Tüm alıştırma soruları

13931 soru

Soru 7501Soru

A container contains 66 red counters, 44 blue counters, and 33 green counters. In how many different ways can a selection of 55 counters be made if the selection must contain at least 22 red counters, at least 11 blue counter, and at most 11 green counter?

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Cevap: 750

Cevap

The total number of different ways to make the selection is 750.
Breaking the problem into disjoint cases based on the number of green counters (0 or 1) and applying the combination formula for red, blue, and green counters in each valid configuration yields 240+510=750240 + 510 = 750 total ways.

Adım Adım Çözüm

1
Determine the allowable counts for Green (GG), Red (RR), and Blue (BB) counters
Green counters can be 0 or 1. If G=0G=0, R+B=5R+B=5 with R2,B1R \geq 2, B \geq 1. If G=1G=1, R+B=4R+B=4 with R2,B1R \geq 2, B \geq 1.
The constraints state G1G \leq 1, R2R \geq 2, and B1B \geq 1 for a total of 5 counters.
2
Calculate combinations for Case 1 (G=0G = 0)
Ways for (2R,3B,0G)=60(2R, 3B, 0G) = 60; (3R,2B,0G)=120(3R, 2B, 0G) = 120; (4R,1B,0G)=60(4R, 1B, 0G) = 60. Total for Case 1 = 240.
Apply the combinations formula (nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!(n-r)!} to each sub-case.
3
Calculate combinations for Case 2 (G=1G = 1)
Ways for (2R,2B,1G)=270(2R, 2B, 1G) = 270; (3R,1B,1G)=240(3R, 1B, 1G) = 240. Total for Case 2 = 510.
Apply combinations to choose 1 Green counter along with the valid Red and Blue combinations.
4
Sum all mutually exclusive cases
Total selection ways = 240+510=750240 + 510 = 750.
According to the addition principle of counting, the totals of mutually exclusive cases are added together.

Anahtar Kavram

Combinations with Multiple Conditional Constraints
Soru 7502Soru

An arithmetic progression has a first term of 55 and a common difference of 44. What is the 12th12^{\text{th}} term of this progression?

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Cevap: 4949

Cevap

4949
The nthn^{\text{th}} term of an arithmetic progression is determined using the formula Tn=a+(n1)dT_n = a + (n - 1)d. Substituting a=5a = 5, d=4d = 4, and n=12n = 12 yields T12=5+11×4=49T_{12} = 5 + 11 \times 4 = 49, making 4949 the correct value.

Adım Adım Çözüm

1
Identify the given values from the problem statement
First term a=5a = 5, common difference d=4d = 4, and position n=12n = 12
These are the standard variables required for calculating terms in an arithmetic progression.
2
Apply the nthn^{\text{th}} term formula for an arithmetic progression
T12=5+(121)×4T_{12} = 5 + (12 - 1) \times 4
The standard formula for the nthn^{\text{th}} term of an AP is Tn=a+(n1)dT_n = a + (n - 1)d.
3
Simplify the expression to find the final value
T12=5+11×4=5+44=49T_{12} = 5 + 11 \times 4 = 5 + 44 = 49
Perform multiplication before addition according to standard order of operations.

Anahtar Kavram

Arithmetic Progression nthn^{\text{th}} Term Formula
Tahmini Süre:45s
Soru 7503Soru

If (1681)x1×(278)x+2=94\left( \frac{16}{81} \right)^{x-1} \times \left( \frac{27}{8} \right)^{x+2} = \frac{9}{4}, find the value of xx.

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Cevap: 88

Cevap

The value of xx is 88.
By writing 1681\frac{16}{81} as (23)4\left(\frac{2}{3}\right)^4, 278\frac{27}{8} as (23)3\left(\frac{2}{3}\right)^{-3}, and 94\frac{9}{4} as (23)2\left(\frac{2}{3}\right)^{-2}, the equation simplifies via exponent addition to (23)4x43x6=(23)2\left(\frac{2}{3}\right)^{4x - 4 - 3x - 6} = \left(\frac{2}{3}\right)^{-2}. Equating indices gives x10=2x - 10 = -2, which yields the correct solution x=8x = 8.

