Find the value of the standard normal random variable z, called z0 such that: (a) P(z≤z0)=0.9645
z0=
(b) P(−z0≤z≤z0)=0.1098
z0=
(c) P(−z0≤z≤z0)=0.2552
z0=
(d) P(z≥z0)=0.4719
z0=
(e) P(−z0≤z≤0)=0.1262
z0=
(f) P(−1.66≤z≤z0)=0.6653

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

definitely a. the others have almost all data points on one side of the line.

step-by-step explanation: