire range of values into a series of intervals—and then count how many values fall into each interval.
Therefore, a graph that uses bars of various heights to represent the frequencies is a <u>Histogram</u>
Answer:
(1, 3) and (4,0)
Step-by-step explanation:
The quadratic function is 
We can rewrite in the form 
The graph is obtained by shifting the parent quadratic function 3 units right and 1 unit down to obtain the parabola shown in the attachment.
The straight line also have equation 
The slope is -1 and y-intercept is 4.
We can easily graph this straight line on the same graph as shown in the attachment.
The two graphs intersected at (4,0) and (1,3).
The solution set is therefore {(x,y)|x=4,y=0 or x=1,y=3}
Answer: The correct option is (B) less than 2.
Step-by-step explanation: Given that Y varies directly with X and has a constant rate of change of 6.
We are to find the possible value of X when the value of Y is 11.
Since Y varies directly with X, so we can write
![Y\propto X\\\\\Rightarrow Y=kx~~~~~~~~~~~~~~~(i),~~~~~~~~~\textup{[where 'k' is the constant of variation]}](https://tex.z-dn.net/?f=Y%5Cpropto%20X%5C%5C%5C%5C%5CRightarrow%20Y%3Dkx~~~~~~~~~~~~~~~%28i%29%2C~~~~~~~~~%5Ctextup%7B%5Bwhere%20%27k%27%20is%20the%20constant%20of%20variation%5D%7D)
Since the rate of change is constant and is equal to 6, so we must have

So, from equation (i), we have

Now, when Y = 11, then from equation (ii), we get

Thus, the value of X is less than 2 when Y = 11.
Option (B) is CORRECT.