Adım Adım Çözüm

1
Express all fractional terms with a common base of 23\frac{2}{3}
1681=(23)4\frac{16}{81} = \left(\frac{2}{3}\right)^4, 278=(32)3=(23)3\frac{27}{8} = \left(\frac{3}{2}\right)^3 = \left(\frac{2}{3}\right)^{-3}, and 94=(32)2=(23)2\frac{9}{4} = \left(\frac{3}{2}\right)^2 = \left(\frac{2}{3}\right)^{-2}
Converting all terms to a single common base allows exponents to be combined using the laws of indices.
2
Substitute the common base expressions back into the original equation
((23)4)x1×((23)3)x+2=(23)2\left( \left(\frac{2}{3}\right)^4 \right)^{x-1} \times \left( \left(\frac{2}{3}\right)^{-3} \right)^{x+2} = \left(\frac{2}{3}\right)^{-2}
Applying the power of a power law (am)n=amn(a^m)^n = a^{mn} to simplify each term.
3
Apply the power law and multiplication law of indices
(23)4(x1)×(23)3(x+2)=(23)2    (23)4(x1)3(x+2)=(23)2\left(\frac{2}{3}\right)^{4(x-1)} \times \left(\frac{2}{3}\right)^{-3(x+2)} = \left(\frac{2}{3}\right)^{-2} \implies \left(\frac{2}{3}\right)^{4(x-1) - 3(x+2)} = \left(\frac{2}{3}\right)^{-2}
When multiplying exponential terms with identical bases, add their exponents: am×an=am+na^m \times a^n = a^{m+n}.
4
Equate the exponents and solve for xx
4(x1)3(x+2)=2    4x43x6=2    x10=2    x=84(x-1) - 3(x+2) = -2 \implies 4x - 4 - 3x - 6 = -2 \implies x - 10 = -2 \implies x = 8
Since the bases on both sides are equal and non-zero, their indices must be equal.

Anahtar Kavram

Laws of Indices: Base Conversion and Exponential Equations
Tahmini Süre:2m 0s
Soru 7504Soru

Let the universal set be U={xZ:1x25}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 25\}. If A={xU:x is a multiple of 3}A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\} and B={xU:x is a perfect square}B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}, what is the cardinality of (AB)(A \cup B)'?

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Cevap: 13

Cevap

13
The total number of elements in the universal set is 25. Set A has 8 elements (multiples of 3) and set B has 5 elements (perfect squares). The element 9 belongs to both sets. Therefore, the union of A and B contains 8 + 5 - 1 = 12 elements. Subtracting this from the universal set size gives 25 - 12 = 13 elements in the complement.

Adım Adım Çözüm

1
Determine the elements of the universal set and its cardinality.
\mathcal{U} = \{1, 2, 3, \dots, 25\},so, so n(\mathcal{U}) = 25$.
The universal set contains all integers from 1 to 25 inclusive.
2
List the elements of set A and set B.
A={3,6,9,12,15,18,21,24}A = \{3, 6, 9, 12, 15, 18, 21, 24\} (so n(A)=8n(A) = 8) and B={1,4,9,16,25}B = \{1, 4, 9, 16, 25\} (so n(B)=5n(B) = 5).
Set A contains multiples of 3 within the domain, and set B contains perfect squares within the domain.
3
Find the intersection ABA \cap B and calculate the cardinality of the union ABA \cup B.
AB={9}A \cap B = \{9\}, so n(AB)=1n(A \cap B) = 1. Thus, n(AB)=n(A)+n(B)n(AB)=8+51=12n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8 + 5 - 1 = 12.
The principle of inclusion-exclusion avoids double-counting the common element 9.
4
Calculate the cardinality of the complement (AB)(A \cup B)'.
n((AB))=n(U)n(AB)=2512=13n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 25 - 12 = 13.
The complement consists of all elements in the universal set that do not belong to the union of A and B.

Anahtar Kavram

Complement of Set Union and Inclusion-Exclusion Principle
Soru 7505Soru

Given that θ\theta is an acute angle such that tanθ=43\tan \theta = \frac{4}{3}, calculate the numerical value of the expression 3sinθ+2cosθ3sinθcosθ\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - \cos \theta}.

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Cevap: 2

Cevap

The exact numerical value of the given expression is 2.
Dividing both the numerator and denominator of 3sinθ+2cosθ3sinθcosθ\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - \cos \theta} by cosθ\cos \theta gives 3tanθ+23tanθ1\frac{3\tan \theta + 2}{3\tan \theta - 1}. Substituting tanθ=43\tan \theta = \frac{4}{3} yields 3(4/3)+23(4/3)1=4+241=63=2\frac{3(4/3) + 2}{3(4/3) - 1} = \frac{4 + 2}{4 - 1} = \frac{6}{3} = 2.

