5. thickness measurements of ancient prehistoric native american pot shards discovered in a hopi village are approximately normally distributed, with a mean of 5.1 millimeters (mm) and a standard deviation of 0.7 mm. what percent (%) of ancient prehistoric native american pot shards discovered in this area has thickness measurement thinner than 3.

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

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3