Basic Theory

Each binary digit carries a weight of 1,2,4,8 from the right. 1011 is 8×1 + 4×0 + 2×1 + 1×1 = 11. 9 is 1001 in binary, 13 is 1101, and 15 is 1111; none of these equals 1011.

Dividing 45 by 2 repeatedly and listing the remainders from the bottom gives 1, 0, 1, 1, 0, 1; read in reverse order it is 101101. Checking: 32 + 8 + 4 + 1 = 45, which matches. 101011 is 43, 110101 is 53, and 111001 is 57; none of them equals 45.

The second hexadecimal digit carries a weight of 16, and F represents 15. 2×16 + 15 = 47. 31 is 1F in hexadecimal, 43 is 2B, and 62 is 3E; none of these equals 2F. Remember that in hexadecimal, A through F represent 10 through 15.

Four binary digits correspond to one hexadecimal digit, so split 11010110 into 1101 and 0110. 1101 is 13, that is D, and 0110 is 6, giving D6. 6D is 01101101, B6 is 10110110, and C6 is 11000110; none matches the original value.

With n bits, 2 to the power n states can be represented, so 8 bits give 2 to the 8th = 256 values, base-10 values 0 through 255. 8 is the number of bits itself, 16 is what 4 bits give, and 128 is what 7 bits give; none is the number of combinations for 8 bits.

With n bits, 2 to the power n values can be distinguished. 2 to the 5th = 32 is not enough for 50 kinds, while 2 to the 6th = 64 suffices, so at least 6 bits are needed. 5 bits can represent only 32 values, which falls short. 7 bits, 128 values, and 8 bits, 256 values, can also represent them but do not satisfy the minimum condition.

The AND is 1 in each position only when both bits are 1. The upper 4 bits become all 0 because the other operand is 0 there, and the lower 4 bits keep 0011, giving 00000011. 00001111 is just the other operand itself, and 10111111 is the result of OR. 10110000 is the result of masking with 11110000 rather than 00001111, extracting the upper 4 bits.

Exclusive OR is 1 in each position only when the bits differ. From the left: 1 and 1 give 0, 1 and 0 give 1, 0 and 1 give 1, and 1 and 1 give 0, so the result is 0110. 1001 is the AND and 1111 is the OR, while 0100 matches the result of none of the basic operations.

Two's complement is formed by inverting every bit of the original number and adding 1. 5 is 00000101; inverted it becomes 11111010, and adding 1 gives 11111011. 11111010 is only the inversion, the one's complement, without the added 1. 10000101 is sign-magnitude with just a sign bit of 1 in the top position, and 11111100 represents −4.

By De Morgan's law, the negation of a conjunction equals the disjunction of the negated terms. Therefore NOT(A AND B) equals NOT A OR NOT B. NOT A AND NOT B is a different expression equal to NOT(A OR B), A OR B contains no negation and gives the opposite result, and A AND NOT B negates only B and does not hold. Checking all 4 rows of a truth table confirms this.

The number using at least one, subtracting the overlap once, is 60 + 45 − 25 = 80. Therefore those using neither number 100 − 80 = 20. 10 and 15 are values from mishandling the overlap, and 25 is the number using both, which is not what is asked.

Exclusive OR is 1 only when the input values differ, and 0 when they are equal. When both are 1 the result is 0; giving 1 there is the AND. When both are 0 the result is also 0. Equal values describe the negation of XOR, the equivalence circuit, whose result is the reverse.

The expected value is the sum of each amount times its probability: (5000×1 + 1000×5) ÷ 100 = 10000 ÷ 100 = 100 yen. 150 and 200 come from multiplying the prize amounts or counts incorrectly, and 600 yen comes from multiplying the winning probability 6÷100 by the total prize money of 10000 yen, an error that fails to weight each amount by its own probability.

There are 6×6 = 36 possible outcomes in total. The sum is 7 in the 6 cases (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), so the probability is 6÷36 = 1/6. 1/12 would correspond to 3 cases, 1/9 to 4 cases, and 1/8 to 4.5 cases; none matches the actual number of combinations.

Choosing without regard to order is a combination: (6×5) ÷ (2×1) = 15 ways. 30 is the permutation 6×5, which counts the same pair twice. 12 is 6×2 and 36 is 6×6; both count incorrectly.

