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swat32
3 years ago
5

Find all of the missing angle measures. Remember you cannot assume right angles or diameters. Also think about how many degrees

are in a triangle. Angle 1: Angle 2: Angle 3: Angle 4: Angle 5: Angle 6: Angle 7: Angle 8: Angle 9: Angle 10: Angle 11: Angle 12: Angle 13: Angle 14: Angle 15:

Mathematics
1 answer:
Gwar [14]3 years ago
8 0

Answer:

See text below or attached figure

Step-by-step explanation:

Given arcs

AC=70

CR=18

therefore AR = 88

RB=80

BE=130

therefor EA = 360-(70+18+80+130) = 360-298 = 62

angles will be denoted (1) for angle 1, etc.

We ASSUME

ARD is a straight line

PFRB is a straight line

FCE is a straight line

Using inscribed angle theorem, angles subtended by chords/arcs equal to half the arc central angle.

Therefore

(4)=80/2=40

(13)=130/2=65

(12)=62/2=31

(11)=70/2=35

(5) = (70+18)/2 = 44

Consider triangle AEG,

(7)=(13)+(11)=65+35=100   [exterior angle]

Consider triangle EGB,

(10)=180-100-31 = 49      [sum of angles of a triangle]

Consider triangle AEH,

(3) = 180-(4)-(13)-(11) = 180-40-65-35 = 40  [sum of angles of a triangle]

From cyclic quadrilateral ARBE,

ARB+AEB=180 =>

ARB=180-AEB=180-(35+49) = 96

By the intersecting secants theorem,

(2) = (130-18)/2 = 56     [secants FE, FB]

(1) = (130+62 - (18+70))/2 = 104/2 = 52  [secants PA,PB]

(8) = (130+62 -80)/2 = 112/2 = 56

ARD is straight line  (see assumptions above)

(9) = 180-96 = 84    [sum of angles on a line]

ARP = (9) = 84     [vertically opposite angles]

Consider triangle ARP

(14) = 180-52-84 = 44

Consider tangent PA

(15) = 180-(44+40+65) = 31    [sum of angles of a triangle]

Consider triangle ABD

(6) = 180 - (40+44+56) = 40   [sum of angles of a triangle]

This completes the search for all sixteen angles, as shown in the diagram, or in the text above.

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Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

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P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

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P(-1.00 < z < 0) = 0.3414

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P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

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P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

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From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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