Answer:
The probability that A occurs given that B occurred is 333/400, that is, 83.25%.
Step-by-step explanation:
Since in an experiment, the probability that event A occurs is 6/7, the probability that event B occurs is 4/5, and the probability that event A and B both occur is 2/3, to determine what is the probability that A occurs given that B occurs the following calculation must be performed:
4/5 = 0.8
2/3 = 0.666
0.8 x X = 0.666
X = 0.666 / 0.8
X = 0.8325
0.8325 = 333/400
Therefore, the probability that A occurs given that B occurred is 333/400, that is, 83.25%.
1. Five-sixths of the students in a math class passed the first test. If there are 36 students in the class, how many did not pass the test? *
Answer:
(1/6) of the students did not pass.
Ans: (1/6)30 = 5 of the students did not pass
To solve this question, we need to first figure out, in fractional form, the number of students who did not pass. The problem tells us that 5/6 of the students passed. Therefore, we know that 1/6 of the students did not pass, because 5/6 + 1/6 = 1. So, all we need to do is multiply 1/6 by the total number of students:
1/6 * 36 = 6
So, 6 students did not pass.
Evaluate the indefinite integral.
integar x4/1 + x^10 dx
Answer:
[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]
Step-by-step explanation:
Given
[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]
Required
Integrate
We have:
[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]
Let
[tex]u = x^5[/tex]
Differentiate
[tex]\frac{du}{dx} = 5x^4[/tex]
Make dx the subject
[tex]dx = \frac{du}{5x^4}[/tex]
So, we have:
[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]
[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}[/tex]
[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du[/tex]
Express x^(10) as x^(5*2)
[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du[/tex]
Rewrite as:
[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du[/tex]
Recall that: [tex]u = x^5[/tex]
[tex]\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du[/tex]
Integrate
[tex]\frac{1}{5} * \arctan(u) + c[/tex]
Substitute: [tex]u = x^5[/tex]
[tex]\frac{1}{5} * \arctan(x^5) + c[/tex]
Hence:
[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]
What is the name for a polygon with 4 sides
Answer: A. Quadrilateral
Step-by-step explanation:
The name for a polygon with 4 sides is a quadrilateral. Quadrilateral is a two-dimensional shape with four straight sides and four angles. The most common type of quadrilateral is the rectangle, which has four right angles. Other types of quadrilaterals include squares, parallelograms, trapezoids, and rhombuses.
Answer:
quadrilateral!!
Step-by-step explanation:
look at the beginning of the word, quad. Quad means 4 and is referring to a 4 sided figure.
In a school containing 360 children, 198 are girls. What percent of all children are girls? What percent of the children are boys? The number of girls is what percent of the number of boys? The number of boys is what percent of the number of girls?
the number of girls is _____% of the number of boys.
Answer:
1st question :55%
2nd ":122.22%
3rd ":81.82%
The expressions A, B, C, D, and E are left-hand sides of trigonometric identities. The expressions 1, 2, 3, 4, and 5 are right-hand side of identities. Match each of the left-hand sides below with the appropriate right-hand side.