Adım Adım Çözüm

1
Express sine and cosine terms in terms of tangent or find individual ratio values
Divide every term in the numerator and denominator by cosθ\cos \theta to obtain 3tanθ+23tanθ1\frac{3\tan \theta + 2}{3\tan \theta - 1}. Alternatively, using a right triangle with opposite side = 4 and adjacent side = 3 gives hypotenuse = 5, so sinθ=45\sin \theta = \frac{4}{5} and cosθ=35\cos \theta = \frac{3}{5}.
Converting to tanθ\tan \theta simplifies the calculation directly without evaluating square roots or hypotenuse.
2
Substitute the value of tanθ=43\tan \theta = \frac{4}{3} into the expression
Numerator: 3(43)+2=4+2=63\left(\frac{4}{3}\right) + 2 = 4 + 2 = 6. Denominator: 3(43)1=41=33\left(\frac{4}{3}\right) - 1 = 4 - 1 = 3.
Simplifies numerical fractions in both parts of the fraction.
3
Divide numerator by denominator
63=2.\frac{6}{3} = 2.
Yields the final integer solution.

Anahtar Kavram

Basic Trigonometric Ratios and Quotients
Soru 7506Soru

The cumulative frequency distribution of the operational lifespans (in hours) for a batch of 100100 precision LED modules tested in a laboratory is summarized below:

Lifespan Interval (hours)Class BoundariesCumulative Frequency
100119100 - 11999.5119.599.5 - 119.51010
120139120 - 139119.5139.5119.5 - 139.52525
140159140 - 159139.5159.5139.5 - 159.56060
160179160 - 179159.5179.5159.5 - 179.58585
180199180 - 199179.5199.5179.5 - 199.5100100

Using linear interpolation for cumulative frequency distributions, calculate the 75th percentile (P75P_{75}) of the lifespan of these modules in hours.

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Cevap: 171.5

Cevap

The 75th percentile of operational lifespan is 171.5 hours.
The 75th percentile rank position is 0.75×100=750.75 \times 100 = 75. The percentile falls within the class boundary 159.5179.5159.5 - 179.5. Substituting lower class boundary L=159.5L = 159.5, preceding cumulative frequency c.f.=60c.f. = 60, class frequency f=25f = 25, and class width c=20c = 20 into P75=L+(75c.f.f)×cP_{75} = L + \left(\frac{75 - c.f.}{f}\right) \times c gives 159.5+(1525)×20=171.5159.5 + \left(\frac{15}{25}\right) \times 20 = 171.5 hours.

Adım Adım Çözüm

1
Determine the rank position of the 75th percentile.
Rank position = 75th value out of 100.
The 75th percentile corresponds to 75% of the total frequency N = 100.
2
Identify the percentile class interval and extract relevant parameters.
Class interval is 159.5 - 179.5, with L = 159.5, c.f. = 60, f = 25, and c = 20.
The cumulative frequency increases from 60 to 85 across the boundary 159.5 to 179.5, which contains the 75th value.
3
Compute the percentile value using ogive linear interpolation.
P_75 = 159.5 + [(75 - 60) / 25] * 20 = 171.5 hours.
Applying the cumulative frequency interpolation formula yields the exact value.

Anahtar Kavram

Calculating Percentiles from Cumulative Frequency / Ogives
Soru 7507Soru

Given the matrices A=(2x13)A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} and B=(1042)B = \begin{pmatrix} 1 & 0 \\ 4 & 2 \end{pmatrix}, if det(A1BT)=15\det(A^{-1} B^T) = \frac{1}{5}, what is the value of xx?

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Cevap: 4

Cevap

The value of xx is 4.
Using the properties det(BT)=det(B)=2\det(B^T) = \det(B) = 2 and det(A1)=1det(A)=16+x\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}, the product rule gives det(A1BT)=26+x\det(A^{-1}B^T) = \frac{2}{6+x}. Equating this to 15\frac{1}{5} gives 6+x=106+x = 10, so x=4x = 4.

Adım Adım Çözüm

1
Calculate the determinants of matrices AA and BB.
\det(A) = (2)(3) - (-1)(x) = 6 + x and and \det(B) = (1)(2) - (0)(4) = 2$.
The determinant of a 2x2 matrix is computed using adbcad - bc.
2
Apply determinant rules for inverse and transpose matrices.
\det(B^T) = \det(B) = 2 and and \det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}$.
The determinant of a transpose equals the original determinant, and the determinant of an inverse is its reciprocal.
3
Express det(A1BT)\det(A^{-1} B^T) using the product rule of determinants.
\det(A^{-1} B^T) = \det(A^{-1}) \cdot \det(B^T) = \frac{1}{6+x} \cdot 2 = \frac{2}{6+x}$.
For square matrices, det(MN)=det(M)det(N)\det(MN) = \det(M)\det(N).
4
Equate the expression to 15\frac{1}{5} and solve for xx.
\frac{2}{6+x} = \frac{1}{5} \implies 10 = 6 + x \implies x = 4$.
Cross-multiplying and simplifying gives x=4x = 4.

Anahtar Kavram

Properties of Determinants (Product, Inverse, and Transpose)
Soru 7508Soru

A box contains 66 red balls and 44 blue balls. A ball is drawn at random, its color is recorded, and it is then returned to the box. A second ball is subsequently drawn at random. What is the probability of selecting a red ball on the first draw and a blue ball on the second draw?

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Cevap: 625\frac{6}{25}

Cevap

625\frac{6}{25}
Because the first ball is replaced after being recorded, the two draws are independent compound events. The probability of drawing a red ball first is 610=35\frac{6}{10} = \frac{3}{5}, and the probability of drawing a blue ball second is 410=25\frac{4}{10} = \frac{2}{5}. Multiplying these probabilities yields P(Red and Blue)=35×25=625P(\text{Red and Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}.

Adım Adım Çözüm

1
Calculate the probability of drawing a red ball on the first draw
P(Red)=66+4=610=35P(\text{Red}) = \frac{6}{6 + 4} = \frac{6}{10} = \frac{3}{5}
There are 66 red balls out of a total of 1010 balls.
2
Determine independence and calculate the probability of drawing a blue ball on the second draw
P(Blue)=410=25P(\text{Blue}) = \frac{4}{10} = \frac{2}{5}
Because the first ball is replaced before the second draw, the total number of balls and their composition remain unchanged.
3
Apply the multiplication law for independent compound events
P(Red and then Blue)=P(Red)×P(Blue)=35×25=625P(\text{Red and then Blue}) = P(\text{Red}) \times P(\text{Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}
For independent events AA and BB, P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

Anahtar Kavram

Multiplication Law of Probability for Independent Events
Tahmini Süre:45s
Soru 7509Soru

Given the universal set U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} and a subset P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\}, which of the following sets represents the complement of PP, denoted as PP'?

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Cevap: {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}

Cevap

The set {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}
The prime numbers in UU are 2,3,5,2, 3, 5, and 77, making P={2,3,5,7}P = \{2, 3, 5, 7\}. The complement PP' consists of all elements in UU that do not belong to PP, which yields {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}.

Adım Adım Çözüm

1
Identify the elements of the universal set UU
U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
The universal set defines the boundary of all possible elements under consideration.
2
Identify the elements belonging to set PP (prime numbers in UU)
P={2,3,5,7}P = \{2, 3, 5, 7\}
Prime numbers have exactly two distinct positive divisors: 1 and themselves. Note that 1 is not a prime number.
3
Compute the complement P=UPP' = U \setminus P
P={1,4,6,8,9,10}P' = \{1, 4, 6, 8, 9, 10\}
The complement of set PP consists of all elements present in UU that are not present in PP.

Anahtar Kavram

Set Complement and Universal Set Boundaries
Tahmini Süre:45s
Soru 7510Soru

By evaluating the limit of the difference quotient limh0f(x+h)f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, determine the value of the derivative of the function f(x)=2x24x+5f(x) = 2x^2 - 4x + 5 at the point where x=3x = 3.

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Cevap: 8

Cevap

The value of the derivative of f(x)=2x24x+5f(x) = 2x^2 - 4x + 5 at x=3x = 3 is 8.
Applying first principles, f(x+h)f(x)=4xh+2h24hf(x+h) - f(x) = 4xh + 2h^2 - 4h. Dividing by hh gives 4x+2h44x + 2h - 4. Taking the limit as h0h \to 0 produces f(x)=4x4f'(x) = 4x - 4. Substituting x=3x = 3 yields 4(3)4=84(3) - 4 = 8.

Adım Adım Çözüm

1
Substitute (x+h)(x+h) into the function f(x)=2x24x+5f(x) = 2x^2 - 4x + 5 and expand
f(x+h)=2(x2+2xh+h2)4x4h+5=2x2+4xh+2h24x4h+5f(x+h) = 2(x^2 + 2xh + h^2) - 4x - 4h + 5 = 2x^2 + 4xh + 2h^2 - 4x - 4h + 5
Apply algebraic expansion to determine the value of the function at x+hx+h.
2
Form the difference f(x+h)f(x)f(x+h) - f(x)
f(x+h)f(x)=4xh+2h24hf(x+h) - f(x) = 4xh + 2h^2 - 4h
Subtract the original function terms to leave only terms containing hh.
3
Divide the difference by hh
\frac{f(x+h) - f(x)}{h} = 4x + 2h - 4
Simplify the difference quotient prior to taking the limit.
4
Compute the limit as h0h \to 0
f(x)=4x4f'(x) = 4x - 4
Taking the limit yields the general derivative function f(x)f'(x).
5
Evaluate f(x)f'(x) at x=3x = 3
f(3)=4(3)4=8f'(3) = 4(3) - 4 = 8
Substitute x=3x = 3 to find the numerical rate of change at the given point.

Anahtar Kavram

Differentiation from First Principles
Soru 7511Soru

If y=ln(cos(3x))y = \ln(\cos(3x)), what is dydx\frac{dy}{dx}?

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Cevap: 3tan(3x)-3\tan(3x)

Cevap

3tan(3x)-3\tan(3x)
Differentiating y=ln(cos(3x))y = \ln(\cos(3x)) requires using the chain rule twice: first for ln(u)\ln(u), giving 1cos(3x)\frac{1}{\cos(3x)}, and then for cos(3x)\cos(3x), giving 3sin(3x)-3\sin(3x). Multiplying these results yields 3sin(3x)cos(3x)\frac{-3\sin(3x)}{\cos(3x)}, which simplifies to 3tan(3x)-3\tan(3x).

Adım Adım Çözüm

1
Apply the chain rule formula for differentiating logarithmic functions ddx[ln(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}
dydx=1cos(3x)ddx(cos(3x))\frac{dy}{dx} = \frac{1}{\cos(3x)} \cdot \frac{d}{dx}(\cos(3x))
The outer function is ln(u)\ln(u) where u=cos(3x)u = \cos(3x).
2
Differentiate the inner trigonometric function cos(3x)\cos(3x)
ddx(cos(3x))=3sin(3x)\frac{d}{dx}(\cos(3x)) = -3\sin(3x)
By the chain rule, ddx[cos(kx)]=ksin(kx)\frac{d}{dx}[\cos(kx)] = -k\sin(kx).
3
Multiply the terms and simplify using trigonometric identities
dydx=3sin(3x)cos(3x)=3tan(3x)\frac{dy}{dx} = \frac{-3\sin(3x)}{\cos(3x)} = -3\tan(3x)
Using the identity sin(θ)cos(θ)=tan(θ)\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta), the derivative simplifies to 3tan(3x)-3\tan(3x).

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule
Tahmini Süre:1m 30s
Soru 7512Soru

In a survey of 200200 agricultural exporters regarding three major commodities—Cocoa (CC), Palm Oil (PP), and Rubber (RR)—it was found that 110110 export Cocoa, 9090 export Palm Oil, and 7575 export Rubber. Exactly 2020 exporters export none of the three commodities, and 4545 export Cocoa only. Furthermore, the number of exporters who export Cocoa and Palm Oil only is twice the number of exporters who export all three commodities; the number who export Palm Oil and Rubber only is equal to the number who export all three; and the number who export Cocoa and Rubber only is 55 more than the number who export all three. Find the total number of exporters who export at least two of the three commodities.

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Cevap: 80

Cevap

The total number of exporters who export at least two of the three commodities is 80.
The correct answer of 80 is obtained by solving for the number of exporters trading in all three commodities (x=15x = 15) using the Cocoa set equation 45+2x+(x+5)+x=11045 + 2x + (x + 5) + x = 110, and then evaluating the region sum for at least two commodities (2x+x+(x+5)+x=5x+5=802x + x + (x + 5) + x = 5x + 5 = 80).

Adım Adım Çözüm

1
Assign a variable to the triple intersection
Let x=n(CPR)x = n(C \cap P \cap R) be the number of exporters of all three commodities.
The intersection of all three sets serves as the common parameter for all double-intersection regions.
2
Write algebraic expressions for the three pairwise-only intersections
n(CP only)=2xn(C \cap P \text{ only}) = 2x, n(PR only)=xn(P \cap R \text{ only}) = x, and n(CR only)=x+5n(C \cap R \text{ only}) = x + 5.
These expressions are derived directly from the relationships given in the problem statement.
3
Formulate and solve an equation using the set of Cocoa exporters
45+2x+(x+5)+x=110    50+4x=110    x=1545 + 2x + (x + 5) + x = 110 \implies 50 + 4x = 110 \implies x = 15.
The set of Cocoa exporters consists of four mutually exclusive regions whose cardinalities sum to 110.
4
Sum the regions corresponding to 'at least two commodities'
(2x)+(x)+(x+5)+x=5x+5=5(15)+5=80(2x) + (x) + (x + 5) + x = 5x + 5 = 5(15) + 5 = 80.
'At least two' encompasses everyone who exports exactly two commodities plus those who export all three.

Anahtar Kavram

Three-set principle of inclusion-exclusion and cardinal region decomposition
Soru 7513Soru

Two independent security systems, XX and YY, operate in a facility. The probability that system YY detects an intrusion is 0.200.20 greater than the probability that system XX detects it. If the probability that at least one of the systems detects an intrusion is 0.920.92, what is the probability that exactly one of the systems detects the intrusion?

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Cevap: 0.440.44

Cevap

The probability that exactly one of the systems detects the intrusion is 0.440.44.
The correct answer is 0.440.44. Setting P(X)=pP(X) = p and P(Y)=p+0.20P(Y) = p + 0.20, we use the independence rule P(XY)=P(X)P(Y)P(X \cap Y) = P(X)P(Y) in the addition law P(XY)=P(X)+P(Y)P(X)P(Y)=0.92P(X \cup Y) = P(X) + P(Y) - P(X)P(Y) = 0.92. Solving p21.80p+0.72=0p^2 - 1.80p + 0.72 = 0 gives P(X)=0.60P(X) = 0.60 and P(Y)=0.80P(Y) = 0.80. The probability of both detecting the intrusion is 0.60×0.80=0.480.60 \times 0.80 = 0.48. Subtracting the probability of both from the probability of at least one (0.920.480.92 - 0.48) gives 0.440.44 for exactly one system detecting the intrusion.

Adım Adım Çözüm

1
Define variables for the individual probabilities
Let P(X)=pP(X) = p. Then P(Y)=p+0.20P(Y) = p + 0.20.
System YY's probability is given as 0.200.20 greater than System XX's probability.
2
Apply the addition law for independent events
P(XY)=P(X)+P(Y)P(XY)=p+(p+0.20)p(p+0.20)=0.92P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + (p + 0.20) - p(p + 0.20) = 0.92
Since XX and YY are independent, P(XY)=P(X)P(Y)P(X \cap Y) = P(X)P(Y).
3
Solve the quadratic equation for pp
2p+0.20p20.20p=0.92    p2+1.80p0.72=0    p21.80p+0.72=02p + 0.20 - p^2 - 0.20p = 0.92 \implies -p^2 + 1.80p - 0.72 = 0 \implies p^2 - 1.80p + 0.72 = 0. Factoring yields (p0.60)(p1.20)=0(p - 0.60)(p - 1.20) = 0. Since p1p \le 1, p=0.60p = 0.60.
Probability values cannot exceed 11, so p=0.60p = 0.60 is the valid root.
4
Calculate individual probabilities and the probability of both occurring
P(X)=0.60P(X) = 0.60, P(Y)=0.80P(Y) = 0.80, and P(XY)=0.60×0.80=0.48P(X \cap Y) = 0.60 \times 0.80 = 0.48.
These are needed to evaluate the compound probability of exactly one event occurring.
5
Calculate the probability that exactly one system detects the intrusion
P(exactly one)=P(XY)P(XY)=0.920.48=0.44P(\text{exactly one}) = P(X \cup Y) - P(X \cap Y) = 0.92 - 0.48 = 0.44.
The probability of exactly one event occurring is the probability of at least one minus the probability of both.

Anahtar Kavram

Probability laws for independent and compound events
Soru 7514Soru

The operating cost CC (in Naira per hour) of a speed boat is partly constant and partly varies directly as the square of its speed vv (in km/h). At a speed of 20 km/h20\text{ km/h}, the operating cost is 18,000₦18,000 per hour, and at a speed of 30 km/h30\text{ km/h}, the operating cost is 33,000₦33,000 per hour. What is the operating cost per hour (in Naira) when the boat travels at a speed of 40 km/h40\text{ km/h}?

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Cevap: 54000

Cevap

The operating cost per hour when the speed boat travels at 40 km/h40\text{ km/h} is 54,00054,000 Naira.
By writing the partial variation relationship as C=k1+k2v2C = k_1 + k_2 v^2 and substituting the given pairs (20,18000)(20, 18000) and (30,33000)(30, 33000), we set up simultaneous equations k1+400k2=18000k_1 + 400k_2 = 18000 and k1+900k2=33000k_1 + 900k_2 = 33000. Solving these yields k2=30k_2 = 30 and k1=6000k_1 = 6000. Evaluating the model at v=40 km/hv = 40\text{ km/h} gives C=6000+30(402)=54,000C = 6000 + 30(40^2) = 54,000 Naira per hour.

Adım Adım Çözüm

1
Set up the general formula for partial variation
C=k1+k2v2C = k_1 + k_2 v^2
The total cost consists of a fixed constant component k1k_1 and a variable component k2v2k_2 v^2 that varies directly with speed squared.
2
Form simultaneous linear equations from the given conditions
18,000=k1+400k218,000 = k_1 + 400 k_2 and 33,000=k1+900k233,000 = k_1 + 900 k_2
Substituting v=20,C=18,000v = 20, C = 18,000 and v=30,C=33,000v = 30, C = 33,000 creates a solvable system of equations in k1k_1 and k2k_2.
3
Solve for the variation constants k1k_1 and k2k_2
k2=30k_2 = 30 and k1=6,000k_1 = 6,000
Subtracting the first equation from the second eliminates k1k_1, giving 500k2=15,000    k2=30500 k_2 = 15,000 \implies k_2 = 30. Substituting k2=30k_2 = 30 into the first equation yields k1=6,000k_1 = 6,000.
4
Calculate the operating cost at v=40 km/hv = 40\text{ km/h}
C=54,000C = 54,000
Evaluating C=6,000+30(40)2=6,000+30(1,600)=6,000+48,000=54,000C = 6,000 + 30(40)^2 = 6,000 + 30(1,600) = 6,000 + 48,000 = 54,000.

Anahtar Kavram

Partial Variation with Simultaneous Equations
Soru 7515Soru

What is the determinant of the matrix P=(5327)P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}?

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Cevap: 29

Cevap

29
For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by adbcad - bc. Substituting the values a=5a = 5, b=3b = 3, c=2c = 2, and d=7d = 7 yields (5×7)(3×2)=356=29(5 \times 7) - (3 \times 2) = 35 - 6 = 29.

Adım Adım Çözüm

1
Identify the values of a,b,c,a, b, c, and dd from the given 2×22 \times 2 matrix P=(5327)P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}.
a=5,b=3,c=2,d=7a = 5, b = 3, c = 2, d = 7
To set up the values for the 2×22 \times 2 determinant formula.
2
Multiply the elements of the main diagonal and the secondary diagonal.
Main diagonal product: 5×7=355 \times 7 = 35; Secondary diagonal product: 3×2=63 \times 2 = 6.
The determinant of a 2×22 \times 2 matrix is the difference between the main diagonal product and the secondary diagonal product.
3
Subtract the product of the secondary diagonal from the product of the main diagonal.
Determinant = 356=2935 - 6 = 29.
Completes the formula det(P)=adbc\det(P) = ad - bc.

Anahtar Kavram

Determinant of a 2x2 Matrix
Soru 7516Soru

When the polynomial P(x)=3x42x3+ax2+bx12P(x) = 3x^4 - 2x^3 + ax^2 + bx - 12 is divided by (x24)(x^2 - 4), the remainder is 5x45x - 4. What is the value of a+ba + b?

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Cevap: 3

Cevap

The value of a+ba + b is 33.
By using the Remainder Theorem for the quadratic divisor (x24)=(x2)(x+2)(x^2 - 4) = (x - 2)(x + 2), evaluating P(2)=6P(2) = 6 gives 2a+b=72a + b = -7, and evaluating P(2)=14P(-2) = -14 gives 2ab=332a - b = -33. Solving these linear equations simultaneously yields a=10a = -10 and b=13b = 13, which sums to a+b=3a + b = 3.