There are 4 choices for the first position, 3 remaining for the second, 2 for the third, and 1 for the fourth, so 4×3×2×1 = 24 arrangements. 12 is 4×3, counting only two positions. 16 is 4×4, counting the same person more than once, and 4 is just the number of people; none is the total number of arrangements.

The median is the middle value when the data is sorted by size. With 5 items it is the 3rd value, 20. 40 is the mean, total 200 ÷ 5, dragged upward by the extremely large value 120. 30 is the 4th value and 120 is the maximum; neither is the median.

The squared differences from the mean are 9, 1, 1, 1, 0, 0, 4, 16, totaling 32. The variance is 32 ÷ 8 = 4, and the standard deviation is its square root, 2. 4 is the variance itself given as the answer, 8 is the number of data points, and 1 is one of the squared differences; none is the standard deviation.

The standard deviation is the square root of the variance and expresses how far the data is scattered from the mean. The smaller it is, the more the data clusters near the mean. The first option is the mean, the second the median, and the fourth the mode; each of these is a single representative value and does not express the size of the spread.

The correlation coefficient ranges from −1 to +1, and the closer to −1, the stronger the negative correlation in which one falls as the other rises. Almost no relationship describes values near 0, and the strong positive description fits values near +1, the opposite sign. The causation claim is wrong: strong correlation does not guarantee a cause-and-effect relationship.

Regression analysis derives, by fitting to real data, an equation that obtains the value of one variable from another, as in predicting sales from advertising spend. The second option is clustering, the third is basket analysis, also called association analysis, and the fourth is sample surveying; none of these derives a prediction equation.

Nano, written n, is the prefix representing 10 to the minus 9th power. 10^-3 is milli, m; 10^-6 is micro, μ; and 10^-12 is pico, p. On the small side the prefixes go milli, micro, nano, pico, each 1000 times smaller than the last.

The digit in the second decimal place is 6, which is 5 or more, so 1 is carried up to the first decimal place, giving 37.5. 37.4 is the result of truncating at the second decimal place, 37.46 comes from rounding the third decimal place, and 38.0 comes from rounding the first decimal place up; none applies the specified rounding at the specified position. Note that rounding off the first decimal place would give 37.

UTF-8 is an encoding that represents Unicode characters in a variable length of 1 to 4 bytes; alphanumerics take 1 byte just like ASCII, giving high compatibility, and it is the most widely used encoding on the Web. The first option is ASCII, and the third is close to a description of Japanese character codes such as Shift JIS. The fourth describes an image format, not a character code, which is something else entirely.

ASCII is the most basic character code, representing 128 kinds of alphanumerics, symbols, and control characters in 7 bits; it cannot handle Japanese. The second option describes Japanese character codes such as Shift JIS and EUC-JP, and the third describes Unicode. The fourth describes A/D conversion, digitization, which has nothing to do with character codes.

First comes sampling, reading the height of the wave at fixed intervals; next quantization, replacing each reading with a value from a fixed set of levels; and finally encoding, turning those values into bit strings of 0s and 1s. The other options shuffle this order, and since levels cannot be assigned nor codes produced before the values are read, they are wrong.

By the sampling theorem, the original waveform can be reconstructed if sampling is done at a frequency exceeding twice the highest frequency in the original signal. About 40kHz, twice 20kHz, is needed. The other three all fall short of twice the highest frequency, and the original waveform cannot be correctly reconstructed.

44100×16×2×60 = 84,672,000 bits; dividing by 8 gives 10,584,000 bytes, about 10.6Mbytes. About 5.3M is the value calculated as monaural, about 2.6M assumes 8-bit quantization and monaural, and about 21.2M is the value calculated with 32-bit quantization.

24 bits is 3 bytes, so 640×480×3 = 921,600 bytes, about 922Kbytes. About 7.4M is the bit count answered without converting to bytes, about 307K assumes 8 bits, 1 byte, per pixel, and about 38K assumes 1 bit per pixel; none matches the given conditions.

