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>
A = 5¹⁴⁹
B = 11⁹⁹
Case 1: A>B
Result: True
Case 2: A=B
Result: False
Case 3: AResult: False
Answer:
1/6 of a cup of milk. Or at least that's my answer.
Step-by-step explanation:
Okay, we need to find out 25% of 2/3.
25% of 2/3 is 1/6.
Answer:
percent (%) of ancient prehistoric thinner than 3 is 0.0013
Step-by-step explanation:
Given data
mean = 5.1 millimeters (mm)
standard deviation = 0.7 mm
to find out
percent (%) of ancient prehistoric thinner than 3
solution
we have given mean and SD
so Z will be = x -mean / SD
Z = ( x - 5.1 ) / 0.7
so P( x < 3 ) = P ( (x - 5.1 ) / 0.7 < ( 3 - 5.1 ) / 0.7 )
P( Z = ( 3 - 5.1 ) / 1.5 )
P( Z = -3 ) = 0.0013
percent (%) of ancient prehistoric thinner than 3 is 0.0013
Answer:
949.2
Step-by-step explanation:
To solve this equation, you must use P.E.M.D.A.S...
First, solve 7x9 and 5x2...
7x9= 63 5x2 = 10
So...
1000+50-36+63+8÷10+1
Next, solve 8÷10 and get 0.8
So...
1000+50-36+63+0.8+1
Then, solve 1000 + 50, and 36+63, and that result with 0.8 and that with 1.
So...
1050-100.8
The Solve!
949.2
Hope this helps!
<em>-kiniwih426</em>
