(MATH) TEAS PRACTICE EXAM B
A DOCTOR SEES EACH PATIENT FOR 20 MINUTES DURING A TYPICAL
APPOINTMENT. HOW MANY PATIENTS CAN THE DOCTOR SEE IN A TYPICAL
7.5 HR DAY ?
A. 23
B. 22
C. 100
D. 150 - -B. 22
1 PATIENT / 20 MINUTES = X / 450 MINUTES.
-WHAT IS THE DECIMAL EQUIVALENT OF 3/5 ? - -0.6
-THE AVERAGE SPEED A NASA SPACE SHUTTLE TRAVELS IS 17,500 MILES
PER HOUR. THE DISTANCE THE SPACE SHUTTLE TRAVELED IS 238,855 MILES.
WHAT IS THE TRAVEL TIME OF THE SPACE SHUTTLE ? - -13.6 HOURS
DIVIDE 238,855 BY 17,500
-R = 5CM
HALF CIRCLE - -25.7CM
HALF CIRCLE IS 10 X 3. ADDED TO THE LENGTH OF THE DIAMETER 5 X
2.
-81 FT SQUARED = ______________ YD SQUARED - -9
THERE ARE THREE FEET TO A YARD AND 9 SQUARE FEET TO A SQUARE
YARD.
81FT SQUARED / X = 9FT SQUARED / 1 YD SQUARED
X = 9YD SQUARED
-WHICH OF THE FOLLOWING IS THE AREA OF THE IRREGULAR POLYGON
ABOVE ? - -44 SQUARE CM.
BREAK DOWN THE FIGURE INTO THREE RECTANGLES.
1 RECTANGLE IS 4CM BY 5 CM.
2 RECTANGLE IS 3CM BY 6CM.
3 RECTANGLE IS 2CM BY 3CM.
A DOCTOR SEES EACH PATIENT FOR 20 MINUTES DURING A TYPICAL
APPOINTMENT. HOW MANY PATIENTS CAN THE DOCTOR SEE IN A TYPICAL
7.5 HR DAY ?
A. 23
B. 22
C. 100
D. 150 - -B. 22
1 PATIENT / 20 MINUTES = X / 450 MINUTES.
-WHAT IS THE DECIMAL EQUIVALENT OF 3/5 ? - -0.6
-THE AVERAGE SPEED A NASA SPACE SHUTTLE TRAVELS IS 17,500 MILES
PER HOUR. THE DISTANCE THE SPACE SHUTTLE TRAVELED IS 238,855 MILES.
WHAT IS THE TRAVEL TIME OF THE SPACE SHUTTLE ? - -13.6 HOURS
DIVIDE 238,855 BY 17,500
-R = 5CM
HALF CIRCLE - -25.7CM
HALF CIRCLE IS 10 X 3. ADDED TO THE LENGTH OF THE DIAMETER 5 X
2.
-81 FT SQUARED = ______________ YD SQUARED - -9
THERE ARE THREE FEET TO A YARD AND 9 SQUARE FEET TO A SQUARE
YARD.
81FT SQUARED / X = 9FT SQUARED / 1 YD SQUARED
X = 9YD SQUARED
-WHICH OF THE FOLLOWING IS THE AREA OF THE IRREGULAR POLYGON
ABOVE ? - -44 SQUARE CM.
BREAK DOWN THE FIGURE INTO THREE RECTANGLES.
1 RECTANGLE IS 4CM BY 5 CM.
2 RECTANGLE IS 3CM BY 6CM.
3 RECTANGLE IS 2CM BY 3CM.