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Step-by-step explanation:

1.

∠ACB=6 x°

∠DCE=30°

6x=30°divide both sides by 6

x=30°/6= 5°

∠BCE=180°-30°=150°sum of angles on a straight line

2.

∠ABD=(4+5x)°

∠CBD=(x+2)°

(4+5x)°+(x+2)°=180°sum of angles on a straight line

4+5x+x+2=180°collect like terms

6x+6=180°

6x=180°-6°

6x=174°divide by 6 both sides

x=174°÷6= 29°

∠ABD=4+5x=4+(5*29)= 4+145=149°

∠CBD= x+2=29°+2°=31°

3.

BE bisects ∠ABE=(3x+1)°

m∠DBA=(8x-14)°

1/2 (m∠DBA)° = m∠ABE

1/2(8x-14)°=(3x+1)°

(4x-7)°=(3x+1)°

4x-7=3x+1

4x-3x=7+1collecting like terms

x=8°

m∠ABE=(3x+1)°=(3*8+1)=24+1=25°

m∠DBA=(8x-14)°= (8*8-14)°=(64-14)=50°

4.

∠ADE=∠CDG

50+3x-y=x+2x-16collect like terms

50+16-y=x+2x-3x

66-y=3x-3x

66°=y

∠ADC +∠ADB+∠BDE

90°+50°+(3x-y)°=180°

140°+(3x-y)°=180°

3x-66°=180°-140°

3x=40°+66°

3x=106°divide both sides by 3

x=106°÷3= 35.33°

∠FDG=(2x-16)°= (2×35.33° - 16° )= 54.67°

∠BDE = (3x-y)°= (3×35.33°-66°)= 105.99-66=39.99°

5.

∠ABD+∠DBC=90°

(6x+4)°+32°=90°

(6x+4)°=90°-32°

6x+4=58°

6x=58°-4°

6x=54°

x=54°÷6=9°

∠ABD= (6x+4)°+32°

∠ABD= (6×9 +4)°+32°

∠ABD= (54°+4°) + 32°

∠ABD=58°+32°=90°

6.

∠AED=∠CEB

(3x+5)°=(4y-15)°form equation of equality

3x+5=4y-15

5+15=4y-3x

20=4y-3x(1)

∠AEC=∠DEB

(y+20)°=(x+15)°

y+20=x+15

20-15=x-y

5°=x-y

5+y=x(2)

Use equation (2) in equation (1)

20=4y-3x

20=4y-3(5+y)

20=4y-15-3y

20+15=4y-3y

35°=y

Solve for x

x=5+y=5°+35°=40°

∠AED=(3x+5)°=(3×40 +5 )=120+5=125°

∠AEC= (y+20)°= 35° + 20° =55°

step-by-step explanation:

answers available in this download

(pearson%2c%25202012)%2520bbs.pdf& usg=aovvaw1obmtxnfghfgxhx0urh_je

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05