Randy Michael NT 1210 Lab 1.1 Professor Chibuzo Onukwufor 4/1/15 Lab 1.1 1: Convert the decimal value 127 to binary. Explain the process of conversion that you used. Decimal Number | Binary Number | Remainder | 127 - | 64 | 63 | 63 - | 32 | 31 | 31 - | 16 | 15 | 15 - | 8 | 7 | 7 - | 4 | 3 | 3 - | 2 | 1 | 1 - | 1 | 0 | Binary | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 | Decimal | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Conversion | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | I took the decimal and divided it by two giving 1 for the remainders and 0 if it did not have a remainder. 2: Explain why the values 102 and 00102 are equivalent. They are equivalent because they represent the powers of 10 3: Based on the breakdown of the decimal and binary systems in this lab, describe the available digit values and the first four digits of a base 5 numbering system.
Now we all the 12 costs per units amount from table 2 and divided by 12 in order to get the average. 38 + 38 + 38 + 35 + 35 + 38 + 35 + 38 + 33 + 39 +35 +35 = 437 Now 437/12 = 36.45 The average materials cost per unit is $36.45 2. Average labor cost per unit I use the same formula as I did above to figure out the average labor cost per unit. Labor Probability Cost Random Numbers Interval 0.18 $ 22.00 0.0 < 0.18 0.08 $ 23.00 0.19 < 0.26 0.25 $ 24.00 0.27 < 0.51 0.19 $ 25.00 0.52 < 0.70 0.3 $ 28.00 0.71 <
It is four octets of numbers from 0-255 represented in decimal form. Class A always has the first bit of their address set to 0. They have an 8 bit network mask that is why they leave it zero so they can have 7 bits for the network portion. They have 128 possible networks. Class B addresses always has the first bit set to 1 and their second bit set to 0.
Ch 3. GAUGE 1 dc in each of next 10 ch. *Ch 10. Skip 13 sc and 14 rows = 4"CLUE PATTERNS10 dc. 1 dc in each of next 10 ch.
Type Your Name in the Report title. C. Page numbers should be RIGHT JUSTIFIED at the page bottom. D. Ensure you have at least 1” spacing all around. 1. Create a report that Sum Sales, Costs, and Profit by Region.
10. Jennifer is twice as old as Ramon. The sum of their ages 7 yrs ago was 13. How old are they now? Module 08 Exercise Answers Faculty Comments | # | Answer | Calculations | | 1 | x=-3 | y=-13x 3y=-5x-12y=-x33y=-5x-123(-x3)=-5x-12-x=-5x-12-x+5x=-12 4x=-12 4x4=-124x=-124x=-3 | | 2 | s=10r=-30 | r=-3s r+4s=10r=-3s (-3s)+4s=10 -3s+4s=10 s=10r=-3(10) s=10r=-30s=10
The aim of this study is to approve or reject the hypothesis being limb length (arm) is equal to 40% of a person’s height. Procedure/Experiment/Materials 1. Materials = Yard Stick, Calculator 2. 13 random subjects selected in Bio Lab 2010-51 3. Standing erect, a measurement was taken from the axillary to the longest finger tip of each subject 4.
Revision Year 12 Maths B, Term 4 1. The function, y=3cosx+5 has an amplitude of 3 and a mean value of 5. Using these values, the range can be found to be 2≤y≤8 . Find the range of the function y=4cosx+7 2. Find the exact value of tan600 (Hint, draw a standard triangle to help) 3.
3-1: Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = 1). | | | | A) | Describe the outcomes of this experiment. List the elements of the sample space. | Smallest sum is 2, this happens if both cards are Aces | Largest sum is 20, this happens if both cards are 10s | The other outcomes are integers between the smallest & largest sum (2 and 20) | | | | | | | B) | What is the probability of obtaining a total of 5 for the two cards? | The following results produce a sum of 5: | Ace+4 | | | 2+3 | | | 3+2 | | | 4+1 | | | 4 suits, 4 outcomes equals 16 different ways to get a sum of 5 | As there are 4 suites, this means these 4 combinations could happen in 4 different ways | 16+16+16+16 = 64 | | | | | This is a reduced deck with only 40 cards | Because the first card picked will be any available cards, it equals 40 | The second card picked will be any of the available cards left, it equals 39 | 40 x 39 = 1560 different ways the 19 outcomes | 64 / 1560 = a probability of 0.410 | | | | C) | Let a be the event “total card value is 5 or less.” Find P ( A ) and P ( A c ).
Hands-On Labs, Inc. LabPaq: CK-S Page 117 The National Academy of Science reports that the Average Daily Intake (ADI) of Blue #1 dye is 16 mg. Its chemical formula is: C37H34N2Na2O9S3, and its molecular weight is: 792.84. The formal chemical name is: Disodium alpha -(4-(N-ethyl-3-sulfonatobenzylamino) phenyl)-alpha-(4-N-ethyl-3- sulfonatobenzylamino, cyclohexa-2,5-dienylidene) toluene-2-sulfonate Typically, the concentration of the FD&C Blue #1 is approximately 0.026 M. You will need to dilute it to make a series of standard solutions of known concentrations. Your starting dilution should be 1:1000 (i.e., 1 part dye and 999 parts water). PROCEDURE: Completely read all instructions and assemble all equipment and supplies before beginning work on this experiment. 1.