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How To Study Effectively For SAT
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Read MoreIf you are looking for a FREE, all-inclusive ALEKS answer key that covers all topics, you have come to the correct place. You will find extensive resources and materials to help you easily pass the ALEKS exam. Whether you are struggling with algebraic concepts, trigonometric functions, or any other difficult topic, our extensive ALEKS answers can provide the guidance and support you need to succeed. Therefore, why seek elsewhere when you can find everything you require here? Examine our ALEKS answer key right away and begin preparing for success.
Assessment and Learning in knowledge spaces, or Alek, is an online educational platform for k-12 and higher education in subjects like chemistry, statistics, accounting, and mathematics used to determine how well a student knows a particular subject.
The platform utilizes adaptive questioning by asking questions that cover the course material, assessing wrong answers, and giving feedback for each problem to determine the areas of coursework you are conversant with and the ones you need to improve on.
This program is used by students worldwide to make learning systematic and fun. Aleks reassesses the candidates to ensure they grasp and retain knowledge in all topics learned. It provides students with information about what they do and don't know to enable them to apply more effort to areas they didn't do well.
Education stakeholders have evaluated and approved Aleks's platform for years to ascertain its suitability for teaching concepts and helping students gain knowledge. Through research and constant improvement on the subjects offered, the platform has introduced several types of assessment ranging from basic recall to deep conceptual understanding that goes into depth regarding the type of subject.
After enrolling for the Aleks course, questions about cheating always come up. Cheating directly while appearing for assessment is impossible. Aleks is an electronic webpage that detects if you are cheating. The surveys conducted about the program have suggested that cheating in Aleks is more difficult than in the actual classroom.
Most students in the Alek courses wonder if they can find answers to Alek questions online or have come across platforms purporting to sell real Alek questions and answers.
You may be tempted to spend your hard-earned money to buy these papers to pass your test. PLEASE DON'T. I hate to break it to you, but you won't find real Aleks's answers online. The Alek system randomly regenerates all the questions and tests, making it impossible to find the questions or answers online. But worry, you can contact our professional tutors at Acemyhomework to help with Aleks's answers.
You can get answers to all questions with a step by step procedures on how they arrived at the answer. They will also assist you in any field or subject of your choice and help you solve whatever question is giving you a headache, including:
Here is a comprehensive list of ALEKS topics, from which you may select any subject and receive free answers.
Choose any of these topics that correspond with your current academic requirements and gain access to the corresponding free answers.
This section will discuss all the math topics, math test questions and answers, and the frequently asked question in ALEKS from K-12 grade to higher education.
Chemistry is a branch of science that deals with substances' properties, composition, and structure.
1. Identify the Lewis acid in K3[Al(C2O4)3].
a) K+
b) Al
c) Al3+
d) [Al(C2O4)3]3-
Answer: c
Explanation: Aluminium ion is the Lewis acid as it can accept 3 electrons from the donor atom to form the complex [Al(C2O4)3]3-.
2. Which of the following is the central atom/ion in [CoCl(NH3)5]2+?
a) Co
b) Co2+
c) Co3+
d) Cl–
Answer: c
Explanation: The central ion in the given complex ion is cobalt, which accepts electrons to bond with the Cl atom and ammonia molecules. Since the primary valence of Co in this compound is +3, the ion in Co3+.
Statistics is the branch of science that studies and develops methods for gathering, analyzing, interpreting, and presenting empirical data.
Descriptive Statistics: Descriptive statistics use data to describe a population through numerical calculations, graphs, and tables. Data is typically arranged and displayed in tables or graphs summarizing details such as histograms, pie charts, bars, or scatter plots.
Inferential Statistics- Inferential Statistics uses a sample of data from the population to form inferences and predictions about the population. It uses probability to derive a conclusion after generalizing a huge dataset.
Mean - It is a metric for calculating the average of all values in a sample set.
Median - It is a measure of a sample set's central value. The data set is sorted from the lowest to the highest value, and then the precise middle is determined.
Mode - The value most frequently in the core set is the mode.
Range measures how values in a sample set or data set are spaced apart.
Range= maximum value-minimum value
Variance expresses how far a random variable deviates from its expected value and can be calculated as a square of deviation.
\[S^2 = \sum_{i=1} ^{n} (x_i - \bar {x})^2 \div n\]
n represents total data points, x represents the mean, and xi represents individual data points in these formulas.
