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

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

pls refer ro the top for answer

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -