Control charts for X and R are maintained on the shear strength in pounds of test spot welds. The subgroup size is 3. The values of X and R are computed for each subgroup. After 30 subgroups, X = 11,346 and ER = 873. (20 points) (X and R charts) A) compute the value of the 3-sigma (30) for the X and R charts and estimate the value of a on the assumption that the process is in statistical control. B) The specified strength for a weld is 529 lb. If the process is in statistical control and normally distributed, what can you conclude regarding its ability to meet this specification?

1/800 is the right answer

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°

statement #3: proof- ang is congruent to ang ecd : angles are congruent.  

step-by-step explanation:

1: the missing angles in statement #3  

a.) acb  

reason- the reason why the answer is acb is because angle ecd and angle acb are both the same size *that's what the word congruent means this is really simple.  

2: the missing word in reason #3  

c.) vertical  

reason- the reason why the answer is vertical is because the angles in that diagram on figure 2 is slanted, which is vertical, and the e and a both lay on vertical lines. along with the other's, but e and a are the 2 main points you want to focus on. and c is kind of like your starting point, because c is for both angles, not just one.  

i hope that was and understanding.