Dispersion measures how far a set of data deviates from its mean.
σ = \[\sqrt{(1 \div n) \sum _{i=1} ^{n} (x_i - \mu)^2}\]
Here are some basic statistics questions and answers to solve and practice.
Question 1: Value of Mean and Mode are Given as 30 and 15, respectively. Value of Median is -
(a) 25
(b) 26
(c) 24.5
(d) 22.5
Solution: The answer is option (a) 25. Mean – Mode = 3 (Mean – Median) is the relation between mean, median, and mode. Substituting the value, the equation becomes 30 – 15 = 3 (30 – Median). On solving, the median comes to 25.
Question 2: Which Among the Following Cannot be Represented Graphically?
(a) Median
(b) Mean
(c) Mode
(d) None of the above option
Solution: The answer is option (b). Mean is a specific value derived from the sum of all values and divided by the number of times values. Given that this value is single and cannot be compared with other values, its graphical representation is not feasible.
Probability is the chance of occurrence in a random event.
Probability Formula: If an event E occurs, then the empirical probability of an event to happen is:
P(E) = number of trials in which the event happened/Total number of trials
The theoretical probability of an event E, P(E), is defined as:
P(E) = (Number of outcomes favourable to E)/(Number of all possible outcomes of the experiment)
Impossible event: The probability of an occurrence/event being impossible is 0. Such an event is called an impossible event.
Sure event: The probability of an event sure to occur is 1. Such an event is known as a certain event or a certain event.
Question 1. Two coins are tossed 500 times, and we get:
Two heads: 105 times
One head: 275 times
No head: 120 times
Find the probability of each event occurring.
Solution: Let us say the events of getting two heads, one head, and no head, by E1, E2, and E3, respectively.
P(E1) = 105/500 = 0.21
P(E2) = 275/500 = 0.55
P(E3) = 120/500 = 0.24
The sum of probabilities of all elementary events of a random experiment is 1.
P(E1)+P(E2)+P(E3) = 0.21+0.55+0.24 = 1
Example 2: A bucket contains 5 blue, 4 green, and 5 red balls. Sudheer is asked to pick 2 balls randomly from the bucket without replacement, and then one more ball will be picked. What is the probability he picked 2 green balls and 1 blue ball?
Algebra is the branch of mathematics that helps represent problems or situations through mathematical expressions. It involves variables like x, y, and z and mathematical operations like addition, subtraction, multiplication, and division to form a meaningful mathematical expression.
An algebraic expression in algebra is formed using constants, variables, and basic arithmetic operations of addition(+), subtraction (-), multiplication(×), and division (/).
The basic rules or properties of algebra for variables, algebraic expressions, or real numbers a, b, and c are as given below,
An example of an algebraic expression is 5x + 6. Here 5 and 6 are fixed numbers, and x is a variable. Further, the variables can be simple, using alphabets like x, y, and z, or complex variables like x2, x3, xn, xy, x2y, etc.
Algebraic expressions are also known as polynomials. A polynomial is an expression consisting of variables (indeterminates), coefficients, and non-negative integer exponents of variables. Example: 5x3 + 4x2 + 7x + 2 = 0
It is an equation that is always true regardless of the values assigned to the variables. These formulae involve squares and cubes of algebraic expressions and help in solving the algebraic expressions in a few quick steps. The frequently used algebraic formulas are listed below.
Let us see the application of these formulas in algebra using the following example,
Question 1: Solve, (x-1)2 = [4√(x-4)]2
Solution: x2-2x+1 = 16(x-4)
x2-2x+1 = 16x-64
x2-18x+65 = 0
(x-13) (x-5) = 0
Hence, x = 13 and x = 5.
Question 2: Solve (2x+y)2
Solution: Using the identity: (a+b)2 = a2 + b2 + 2 ab, we get;
(2x+y) = (2x)2 + y2 + 2.2x.y = 4x2 + y2 + 4xy
Question 3: Solve (99)2 using the algebraic identity.
Solution: We can write 99 = 100 -1
Therefore, (100 – 1 )2
= 1002 + 12 – 2 x 100 x 1 [By identity: (a -b)2 = a2 + b2 – 2ab
= 10000 + 1 – 200
= 9801
Question 4: There are 47 boys in the class. This is three more than four times the number of girls. How many girls are there in the class?
