Suppose we have 999 records of employees Emp1, Emp2, Emp5, Emp6 and so on up to Emp998 and Emp999 stored in a sequence. Hence, records are nothing but a row in the table. In the employee table, we want to insert new records Emp3 and Emp4 in the sequence, and once we have done insertion we need to update and retrieve the record efficiently to optimize the performance of a database by minimizing the access time when query is being executed.
The records in the table represents the employee ID of each employee. Assume the table name is EMPLOYEES and the field name of employee IDs is ID, the following SQL query inserts Emp3 and Emp4 into the employees table.
INSERT INTO EMPLOYEES (ID) values (Emp3), (Emp4)
The complete question requires that we list the pros and cons of solving the problem using Index sequential access method (ISAM).
The advantage of using ISAM is that priorities are given to the indices of the records being inserted. So, Emp3 and Emp4 will be inserted immediately after Emp2.
However, a major disadvantage of using ISAM is that it is very expensive to keep the indices in a sequential order;
See attachment for the illustration of Index sequential access method (ISAM).
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Hydraulic fracturing is: Select one: a. a first step in storage of high-level radioactive waste b. a way of extracting natural gas shale formations c. a step in oil refining where oil is separated into different viscosities d. none of the above
Shale and other types of "tight" rock, or impermeable rock formations that lock in oil and gas and complicate the production of fossil fuels, can be extracted of their natural gas or oil using the modern high-volume hydraulic fracturing process. Thus, option B is correct.
What are the factor involving in Hydraulic fracturing?Shale gas is extracted through a procedure called hydraulic fracturing, also referred to as “fracking.” Due to the fact that shale reserves are often dispersed horizontally rather than vertically, deep holes are first drilled into the shale rock, then horizontal drilling is used to reach more of the gas.
A mixture of sand, chemicals, and water is pushed at high pressure down a borehole during hydraulic fracturing. The released gas can flow out of the rocks and back up the borehole because the water pressure causes cracks in the rock to widen, and the sand particles settle into the openings to keep them open.
Therefore, a way of extracting natural gas shale formations.
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What is the following diagram called?
Answer:
business
Explanation:
because shopping has great income
Answer:
is ur question right
Explanation:
is ur question right
is ur question right
Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate
records are kept. He used 13 pounds on a systein for Ms. Jones and 9 pounds on a system for a commercial
client. How many pounds should he have left in the cylinder?
tof
Troy should have
pounds of refrigerant left in the cylinder.
baon naid Thamar basic
محمود احمد مجد
12
اهداء ما در
Answer:
Amount of gas still in cylinder = 28 pound
Explanation:
Given:
Amount of gas in cylinder = 50 pound
Amount of gas used in Ms. Jones system = 13 pound
Amount of gas used in client system = 9 pound
Find:
Amount of gas still in cylinder
Computation:
Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system
Amount of gas still in cylinder = 50 - 13 - 9
Amount of gas still in cylinder = 28 pound
Why is logging done during drilling?
Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible
an atom that gained an electron is called
Answer:
Hey mate......
Explanation:
This is ur answer.....
When an atom gains/loses an electron, the atom becomes charged, and is called an ion.Hope it helps!
Brainliest pls!
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hỗ trợ mình với được không các bạn
Answer:
Explanation:
Be bop
A resistor has code 104 printed on it .What is the resistive value of this resistor
Answer:
X = 1 (1st digit in the code)
Y = 0 (2nd digit)
Z = 4 (3rd multiplier digit)
104 → 10 × 10^4 Ω
→ 10 × 10000Ω
→ 100 kΩ
resistors are marked 104, 105, 205, 751, and 754. The resistor marked with 104 should be 100kΩ (10x10^4), 105 would be 1MΩ (10x10^5), and 205 is 2MΩ (20x10^5). 751 is 750Ω (75x10^1), and 754 is 750kΩ (75x10^4).
Here we need to understand how a code in a resistor gives us information on the resistor. Here we will see that the code means that the resistance is 100,000 Ω.
When we use numbers, let's assume that we have 3 single-digit numbers abc.
