MA 1135 Practice Final, Part I (answers on last page)
Name
Tuesday, April 28, 2015
Note: Final Exam is Thursday, May 7th at 10:45am.
We’ll be taking the final in our regular classroom, so we won’t be using the computers. I’ll ask you questions
about the Excel spreadsheets instead of actually having you use Excel.
You may have four formula sheets (so you can use your old ones), but I would think that less is more in this
case. You’ll also need your normal table and t table. I’ll give you conversion equations.
Stuff From After Spring Break
For the conversions, to convert 107 inches into feet, I’d like to see
(1)
107 6 inches ·
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1 foot
= 8.92 feet.
12 6 inches
5280 feet = 1 mile
12 inches = 1 foot
60 minutes = 1 hour
60 seconds = 1 minute
16 ounces = 1 pound
16 fluid ounce = 1 pint
2 pints = 1 quart
4 quarts = 1 gallon
1 foot3 = 7.48 gallons
1 foot = 0.3048 meters
2.54 centimeters = 1 inch
1000 millimeters = 1 meter
100 centimeters = 1 meter
2.2 pounds = 1 kilogram
1000 grams = 1 kilogram
2000 pounds = 1 ton
1 quart = 0.946 liters
1 cubic meter = 1 meter3 = 1000 liters
1 centimeter3 = 1 milliliter
1 acre = 43560 square feet
Lecture 11
1.
Convert 1600 meters into miles.
2.
Convert a speed of 30 miles per hour into meters per minute.
3.
Gasoline has a density of 0.759 pounds per pint. Convert this to grams per liter.
Lecture 12
1.
I’ve got some gravel, which when piled up, has a dry density of ρdry = 2.24 grams per cubic centimeter
and a porosity of n = 0.20.
1
MA 1135 Practice Final, Part I (answers on last page)
2
a.
The dry density tells you how much a cubic centimeter of dry gravel weighs in grams. Convert the dry
density to kilograms per cubic meter.
b.
If you filled the swimming pool with volume 2000 meters3 with gravel, how much would that gravel
weigh in kilograms?
c.
How much is that in pounds?
d.
How much is that in tons?
e.
With the pool filled with dry gravel, how much water could you put in? Express your answer in cubic
meters.
Lecture 13
1.
Find the following values on your calculator.
a.
sin(30◦ ).
b.
cos(15◦).
c.
tan(2◦ ).
2.
Find the slope and percentage grade.
a.
Horizontal distance: 150 feet. Vertical distance: 12 feet.
b.
Angle of inclination: 3◦.
Lecture 14
In the following, you’re given two of x, y, R, θ, and m. Find the other three.
1.
x = 77 feet and y = 12 feet.
2.
m = 6% and x = 145 feet.
Lecture 15
Find the angle of inclination, if necessary, the perpendicular, parallel, and slide forces, and then answer the
question posed.
1.
Can a 3500 pound car be parked on a wet 25% grade with coefficient of friction µ = 0.20?
Lecture 16
1.
Convert the number in scientific notation into normal decimal notation.
a.
7.943 × 10−8 .
b.
1.23 × 107 .
2.
Convert the given number into scientific notation.
a.
74,324,450,000
MA 1135 Practice Final, Part I (answers on last page)
b.
3
0.0000005691
3.
Multiply or divide, and express answer in correct scientific notation (the decimal part should have exactly
one non-zero digit to the left of the decimal point).
a.
(3.546 × 104 ) · (2.113 × 10−2 )
b.
(5.743 × 1042) · (8.425 × 10−3 )
c.
7.75×1015
2.05×103
d.
3.75×10−5
6.20×10−7
Lecture 17
1.
Find the number between 2.5 and 8.3 that is three-quarters of the way from 2.5.
2.
Find the point halfway between (1, 3) and (5, 11).
3.
Find the point between (2.3, 9.2) and (7.5, 3.1) that is two-fifths of the way from (2.3, 9.2).
Lecture 18
1.
Find an equation for the line through (1, 3) and (4, 15), and interpolate a y value for x = 2.
2.
