Answer:
false
Step-by-step explanation:
Answer: Option B.
Step-by-step explanation:
For this exercise it is important to remember that, by definition, a relation is a function if and only if each input value has an unique output value.
The input values are the values of the variable "x" and the output values are the values of the variable "y".
You can observe in the table shown in the picture, that the input value -2 has two different output values:
Notice that the input vale -1 has two different output values too:
Therefore, based on this, you can conclude that the information given in the table is a relation but not a function.
From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
Answer:
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.
Step-by-step explanation:
Given that,
The length of fencing of the rectangular plot is = 108 ft.
Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.
Since the fence along the river does not need.
So the total perimeter of the rectangle is =2(x+y) -x
=2x+2y-y
=x+2y
So,
x+2y =108
⇒x=108 -2y
Then the area of the rectangle plot is A = xy
A=xy
⇒A= (108-2y)y
⇒ A = 108y-2y²
A = 108y-2y²
Differentiating with respect to x
A'= 108 -4y
Again differentiating with respect to x
A''= -4
For maximum or minimum, A'=0
108 -4y=0
⇒4y=108
⇒y=27.
Since at y= 27, A''<0
So, at y=27 ft , the area of the rectangular plot maximum.
Then x= (108-2.27)
=54 ft.
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.