So if the code in our resistor is abc, this will mean that the resistance of the resistor is:
ab×10^c Ω
Using this general rule we can see that if the code is 104, then the resistance will be:
r = 10×10^4 Ω
= 100,000 Ω
Then we can conclude that the resistive value of this resistor is 100,000 Ω
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Question in Statistics and Probability 2
Petrol samples were taken at three different stations along a national highway to determine whether the octane rating varied from one station to another. Six petrol specimens of the same category were randomly selected at each station in different days. We are interested to know if there is sufficient evidence to indicate a difference in the mean octane rating at the different stations.
State the Null and the Alternative Hypotheses.
once scientist has asked a question conducted an investigation and analyzed the data his or hwr goalis to
Answer:
hola
Explanation:
Explanation:
so its an question?..........
Tech A says that the voltage regulator controls the strength of the rotor s magnetic field. Tech B says that the voltage regulator is installed between the output terminal of the alternator and the positive terminal of the battery. Who is correct?
Answer:
Voltage Regulator
Technician A is correct.
Explanation:
Technician B is not correct. The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B. Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct. The computer can also be used to control the output of the alternator by controlling the field current.
If OSHA determines that an employer's response to a non-formal complaint is adequate, what options does the employee filing the non-formal complaint have?
Solve using Matlab the problems:
One using the permutation of n objects formula
One using the permutation of r objects out of n objects
You can pick these problems from the textbook or you can make up your own questions.
Help me pleaseeeee
Answer:
Explanation:
% Clears variables and screen
clear; clc
% Asks user for input
n = input('Total number of objects: ');
r = input('Size of subgroup: ');
% Computes and displays permutation according to basic formulas
p = 1;
for i = n - r + 1 : n
p = p*i;
end
str1 = [num2str(p) ' permutations'];
disp(str1)
% Computes and displays combinations according to basic formulas
str2 = [num2str(p/factorial(r)) ' combinations'];
disp(str2)
=================================================================================
Example: check
How many permutations and combinations can be made of the 15 alphabets, taking four at a time?
The answer is:
32760 permutations
1365 combinations
==================================================================================
Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Answer:
p q output ¬(p ∧ q)
0 0 1
0 1 1
1 0 1
0 0 0
p q output ¬p ∨ ¬q
0 0 1
0 1 1
1 0 1
0 0 0
Explanation:
We'll create two separate truth tables for both sides of the equation, and see if they match.
The expressions in the question use AND, OR and NOT operators.
The AND operation needs both inputs to be 1 to return a 1.The OR operation needs at least 1 of the inputs to be 1 to return a 1. The NOT operation takes a 1 and turns it into a 0, or takes a 0 and turns it into a 1.Let's start with ¬(p ∧ q)
NOT (0 AND 0) = NOT (0) = 1NOT (0 AND 1) = NOT (0) = 1NOT (1 AND 0) = NOT (0) = 1NOT (1 AND 1) = NOT (1) = 0Now let's move on to the second expression ¬p ∨ ¬q
NOT(0) OR NOT(0) = 1 OR 1 = 1NOT(0) OR NOT(1) = 1 OR 0 = 1NOT(1) OR NOT(0) = 0 OR 1 = 1NOT(0) OR NOT(0) = 0 OR 0 = 0Therefore we can say the two expressions are equivalent.
Attached the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:
What is the explanation of the truth table?As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.
De Morgan's law is a fundamental principle in propositional logic.
It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.
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Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium
at temperature T. The overall mole fraction of species 1 in the system is z1 = 0.65. At
temperature T, lnγ1 = 0.67 x2
2; lnγ2 = 0.67 x1
2; P1
sat = 32.27 kPa; and P2
sat = 73.14 kPa.
Assuming the validity of Eq. (13.19),
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13.10. Problems 511
smi96529_ch13_450-523.indd 511 01/06/17 03:27 PM
(a) Over what range of pressures can this system exist as two phases at the given T and z1?
(b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar
fraction of the system is vapor?
(c) Show whether or not the system exhibits an azeotrope