A. tan(x)
B. cos(x)
C. sec(x)csc(x)
D. 1â(cos(x))^2/ cos(x)
E. 2sec(x)
1. sin(x)tan(x)
2. sin(x)sec(x)
3. tan(x)+cot(x)
4. cos(x)/1âsin(x)+1âsin(x)/cos(x)
5. sec(x)âsec(x)(sin(x))2
Answer:
[tex]A.\ \tan(x) \to 2.\ \sin(x) \sec(x)[/tex]
[tex]B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)[/tex]
[tex]C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)[/tex]
[tex]D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)[/tex]
[tex]E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex]
Step-by-step explanation:
Given
[tex]A.\ \tan(x)[/tex]
[tex]B.\ \cos(x)[/tex]
[tex]C.\ \sec(x)csc(x)[/tex]
[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]
[tex]E.\ 2\sec(x)[/tex]
Required
Match the above with the appropriate identity from
[tex]1.\ \sin(x) \tan(x)[/tex]
[tex]2.\ \sin(x) \sec(x)[/tex]
[tex]3.\ \tan(x) + \cot(x)[/tex]
[tex]4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}[/tex]
[tex]5.\ \sec(x) - \sec(x)(\sin(x))^2[/tex]
Solving (A):
[tex]A.\ \tan(x)[/tex]
In trigonometry,
[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]
So, we have:
[tex]\tan(x) = \frac{\sin(x)}{\cos(x)}[/tex]
Split
[tex]\tan(x) = \sin(x) * \frac{1}{\cos(x)}[/tex]
In trigonometry
[tex]\frac{1}{\cos(x)} =sec(x)[/tex]
So, we have:
[tex]\tan(x) = \sin(x) * \sec(x)[/tex]
[tex]\tan(x) = \sin(x) \sec(x)[/tex] --- proved
Solving (b):
[tex]B.\ \cos(x)[/tex]
Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1
So, we have:
[tex]\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}[/tex]
[tex]\cos(x) = \frac{\cos^2(x)}{\cos(x)}[/tex]
In trigonometry:
[tex]\cos^2(x) = 1 - \sin^2(x)[/tex]
So, we have:
[tex]\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}[/tex]
Split
[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}[/tex]
Rewrite as:
[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)[/tex]
Express [tex]\frac{1}{\cos(x)}\ as\ \sec(x)[/tex]
[tex]\cos(x) = \sec(x) - \sec(x) * \sin^2(x)[/tex]
[tex]\cos(x) = \sec(x) - \sec(x)\sin^2(x)[/tex] --- proved
Solving (C):
[tex]C.\ \sec(x)csc(x)[/tex]
In trigonometry
[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]
and
[tex]\csc(x)= \frac{1}{\sin(x)}[/tex]
So, we have:
[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}[/tex]
Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1
[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}[/tex]
[tex]\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}[/tex]
Express [tex]\frac{1}{\cos^2(x)}\ as\ \sec^2(x)[/tex] and [tex]\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}[/tex]
[tex]\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}[/tex]
[tex]\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}[/tex]
In trigonometry:
[tex]tan^2(x) + 1 =\sec^2(x)[/tex]
So, we have:
[tex]\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}[/tex]
Split
[tex]\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}[/tex]
Simplify
[tex]\sec(x)csc(x) = \tan(x) + \cot(x)[/tex] proved
Solving (D)
[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]
Open bracket
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}[/tex]
[tex]1 - \cos^2(x) = \sin^2(x)[/tex]
So, we have:
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}[/tex]
Split
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}[/tex]
[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]
So, we have:
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)[/tex]
[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)[/tex] --- proved
Solving (E):
[tex]E.\ 2\sec(x)[/tex]
In trigonometry
[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]
So, we have:
[tex]2\sec(x) = 2 * \frac{1}{\cos(x)}[/tex]
[tex]2\sec(x) = \frac{2}{\cos(x)}[/tex]
Multiply by [tex]\frac{1 - \sin(x)}{1 - \sin(x)}[/tex] --- an equivalent of 1
[tex]2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}[/tex]
[tex]2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}[/tex]
Open bracket
[tex]2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]
Express 2 as 1 + 1
[tex]2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]
Express 1 as [tex]\sin^2(x) + \cos^2(x)[/tex]
[tex]2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]
Rewrite as:
[tex]2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]
Expand
[tex]2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]
Factorize
[tex]2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]
Factor out 1 - sin(x)
[tex]2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]
Express as squares
[tex]2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]
Split
[tex]2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]
Cancel out like factors
[tex]2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex] --- proved
Johnny earns $15 per hour dog walking. He wants to earn at least $105 to buy a new
bike. At least how many hours does he need to work?
Answer:
it will take Johnny 7 hours
Step-by-step explanation:
It was easy, i divided 105 by 15 and got 7
Have a wonderful day :)
Answer:
7 hours
Step-by-step explanation:
15 * x = 105
--- ---
15 7
x=7
a geometry class has a total of 25 students. the number of males is 5 more than the number of females. how many males and how many females are in the class
Answer:
10 females, 15 males
Step-by-step explanation:
Let the number of females in the class be x. As the number of males is 5 more than the number of females, which is x, the number of males would be shown as x+5.
The number of females plus the number of males equal 25, so we get this equation:
x + (x+5) = 25
x + x + 5 = 25
2x + 5 = 25
2x = 20
x = 10.
So, there are 10 females, and since there are 5 more males than females, there are 10 + 5 = 15 males.
Evaluate the following expression 12xy=2,y=4
Answer:
1/24
Step-by-step explanation:
Solve for y =4
48x=2
Jasmine claims that the two triangles are congruent since the Side-Angle-Side (SAS) triangle congruence criterion ensures that a sequence of transformations will carry one triangle onto the other.
Which of the following statements best describes Jasmine’s claim?
1. The claim is correct since a translation carries one triangle onto the other.
2. The claim is correct since SAS is an acceptable triangle congruence criterion.
3. The claim is incorrect since SAS is not an acceptable triangle congruence criterion.
4. The claim is incorrect since the triangles do not meet the SAS triangle congruence criterion.
Image of triangles:
The claim is not correct as both the triangles do not meet the SAS triangle congruence criterion.
What is SAS Conguency ?If any two sides and angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule.
As from the definition above we can understand that for the triangles to be congruent the sides and the angle included between them should be same.
In the given triangles
The two sides of both the triangle has same length but the angle 30 degree is not the angle included for both the traingles
Therfore we can say that the triangles are not congruent with SAS congruency.
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(-3 squared +3) divided by
(10 – 6)
Answer:
research it many my answer will be wrong
3
Step-by-step explanation:
-3 square is 9, adding 3 you will have 12. 10-6 is 4. so dividing 12 by 4 would give you 3.
800 miles in 5 hours miles per hour
convert logx8=2 to exponential form
Answer:
8 = x²
Step-by-step explanation:
Logarithmic form: logₓ(8) = 2
Exponential form: 8 = x²
Find the cost of 12 items if the cost function is C(x) = 2x +12
Answer:
$36
Step-by-step explanation:
we need to find the cost of 12 items, and the function of determining the cost is C(x)=2x+12, where x is the number of items (the input value of the function)
we know that there are 12 items, so in this function, x=12
substitute x as 12 in the function; the x in C(x) also gets substituted as 12, as 12 is the input value and the value of C(x) is the output
C(12)=2(12)+12
multiply
C(12)=24+12
add
C(12)=36
that means, for 12 items, the cost will be $36
can someone help me solving for x?
Answer:
8, 45, 18, 8.75 pretty sure
Step-by-step explanation:
cross multiplying: when a/b=x/c you type in a*c/b to get x in calculator
what is the generalised from of 85
Answer:
8 × 10 + 5 × 1. is the correct answer .STAY BRAINLY
Activity: Factor Theorem
Use the Remainder Theorem and Factor Theorem to determine whether the given
binomial is a factor of P(x).
P(x) = 9x³+ 6x - 40 - 2x² + 2x4; binomial: x + 5
the given problem.
The given polynomial is a factor of P(x) because P(-5)=0.
From the remainder theorem, how do you derive the factor theorem?The Remainder Theorem states that if a synthetic division of a polynomial by x = a results in a zero remainder, then x = an is a zero of the polynomial (thanks to the Remainder Theorem), and x an is also a factor of the polynomial (courtesy of the Factor Theorem).
How can you tell if P(x) is a factor of X C?The Factor Theorem states that x - c is only a factor of P(x) when P(c) = 0.
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If 4 people share a 1/2 pound chocolate bar, how much chocolate will each person receive?
Answer:
1/8 pound per person
Step-by-step explanation:
Huilan is 7 years older than Thomas. The sum of their ages is 79. What is Thomas’s age?
Answer:
To solve this problem, we can set up the following equation:
Huilan's age + Thomas's age = 79
Let H be Huilan's age and T be Thomas's age. We can then rewrite the equation as:
H + T = 79
We are told that Huilan is 7 years older than Thomas, so we can write the following equation:
H = T + 7
Substituting the second equation into the first equation, we get:
(T + 7) + T = 79
Combining like terms, we get:
2T + 7 = 79
Subtracting 7 from both sides, we get:
2T = 72
Dividing both sides by 2, we get:
T = 36
Therefore, Thomas's age is 36 years.
Step-by-step explanation:
Answer:
let Thomas age be a
there fore, hullans age is a+7
(a+7)+a=79
2a=79-7
2a=72
a=36
Find, rounded to the nearest hundredth, the diagonal of a rectangle whose sides are 6 and 11.
WHAT IS THE ANSWER. I NEED IT ASAP
Answer:
The length of the diagonal is 12.53
Step-by-step explanation:
Since we have a rectangle, the diagonal is the hypotenuse of a right triangle. So we can use Pythagorean Theorem.
a^2 + b^2 = c^2
We know the two legs are 6 and 11, so we can find the hypotenuse (which is the diagonal) see image.
See image.
what is the area of 12 1/2 feet long and 11 3/4 feet wide
now, we're assuming this is some rectangular area, and thus is simply the product of both quantities, let's change all mixed fractions to improper fractions firstly.
[tex]\stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} ~\hfill \stackrel{mixed}{11\frac{3}{4}} \implies \cfrac{11\cdot 4+3}{4} \implies \stackrel{improper}{\cfrac{47}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{25}{2}\cdot \cfrac{47}{4}\implies \cfrac{1175}{8}\implies 146\frac{7}{8}~ft^2[/tex]
Evaluate 3a for a = 5
Answer:
Hello There!!
Step-by-step explanation:
The answer is 15 as 3a=3×a so 3×5=15.
hope this helps,have a great day!!
~Pinky~
A company employing 10,000 workers offers deluxe medical coverage (D), standard medical coverage (S) and economy medical coverage (E) to its employees. Of the employees, 30% have D, 60% have 5 and 10% have E. From past experience, the probability that an employee with D, will submit no claims during next year is 0.1. The corresponding probabilities for employees with S and E are 0.4 and 0.7 respectively. If an employee is selected at random;
a) What is the probability that the selected employee has standard coverage and will submit no
claim during next year? b) What is the probability that the selected employee will submit no claim during next year?
c) If the selected employee submits no claims during the next year, what is the probability that the employee has standard medical coverage (S)?
please give full answer
Find the mean absolute deviation (MAD) of the data set of 0.4, 0.2, 0.4, 0.6
Answer:
Assuming data is from population, 0.1
Step-by-step explanation:
Which is the equivalent of 145.12° written in DMS form?
A. 145° 7' 2"
B. 145° 12' 0"
C. 145° 2'36"
D. 145° 7' 12"
Answer:
the answer is
B. 145° 12' 0"
X = ?°
please help :)
Answer:
x=17 degrees
Step-by-step explanation:
We know that a right angle is equal to 90 degrees. Therefore, two angles between a right angle have to equal 90. Therefore, 73+x=90. Isolate x and get x=90-73. x=17. Therefore, x=17 degrees.
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When you are solving a compound inequality, how can you express the solution?
Please give your rationale.
Answer:
When solving a compound inequality, you can express the solution in two different ways: using a union or intersection.
