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Similar to the other problem.....82  and 88 are within 1 std dev from the mean....about 68% of the bags are clustered here

So

.68 * 120  ≈  81. 6 bags  =  82 bags

One standard deviation above and below the mean encompasses 68%

68% x 97=~~66

When I was in school, we were taught to remember the standard bell curve values like the one below:

Both have the same y intercept  at y = 0

f(x) <  g(x)  on (0, 1)  is true

g(1) = 2      and  f(1)  = 2 ....so  f(1)  = g(1)

The rate of change on [0, 1 ]  is the same....both functions pass through (0, 0) and (1, 2)

First one : False

Second one    g(x) max is 5   f(x) max is 2

Third one :    g(x) x-intercept s   -2 and 2     f(x) x-intercepts  -2 and 2

Fourth one:   g(x)  y intercept (where x=0) is -4     f(x) where x= 0 is 2

-2 cos^2 = -1     divide both sides of the equation by  -2

cos^2 = 1/2

cos = +- sqrt 2 / 2          pi/4   3 pi/4    5pi/4      7 pi/4

Value at 0  = 1

Value at 2 =~~4

(1-4)/(0-2) = 3/2 = 1  1/2

Rate of change  9 to 10  =    [ 4052 - 11014 ] / [ 9 - 10 ]  =  6962

Rate of change 5 to 8  =  [ 75 - 1491 ] / [ 5 - 8 ]  =  472

So  [  6962 - 472 ]  = 

6490

Rate  of change from  a to b  =

f(a) - f(b)            log(4) - log (6)           

_______  =      _____________   =   0.088  = 0.09

  a  - b                   4  -  6 

The range consists of the set of all possible heights.....this is any height between 0 and 64, inclusive

So

[ 0, 64 ]

This is a binomial probability

P( no 3's)   =  (5/6)^10

P( one 3)  =  C(10, 1) * ( 1/6) (5/6)^9

P( two 3's)  = C(10, 2) *(1/6)^2 (5/6)^8

P( three 3's)  = C(10, 3) *(1/6)^3 (5/6)^7

P(four 3's) = C(10, 4) * (1/6)^4 (5/6)^6

So....the sum of these  ≈ .9845  ≈   98.45 %