1Gbyte is 8×10^9 bits. With 50% transmission efficiency the effective speed is 50×10^6 bits per second, so 8×10^9 ÷ (50×10^6) = 160 seconds. 80 seconds ignores efficiency and assumes 100%, while 320 and 640 stack up errors in the byte-to-bit conversion and the handling of efficiency.

Supervised learning learns the relationship between inputs and answers from labeled data so that predictions can be made for unseen inputs. The second option describes unsupervised learning and the third reinforcement learning. The fourth is a hand-written rule-based approach, which does not count as machine learning from data.

Unsupervised learning finds structure and clusters in data without being given answers, and grouping customers, clustering, is a representative example. The first and second options are supervised learning, which learns from labeled data, and the third is reinforcement learning, which learns actions from rewards; none is unsupervised learning.

Reinforcement learning learns, by trial and error, how to choose actions so that total reward is maximized, based on the rewards received for actions taken in a given state; it is used in Go-playing AI and robot control. The first option is supervised learning and the third unsupervised learning. The fourth is data preprocessing work, not a category of learning method.

Deep learning is a method that deeply stacks the intermediate layers of a neural network, which mimics the connections of neurons in the brain, and can automatically extract the features worth attending to from data without humans specifying them. The first option is spreadsheet analysis, the second rule-based inference, and the fourth encryption technology; none of these is deep learning.

Overfitting is the phenomenon of memorizing even the fine details of the training data, giving high accuracy on the training data but degraded accuracy on unseen data. The second option is a data shortage problem, the third describes transfer learning, and the fourth is a computing resource problem; none refers to excessive fitting to the training data.

If accuracy is measured on the data used for training, a high score can result from mere memorization of that data. Evaluating on test data not used in training measures true capability on unseen data. Speed is not the main purpose of splitting, privacy protection is handled by other measures such as anonymization, and error correction is not something the split accomplishes.

Since the training uses data with the answers, the labels, attached in advance, this is supervised learning, and since the output is a category, it is classification. Unsupervised learning gives no answers, and reinforcement learning learns actions from rewards. Deep reinforcement learning combines reinforcement learning with neural networks and does not fit this problem, which uses correct labels.

x starts at 1 and becomes 2 after the first iteration, 4 after the second, and 8 after the third, so the answer is 8. 2 is the value after only the first iteration and 4 after only the second. 6 is the result of adding 2 three times, not of repeated doubling.

Each comparison in a binary search halves the candidates. 2 to the 9th is 512, not enough for 1000 items, while 2 to the 10th is 1024, exceeding 1000, so the target can be found in at most 10 comparisons. 500 is close to the average comparison count of a linear search, and 7 and 9 provide too few halvings.

The cases of finding it on the 1st comparison through the nth are equally likely, so the average is (1+2+…+n)÷n = (n+1)÷2. n is the worst-case comparison count, and log2 n is the rough comparison count of a binary search. n÷4 is a value with no basis.

Comparing 3 and 1 and swapping gives 1,3,4,1,5; 3 and 4 stay; swapping 4 and 1 gives 1,3,1,4,5; and 4 and 5 stay. The fully sorted state cannot be reached in one pass. The fourth option is the intermediate state after only the first swap, and the third is the arrangement that would result if the pass proceeded without swapping the leading 3 and 1.

A stack is last in, first out, so the first pop yields 3, the last value pushed. Pushing 4 and popping yields 4, and the next pop yields 2, giving 3, 4, 2. The first option is close to the order for a first-in, first-out queue, and the other two mix up the order of removal.

A list is a structure in which each element holds the location of the next element, a pointer, so insertion and deletion in the middle require only relinking the pointers. The second option describes an array, the third a stack, last in first out, and the fourth a queue, first in first out; none is a characteristic of the list structure.

JSON is a lightweight data description format that expresses pairs of item names and values enclosed in curly braces and is widely used for data exchange on the Web. CSV is a tabular format separating values with commas, XML is a markup language expressing the meaning and structure of data with custom tags, and HTML is the markup language describing the structure of Web pages.

R is a programming language specialized for statistical analysis and data analysis, with abundant statistical methods and graph-plotting features. HTML is a markup language for describing the structure of Web pages, SQL is a query language for operating databases, and XML is a markup language expressing the meaning and structure of data; none of these is a programming language for statistical analysis.