Answer:
5 in the 85327= five thousand units
5 in the 435= five units
Step-by-step explanation:
85,327= eighty five thousand, three hundred and twenty seven
Value of five or the position if five in the above number is the thousand position
5 in the above = five thousand unit
435= four hundred and thirty five
The position of five in the above number is the unit position
Five in the above= five unit
The <em>correct answers</em> are:
5x²+70x+245 ≥ 1050; and
Yes.
Explanation:
Let x be the width of the tablet. Since the width of the TV is 7 inches more than the tablet, the width of the TV would be x+7.
The length of the TV is 5 times the width; this makes the length 5(x+7) = 5x+35.
The area of the TV would be given by
(x+7)(5x+35).
Since Andrew wants the area to be at least 1050, we set the expression greater than or equal to 1050:
(x+7)(5x+35) ≥ 1050
Multiplying this, we have:
x*5x+x*35+7*5x+7*35 ≥ 1050
5x²+35x+35x+245 ≥ 1050
Combining like terms,
5x²+70x+245 ≥ 1050
To see if 8 is a reasonable width for the tablet, we substitute 8 for x:
5(8²)+70(8)+245 ≥ 1050
5(64)+560+245 ≥ 1050
320+560+245 ≥ 1050
1125 ≥ 1050
Since this inequality is true, 8 is a reasonable width.
the data represents the heights of fourteen basketball players, in inches. 69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 8
Daniel [21]
If you would like to know the interquartile range of the new set and the interquartile range of the original set, you can do this using the following steps:
<span>The interquartile range is the difference between the third and the first quartiles.
The original set: </span>69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 82
Lower quartile: 72
Upper quartile: 76.25
Interquartile range: upper quartile - lower quartile = 76.25 - 72 = <span>4.25
</span>
The new set: <span>70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77
</span>Lower quartile: 72.5
Upper quartile: 76
Interquartile range: upper quartile - lower quartile = 76 - 72.5 = 3.5
The correct result would be: T<span>he interquartile range of the new set would be 3.5. The interquartile range of the original set would be more than the new set.</span>
