Determine the area under the standard normal curve that lies to the left of (a) z equals 1.41 commaz=1.41, (b) z equals 1.19z=1.19, (c) z equals 1.34z=1.34, and (d) z equals 1.11z=1.11.(a) The area to the left of zequals=1.411.41 is .(Round to four decimal places as needed.)(b) The area to the left of zequals=1.191.19 is. (Round to four decimal places as needed.)(c) The area to the left of zequals=1.341.34 is .(Round to four decimal places as needed.)(d) The area to the left of zequals=1.111.11 is .(Round to four decimal places as needed.)

a)0.9207

b)0.883

c)0.9099

d)0.8665

Step-by-step explanation:

a)

The area lies to left of z=1.41 is P(z<1.41).

P(z<1.41)=P(-∞<z<0)+P(0<z<1.41)

P(z<1.41)=0.5+0.4207

P(z<1.41)=0.9207

b)

The area lies to left of z=1.19 is P(z<1.19).

P(z<1.19)=P(-∞<z<0)+P(0<z<1.19)

P(z<1.19)=0.5+0.383

P(z<1.19)=0.883

c)

The area lies to left of z=1.34 is P(z<1.34).

P(z<1.34)=P(-∞<z<0)+P(0<z<1.34)

P(z<1.34)=0.5+0.4099

P(z<1.34)=0.9099

d)

The area lies to left of z=1.11 is P(z<1.11).

P(z<1.11)=P(-∞<z<0)+P(0<z<1.11)

P(z<1.11)=0.5+0.3665

P(z<1.11)=0.8665

29.5977692753

step-by-step explanation:

i believe that the slope of the line is -5

step-by-step explanation: