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

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !