Week 6 CheckPoint



|NAME: | |BR1 |

Instructions: In order to complete your assignment in an organized and efficient manner, please use this template to type your work and answers. Save this template to your computer, complete the problems (showing all work as if you did not have a calculator, unless otherwise stated), and post as an attachment to your forum. If you cannot read the problems on this template, refer to the eTextbook on the course aXcess page.

REMEMBER: NO WORK = NO CREDIT

|Problems (1 point each) |TYPE YOUR SOLUTION(S) HERE |

|Section 6.1 #58a (pg. 441) |9% |

| | |

|Find decimal and fractional notation for the percent notation |Divide by 100 to convert to decimal |

|in each sentence. | |

|58 (a). 1 ounce of Tostitos provides 9% of the MDV of fat. |Decimal: 0.09 |

| | |

| |Put it over 100: |

| |Fractional notation: 9/100 |

|Section 6.1 #58c (pg. 441) | |

| |6% |

|Find decimal and fractional notation for the percent notation | |

|in each sentence. |Divide by 100 to convert to decimal |

|58 (c). 1/2 cup of Campbells’ New England clam chowder provides| |

|6% of the MDV of iron. |Decimal: 0.06 |

| | |

| |Put it over 100: |

| |Fractional notation: 6/100 = 3/50 |

| | |

|Section 6.2 #34 (pg. 448) | |

| |Convert to a decimal |

|Write as a percent. |= 1.2 |

|34. [pic] |Multiply by 100% |

| |= 120% |

| | |

|Section 6.3 #12 (pg. 456) |(no work to be shown) |

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|Identify the rate, base, and amount. Do not solve the exercise |Rate: the percentage = 75% |

|at this point. | |

|12. 150 is 75% of what number? |Base: the number we’d have to solve for |

| | |

| |Amount: 150, since that is 75% of something |

|Section 6.3 #18 (pg. 457) |(no work to be shown) |

| | |

|Identify the rate, base, and amount. Do not solve the exercise |Rate: the unknown percentage |

|at this point. | |

|18. In a shipment of 750 parts, 75 were found to be defective. |Base: 750, the total number of parts |

|What percent of the parts were faulty? | |

| |Amount: 75, the defective parts |

| | |

|Section 6.4 #46 (pg. 465) |0.065x = 325 |

| |Divide by 0.065: |

|Solve each of the following problems involving percent. |x = 5000 |

|46. 6.5% of what number is 325? | |

|Section 6.4 #50 (pg. 465) |That is about half of 1500: |

| |750 |

|Estimate the amount in each of the following problems. | |

|50. What is 48.3% of 1,500? | |

| | |

|Section 6.5 #6 (pg. 477) |Divide to get the rate: |

| |275/2500 |

|6. Ms. Jordan has been given a loan of $2,500 for 1 year. If |= 0.11 |

|the interest charged is $275, what is the interest rate on the |Multiply by 100% |

|loan? |= 11% |

|Section 6.5 #26 (pg. 479) |Multiply: |

| |7% times 900 |

|26. A school had 900 students at the start of a school year. If|7% = 0.07 |

|there is an enrollment increase of 7% by the beginning of the |0.07 times 900 |

|next year, what is the new enrollment? |= 63 |

| |That is the increase, so add to the original amount: |

| |900 + 63 |

| |= 963 |

|Section 6.5 #42 (pg. 481) |40% = 0.4 |

| |Divide the 300 by that: |

|42. A virus-scanning program is checking every file for |300/0.4 |

|viruses. It has completed checking 40% of the files in 300 |= 750 seconds |

|seconds. How long should it take to check all the files? | |

|Section 6.5 #56 (pg. 483) |85% = 0.85 |

| |Multiply by the total: |

|56. Gasoline accounts for 85% of the motor fuel consumed in the|0.85*8882 |

|United States every day. If 8,882 thousand barrels of motor |= 7549.7 |

|fuel is consumed each day, how much gasoline is consumed each | |

|day in the United States? Round to the nearest barrel. |Rounds to: |

| |7550 bbl |

|Section 6.5 #68 (pg. 485) |The increase is: 0.105 times 18500 = 1942.5 |

| |Add that to the original amount: |

|68. Jerry earned $18,500 one year and then received a 10.5% |1942.5+18500 |

|raise. What is his new yearly salary? |= $20442.50 |

|Section 6.5 #70 (pg. 485) |The new percentage is: |

| |100-8.2 = 91.8% |

|70. Yi Chen made a $6,400 investment at the beginning of a |Multiply that by the 6400: |

|year. By the end of the year, the value of the investment had |6400 times 0.918 |

|decreased by 8.2%. What was its value at the end of the year? |= $5875.20 |

| | |

|Section 7.1 #50 (pg. 508) |Add the hours: |

| |3+4+5 = 12 |

|Add. |Add the minutes: |

|50. 3 hours 20 minutes |20 + 25 + 35 = 80 min = 1 hr 20 min |

|4 hours 25 minutes |Total: |

|+ 5 hours 35 minutes |13 hr 20 min |

|Section 7.1 #54 (pg. 508) |Subtract the pounds: 9-5 = 4 |

| |Subtract the oz: 15-8 = 7 |

|Subtract. |4 lb 7 oz |

|54. 9 pounds 15 ounces | |

|- 5 pounds 8 ounces | |

|Section 7.1 #74 (pg. 509) |3 quarts = 6 pints |

| |1 pint = 16 oz |

|74. Mark uses 1 pint 9 fluid ounces and then 2 pints 10 fluid |3 quarts = 96 oz |

|ounces from a container of film developer that holds 3 quarts. |96 oz – (16+9) – (32+10) |

|How much of the developer remains? |= 29 oz left over |

| | |

|Section 7.2 #50 (pg. 520) |(no work to be shown) |

| | |

|50. What unit in the metric system would you use to measure |(a) km |

|each of the following quantities? |(b) cm |

|(a) Distance from Los Angeles to New York |(c) micrometers |

|(b) Your waist measurement |(d) m |

|(c) Width of a hair | |

|(d) Your height | |

| | |

|Section 7.3 #70 (pg. 528) |19.5 million = 19,500,000 |

| |Multiply by 1000 kg: |

|70. The United States emitted 19.5 million tons of nitrogen |19,500,000,000 kg |

|oxides into the atmosphere in 1987. One metric ton equals 1,000| |

|kilograms. How many kilograms of nitrogen oxides were emitted | |

|to the atmosphere in the United States during 1987? | |

| | |

|Section 7.4 #6 (pg. 533) |1 in = 2.54 cm, so 72 multiplied by 2.54 |

| |= 182.88 cm |

|Complete this statement. (Round to the nearest hundredth) | |

|6. 72 inches = ____ centimeters | |

|Section 7.4 #12 (pg. 533) |1 L = 1.05668821 qt |

| |Multiply by 7: |

|Complete this statement. (Round to the nearest hundredth) |= 7.39681747 qt |

|12. 7 liters = ____ quarts |Rounds to: |

| |7.40 quarts |

|Section 7.4 #18 (pg. 533) |1 mi/hr = 1.609344 km/hr |

| |55 times 1.609344 |

|18. Samantha’s speedometer reads in kilometers per hour. If the|= 88.51392 km/hr |

|legal speed limit is 55 miles per hour, how fast can she drive?|Round: |

| |88.51 km/hr |

| | |

|Section 7.5 #6 (pg. 546) |(no work to be shown) |

| | |

|Identify each object as a line or line segment. |It is a line, since it goes infinitely in each direction (assuming that you aren’t asking |

|[pic] |specifically about the part between u and v – that is just a line segment) |

|Section 7.5 #26 (pg. 547) |(no work to be shown) |

| | |

|Give an appropriate name for each indicated angle. |Acute |

|[pic] | |

|Section 7.5 #52 (pg. 550) |(Show or explain why) |

| | |

|Find m(a, m(b, and m(c. |Angle a = 56 |

|[pic] |because of alternate exterior angles |

| |Angle b = 124 because straight lines add to 180 |

| |Angle c = 124 because straight lines add to 180 |

| | |

| | |

|Section 7.6 #2 (pg. 557) |(no work to be shown) |

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|Label the triangle as acute or obtuse. |Obtuse |

|[pic] | |

|Section 7.6 #10 (pg. 557) |(no work to be shown) |

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|Label the triangle as equilateral, isosceles, or scalene. |Scalene |

|[pic] | |

|Section 7.6 #24 (pg. 559) | |

| |mD = mF = (180-72)/2 = 54 degrees |

|Assume that the given triangle is isosceles. |They have to be equal, and the whole thing must add to 180 degrees |

|[pic] | |

|Section 7.6 #26 (pg. 559) |Triangles “a” and “b” are similar since they are both 30-60-90 triangles. The angles must add to |

| |180 degrees, and the non right angle in triangle C is not 30 or 60. |

|26. Which two triangles are similar? | |

|[pic] [pic] | |

| | |

|[pic] | |

|Section 7.6 #44 (pg. 563) |Set up a proportion: |

| |[pic] |

|44. A tree casts a shadow that measures 5 meters. At the same |Cross multiply: |

|time, a meter stick casts a shadow that is 0.4 meters long. How|0.4x = 5 |

|tall is the tree? |Divide by 0.4: |

| |x = 12.5 meters |

| | |

|Section 7.7 #4 (pg. 570) |(using a calculator is OK, but show work to check) |

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|Find the square root. |What number multiplied by itself equals 196? |

|4. [pic] |14 times 14 = 196 |

| |= 14 |

|Section 7.7 #16 (pg. 570) |(using a calculator is OK, but show work to check) |

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|Find the missing length for the given triangle. |Use the Pythagorean Theorem: |

|[pic] |[pic] |

| |[pic] |

| |[pic] |

| |[pic] |

| |[pic] |

Finished? I encourage you to go back and check each answer. Did you show all your work? Did you label every answer?

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