9/16 plus 1/2 = 9 sixteenths plus one half answer

Mathematics

2 answers:

klemol [59]3 years ago

7 0

$\frac{9}{16} + \frac{1}{2} = \frac{9}{16} + \frac{8}{16} = \frac{17}{16} = 1 \frac{1}{16}$

fgiga [73]3 years ago

4 0

$\rm \frac{9}{16}+^{8)}\frac{1}{2}= \frac{9}{16}+ \frac{8}{16}=\bold{\frac{17}{16}}$

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Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Sea-hawks when they won Su

weeeeeb [17]

Answer:

No. These measures show a thinner team of NFL players according to the mean, variance, standard deviation, and quartiles.

Step-by-step explanation:

1) The measures of variation, namely The Range, Variance, Quartiles, Interquartiles, Sum of Squares, etc. shows us how the data are dispersed.

The Range Δ is calculated:

$\Delta =305-189=116$ Maximum value - Minimum value for weight

Mean:

$\bar{x}=\frac{189+254+235+225+190+305+202+190+252+305}{11}\approx 231.09 \:pounds$

Variance:

$s^{2}=\frac{\sum_{i=1}^{n} (x_i-\bar{x})^{2}}{n-1}\Rightarrow s=1923.69\\$

The Standard Deviation of the sample

$s=\sqrt{s^{^{2}}}\Rightarrow s\approx 43.86\\$

2) Since there is no preceding exercise, the comparison was made to a recent study in which a NFL player average weight is about 245 pounds (average),

Since 25% of this list are player whose weight is 192.5 lbs and 50% (2nd Quartile) =225 lbs , finally only at the 3rd Quartile we have players above the regular NFL average with 253. This, added with the other data, allow us to say that this list is not a typical of all NFL players.