4 = - 3(2) + c
4 = - 6 + c
c = 10
y = - 3x + 10
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>
The volume for any cylinder, right or oblique, would be base x height. Even though an oblique cylinder looks quite different from a right cylinder, their volumes would be equal (given that their radius and height are equal). Think about it, the area of a parallelogram would equal the area of the rectangle if their heights and bases were the same, so that would apply for this also.
V= Bh
The base would stand for that top and bottom of the cylinder, or the circles. The volume for circle is pi x radius squared.
V =
h
V =
256 x 16
V =
4096
Answer:
<h2>-3n = 27</h2>
Step-by-step explanation:
The product of a number and negative three is twenty-seven.
<em>n - </em><em>a number</em>
the product of a number and negative three: <em>n · (-3) = -3n</em>
What I like to do to solve these is to divide the 175 by 100 because that's kind of what it relates to a percent... and then I multiply by 96. So in this problem:
175/100= 1.75
1.75 x 94 and then that equals your answer 164.50 euros is the new cost of the ticket.
Hope this Helps!
P.S a thanks or brainliest would mean a lot. :)
