Need asap suppose that quiz scores in a beginning statistics class have a mean of 6.4 with a standard deviation of 0.7. using chebyshev's theorem, state the range in which at least 88.9% of the data will reside. do not round your answers.

Given that mean of quiz scores = 6.4 and standard deviation = 0.7

And we need to use Chebyshev's theorem to find the range in which 88.9% of data will reside.

Chebyshev's theorem states that "Specifically, no more than of the distribution's values can be more than k standard deviations away from the mean".

That is

k = 3

So, we want the range of values within 3 standard deviations of mean.

Hence range is [mean -3*standard deviation, mean +3*standard deviation]

                      = [6.4 - 3*0.7 , 6.4+3*0.7]

                      = [6.4 - 2.1 , 6.4+2.1] = [4.3,8.5]

answer:      

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.