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Amanda [17]
3 years ago
15

There is a set of 100 obserations with a mean of 46 and a standard deviation of 0. What is the value of smallest obserstion in a

set?
Mathematics
1 answer:
VashaNatasha [74]3 years ago
6 0

Answer:

Solution = 46

Step-by-step explanation:

I believe you meant standard deviation. Standard deviation is defined as the variation of the data set, or the differences between the values in this set. In order for the standard deviation to be 0, all values should be the same.

Now if the mean is 46, the smallest possible number of each value in the data set should be 46 as well.  This is considering the mean is the average of the values, and hence any number of values in the data set being 46 will always have a mean of 46. Let me give you a demonstration -

Ex. [ 46, 46, 46 ], and, [46, 46, 46, 46, 46]\\Average = 46 + 46 + 46 / 3 = 46,\\Average = 46 + 46 + 46 + 46 + 46 / 5 = 46

As you can see, the average is 46 in each case. This proves that a data set consisting of n number of values in it, each value being 46, or any constant value for that matter, always has a mean similar to the value inside the set, in this case 46.<u><em> And, that the value of the smallest standard deviation is 46. </em></u>

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Read 2 more answers
Please explain thanks
Gre4nikov [31]
<h3><u>Answer:</u></h3>

\large\boxed{\pink{\sf \leadsto Value \ of \ LM \ is \ 11 \ units . }}

<h3><u>Step-by-step explanation:</u></h3>

A figure is given to us in which we can see two triangles one is ∆ MPL and other is ∆MPN .

<u>Figure</u><u> </u><u>:</u><u>-</u><u> </u>

\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\linethickness{0.25mm}\put(0,0){\line(1,2){2}} \put(0.001,0){\line(1,0){4}}\put(2,0){\vector(0,1){5}}\put(4,0){\line( - 1,2){2}}\put(0,-0.4){$\bf L $}\put(2,-0.4){$\bf P$}\put(4,-0.4){$\bf N $}\put(2.2,4){$\bf M $}\put(2.8, - .4){$\bf 5 $}\put(1, - .4){$\bf 5 $}\put(3.4, 2){$\bf 11 $}\put(2.3,0){\line(0,1){.3}}\put(2.3,.3){\line( - 1,0){.3}}\end{picture}

\underline{\blue{\sf In\: \triangle MPL \ \& \ \triangle MPN :- }}

\qquad \bullet LP = PN = 5 \:\:(given) \\\\\qquad \bullet MP = MP \:\:(Common) \\\\\qquad \bullet \angle MPN = \angle MPL = 90^{\circ} \:\: (given)

Hence by SAS congruence condition ,

\orange{\bf \triangle MPL \cong \triangle MPN }

Hence by cpct ( Corresponding parts of congruent triangles ) we can say that , LM = NM = 11 units .

<h3><u>Hence </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>LM</u><u> </u><u>is</u><u> </u><u>1</u><u>1</u><u> </u><u>units</u><u> </u><u>.</u></h3>
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