Interpolate a y value for x = 5 between the points (3, 2) and (8, 7).
3.
When I “curve” your test scores, I’ll base it on two particular scores, and interpolate the rest. For
example, on Test III, there were 55 points possible, and I decided that a perfect 55 points would get a curved
score of 101, and 28 points would get a curved score of 63. Interpolate a curved score for a raw score of 40
points.
Answers on next page.
MA 1135 Practice Final, Part I (answers on last page)
11-1) 1600 meters ·
1 foot
0.3048 meters
·
1 mile
5280 feet
4
= 0.994 miles.
hour = 804.67 meters per minute
11-2) 30 miles · 5280 feet · 0.3048 meters · 60 1minutes
hour
1 mile
1 foot
pounds 1 kilogram 1000 grams 2 pints
1 quart
11-3) 0.759 pint ·
·
·
·
= 729.39 grams per liter.
2.2 pounds 1 kilogram 1 quart 0.946 liters
grams
centimeters)3 · 1 kilogram = 2.24 grams ·
· (100 (1
12-1a) 2.24
1000 grams
meter)3
centimeter 3
centimeter3
kilograms
2240
.
meter3
2240 kilograms
b) 2000 meters3 ·
= 4480000 kilograms.
3
1 meter
2.2 pounds
c) 4480000 kilograms ·
= 9856000 pounds.
1 kilogram
d) 9856000 pounds ·
e) Porosity: n =
1 ton
2000 pounds
Vpore
space
2000 meters
1003
centimeters 3 · 1 kilogram =
3
1000 grams
1 meter
= 4928 tons.
= 0.20, so Vpore space = 0.20 · 2000 meters3 = 400 meters3 .
3
13-1a) 0.5, b) 0.966, c) 0.035.
12
13-2a) m = 150
= 0.08 = 8%. b) m = tan(3◦ ) = 0.052 = 5.2%.
√
12
= 0.156 = 15.6%. θ = tan−1 (0.156) = 8.87◦.
14-1) R = 772 + 122 = 77.93 feet. m = 77
√
14-2) y = 0.06 · 145 = 8.7 feet. R = 1452 + 8.72 = 145.26 feet. θ = tan−1 (0.06) = 3.43◦.
15-1) m = 0.25. θ = tan−1 (0.25) = 14.04◦. Fperp = 3500 · cos(14.04◦) = 3395.44 pounds, Fpar =
3500 · sin(14.04◦) = 849.10 pounds, Fslide = 0.20 · 3395.44 = 679.09 pounds. It will slide, so no, we can’t
park the car there.
16-1a) 0.00000007943; b) 12,300,000;
16-2a) 7.432445 × 1010 ; b) 5.691 × 10−7 ;
16-3a) (3.546 × 104 ) · (2.113 × 10−2 ) = 7.4927 × 102
b) (5.743 × 1042 ) · (8.425 × 10−3 ) = 48.385 × 1039 = 4.8385 × 1040
c)
7.75×1015
2.05×103
d)
3.75×10−5
6.20×10−7
= 3.7805 × 1012
= 0.6048 × 102 = 6.048 × 101
17-1) 14 (2.5) + 34 (8.3) = 6.85
17-2) x = 12 (1) + 12 (5) =
1
2
+
5
2
=
6
2
= 3, y = 12 (3) + 12 (11) =
3
2
+
11
2
=
14
2
= 7, so (3, 7).
17-3) x = 35 (2.3) + 25 (7.5) = 4.38, y = 53 (9.2) + 25 (3.1) = 6.76, so (4.38, 6.76).
18-1) y =
y = 7.
12
3 (x − 1) + 3
or y = 4(x − 1) + 3 or y = 4x − 1. Plugging x = 2 into any of these equations yields
18-2) I didn’t ask you for an equation, so you can do everything at once
7−2
(2)
y=
((5) − 3) + 2 = 4.
8−3
18-3) If you let (x0 , y0 ) = (55, 101) and (x1 , y1 ) = (28, 63), and did everything at once, you would get
63 − 101
(3)
y=
((40) − 55) + 101 = 79.89
28 − 55