Using a union:
A union represents the solutions that belong to either one inequality or the other, or both. You can express a union using the symbol "or," or by writing the solutions for each inequality separately and combining them using a "union" symbol (∪). For example, if you have the compound inequality "x > 2 or x < -3," you can express the solution as "x ∈ (-3, 2) ∪ (-∞, -3) ∪ (2, ∞)."
Using an intersection:
An intersection represents the solutions that belong to both inequalities. You can express an intersection using the symbol "and," or by writing the solutions for each inequality separately and combining them using an "intersection" symbol (∩). For example, if you have the compound inequality "x > 2 and x < -3," you can express the solution as "x ∈ (-3, 2) ∩ (-∞, -3) ∩ (2, ∞)."
Which method you use to express the solution will depend on the specific inequalities and the type of solutions you are looking for. It is important to carefully consider the inequality symbols (>, <, ≥, ≤) and the use of "and" or "or" in order to determine the correct solution.
When solving a compound inequality, the solution can be expressed in one of two ways: as an interval or as a union of intervals.
What is inequality?Inequality is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are not equal.
An interval is a set of all real numbers between two given numbers, and is written in the form (a, b) or [a, b], where a and b are real numbers, and a < b. If a and b are included in the interval, then the interval is written as [a, b]; if a and b are not included in the interval, then the interval is written as (a, b).
A union of intervals is a combination of two or more intervals, and is written in the form (a, b) ∪ (c, d) or [a, b] ∪ [c, d], where a, b, c, and d are real numbers, and a < b and c < d.
When solving a compound inequality, it is important to first simplify the inequality by combining like terms, isolating the variable on one side of the inequality, and combining any separate inequalities into a single inequality. Then, the solution can be found by graphing the inequality on a number line, and shading the portion of the number line that represents the solution.
Once the solution is found the solution can be expressed either as an interval or as a union of intervals depending on the inequality. For example, if the inequality is x>1 and x<3, then the solution would be expressed as (1,3) and this is an example of an interval. If it's x>1 and x≥3 then the solution would be expressed as (1,3] U [3,∞) and this is an example of union of intervals.
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2) Insert the smallest
digit to make the
number divisible by 8
.
1039_
Answer:
it would be 10392
Step-by-step explanation:
10392÷8=1299
Answer:
10392
Step-by-step explanation:
10391 ÷ 8 = 1298.875
Not divisible by 8
10392 ÷ 8 = 1299
Divisible by 8
Find the distance between the points
(-5/2, 9/4) and(-1/2,-3/4)
Give the exact distance and an approximate distance rounded to at least 2 decimal places.
Make sure to fully simplify any radicals in first answer
Answer:
Step-by-step explanation:
distance = [tex]\sqrt{(-5/2+1/2)^{2} +(9/4+3/4)^{2} }=\sqrt{13}[/tex]=3.61
if i'm at $567,811 how much more do i need to get to 1,000,000
Answer:
$432,189
Step-by-step explanation:
$1,000,000 - $567,811 = $432,189
Ian is taking a true/false quiz that has four questions on it. The correct sequence of answers
is T, T, F, T. Ian has not studied and yet guesses all answers correctly- He decides he wants
to determine how likely this is to happen using a simulation. He decides to use a fair coin to
do the simulation.
Describe the design of a simulation that would allow Ian to determine how likely it would
be to guess all answers correctly on this quiz.
The design of a simulation that would allow Ian to determine how likely it would
be to guess all answers correctly on this quiz can be Illustrated through a coin.
How to design the simulation?Ian can design a simulation to determine the likelihood of guessing all answers correctly on the quiz by using a fair coin. He can flip the coin four times, with each flip representing a guess on the quiz. If the coin lands heads up, he marks it as a true answer, and if it lands tails up, he marks it as a false answer.
He can repeat this process a large number of times (e.g. 1000) and count the number of times that the sequence T, T, F, T is generated. The proportion of times that this sequence is generated out of the total number of simulations will be an estimate of the probability of guessing all answers correctly on the quiz by chance.
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