Solution: Let the number of girls be x
As per the given statement,
4 x + 3 = 47
4x = 47 – 3
x = 44/4
x = 11
Geometry is a discipline of mathematics that studies various forms of shapes and sizes of real-world objects. We study different angles, transformations, and similarities of figures in geometry.
The fundamentals of geometry are based on point, line, angle, and plane concepts. These fundamental geometrical concepts govern all geometrical shapes.
Flat shapes like squares, circles, and triangles are a part of flat geometry. These shapes have only 2 dimensions, the length, and the width.
A 3D can be defined as a solid figure, object, or shape with three dimensions: length, width, and height. A cube, rectangular prism, sphere, cone, and cylinder are the basic 3-dimensional shapes.
An angle can be defined as the figure formed by two rays meeting at a common endpoint. The symbol represents an angle ∠. Angles are measured in degrees (°) using a protractor. For example, 45 degrees is represented as 45°.
Classification of angles based on measurements
Example
Question 1: The length of a rectangle is 3 more inches than its breadth. The area of the rectangle is 40 in2. What is the perimeter of the rectangle?
Solution:
Given: Area = 40 in2.
Let "l" be the length and "b" be the breadth of the rectangle.
According to the given question,
b = b and l = 3+b
We know that the area of a rectangle is lb units.
So, 40 = (3+b)b
40 = 3b +b2
This can be written as b2+3b-40 = 0
On factoring the above equation, we get b= 5 and b= -8.
Since the length value cannot be negative, we have b = 5 inches.
Substitute b = 5 in l = 3 + b, we get
l = 3 + 5 = 8 inches.
As we know, the perimeter of a rectangle is 2(l+b) units
P = 2 ( 8 + 5)
P = 2 (13) = 26
Hence, the perimeter of a rectangle is 26 inches.
Question 2: What is the area of a circle in terms of π, whose diameter is 16 cm?
Solution:
Given: Diameter = 16 cm.
Hence, radius, r = 8 cm
We know that the area of a circle = πr2 square units.
Now, substitute r = 8 cm in the formula, and we get
A = π(8)2 cm2
A = 64π cm2
Hence, the area of a circle whose diameter is 16 cm = 64π cm2.
Question 3: Find the curved surface area of a hemisphere whose radius is 14 cm.
Solution:
Given: Radius = 14 cm.
As we know, the curved surface area of a hemisphere is 2πr2 square units.
CSA of hemisphere = 2×(22/7)×14×14
CSA = 2×22×2×14
CSA = 1232
Hence, the curved surface area of a hemisphere is 1232 cm2.
Question 4: Find the volume of a cone in terms π, whose radius is 3 cm and height is 4 cm.
Solution:
Given: Radius = 3 cm
Height = 4 cm
We know that the formula to find the volume of a cone is V = (⅓)πr2h cubic units.
Now, substitute the values in the formula, and we get
V = (⅓)π(3)2(4)
V = π(3)(4)
V = 12π cm3
Hence, the volume of a cone in terms of π is 12π cm3.
This is a branch of math that deals with saving and spending money.
Millicent is planning to travel to the United States of America in 3 years. She estimated that her vacation would cost $1,236,000. Given the current interest rate of 6%, how much money should she invest now?
Solution
Let's use the formula to determine the future value of compound interest:
FV = PV*(1+r)^n
where FV is the future value;
PV is the present value;
r is the interest rate.
In our case, the FV is $1,236,000, r is 6%, n is 3 years, and we need to find the present value (PV)
PV = FV / (1+r)^n = 1,236,000 / (1+0,06)^3 = $1,037,769.43
Millicent must invest $1,037,769.43 now to have $1,230,000 in 3 years at 6% per annum
Answer: C. $1,037,769.43
Everything you need to know about graphs
Aleks platform has been evaluated and approved by education stakeholders for many years to ascertain whether it is successful in teaching concepts and helping students gain knowledge. If you are struggling with Alek questions, you can ask a tutor directly or share your question on the site question page, and a tutor from all across the globe will answer them. Tutors not only help students understand their homework but also offers extra help to students to enable them to answer frequently asked questions about the subject.
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Discover the ultimate guide to studying effectively for the SAT with expert tips, tricks, and techniques from AceMyHomework. Boost your score and secure your dream college spot!
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