Adım Adım Çözüm

1
Set up the polynomial division relation.
P(x)=(x2)(x+2)Q(x)+(5x4)P(x) = (x - 2)(x + 2)Q(x) + (5x - 4)
By the Remainder Theorem and Division Algorithm, dividing by (x24)(x^2 - 4) yields a remainder of R(x)=5x4R(x) = 5x - 4.
2
Find the values of P(2)P(2) and P(2)P(-2) from the remainder.
P(2)=6P(2) = 6 and P(2)=14P(-2) = -14
Substituting the roots of the divisor x=2x = 2 and x=2x = -2 eliminates the quotient term (x24)Q(x)(x^2 - 4)Q(x).
3
Substitute x=2x = 2 into the polynomial P(x)P(x) and set equal to 66.
2a+b=72a + b = -7
3(16)2(8)+4a+2b12=20+4a+2b=6    4a+2b=143(16) - 2(8) + 4a + 2b - 12 = 20 + 4a + 2b = 6 \implies 4a + 2b = -14.
4
Substitute x=2x = -2 into the polynomial P(x)P(x) and set equal to 14-14.
2ab=332a - b = -33
3(16)2(8)+4a2b12=52+4a2b=14    4a2b=663(16) - 2(-8) + 4a - 2b - 12 = 52 + 4a - 2b = -14 \implies 4a - 2b = -66.
5
Solve the system of equations for aa and bb.
a=10a = -10, b=13b = 13, and a+b=3a + b = 3
Adding the two linear equations gives 4a=40    a=104a = -40 \implies a = -10. Substituting a=10a = -10 into 2a+b=72a + b = -7 gives b=13b = 13.

Anahtar Kavram

Polynomial Remainder Theorem for Non-Linear Divisors
Tahmini Süre:2m 30s
Soru 7517Soru

An arithmetic progression (A.P.) has a first term of 77 and a common difference of 55. What is the 12th12^{\text{th}} term of the progression?

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Cevap: 62

Cevap

The 12th12^{\text{th}} term of the progression is 6262.
Using the formula for the nthn^{\text{th}} term of an arithmetic progression, Tn=a+(n1)dT_n = a + (n - 1)d, with a=7a = 7, d=5d = 5, and n=12n = 12, we calculate T12=7+(121)×5=7+55=62T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62.

Adım Adım Çözüm

1
Identify known parameters from the question
a=7a = 7, d=5d = 5, n=12n = 12
These parameters are given in the problem statement.
2
Use the general formula for the nthn^{\text{th}} term of an arithmetic progression
Tn=a+(n1)dT_n = a + (n - 1)d
This formula connects the first term, common difference, and term number to the value of the term.
3
Substitute values and evaluate
T12=7+(121)×5=7+55=62T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62
Performing basic arithmetic gives the final result.

Anahtar Kavram

Finding the nthn^{\text{th}} term of an Arithmetic Progression
Tahmini Süre:45s
Soru 7518Soru

Which of the following types of waves requires a physical material medium for its propagation?

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Cevap: Sound waves

Cevap

Sound waves are mechanical waves and require a material medium to propagate.
Sound waves are mechanical longitudinal waves. They propagate through compressions and rarefactions of particles in a medium, meaning they cannot travel through a vacuum.

Adım Adım Çözüm

1
Classify waves based on their propagation requirements.
Waves are classified broadly into mechanical waves (which require a medium) and electromagnetic waves (which do not require a medium).
Understanding the fundamental distinction between mechanical and electromagnetic propagation.
2
Identify the nature of each given wave option.
Radio waves, infrared rays, and gamma rays are electromagnetic waves. Sound waves are mechanical waves.
Mechanical waves depend on the elastic interactions of physical particles to transmit energy.

Anahtar Kavram

Classification of waves into mechanical and electromagnetic based on medium requirement
Tahmini Süre:45s
Soru 7519Soru

An isosceles triangle has a perimeter of 36 cm36\text{ cm} and a base of length 16 cm16\text{ cm}. Calculate the area of the triangle in cm2\text{cm}^2.

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Cevap: 48

Cevap

The area of the isosceles triangle is 48 cm248\text{ cm}^2.
The two equal sides of the isosceles triangle measure 36162=10 cm\frac{36 - 16}{2} = 10\text{ cm} each. An altitude dropped perpendicularly to the base bisects the 16 cm16\text{ cm} base into two 8 cm8\text{ cm} segments. By the Pythagorean theorem, the perpendicular height is h=10282=6 cmh = \sqrt{10^2 - 8^2} = 6\text{ cm}. Therefore, the area is 12×16×6=48 cm2\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2.

Adım Adım Çözüm

1
Determine the length of the two equal sides
Each equal side is 10 cm10\text{ cm}
Subtract the base length from the total perimeter (3616=20 cm36 - 16 = 20\text{ cm}) and divide by 22.
2
Calculate the perpendicular height (altitude) to the base
Height h=6 cmh = 6\text{ cm}
The perpendicular altitude bisects the base into two 8 cm8\text{ cm} segments, creating right-angled triangles with hypotenuse 10 cm10\text{ cm}. Use Pythagoras' theorem: h=10282=6 cmh = \sqrt{10^2 - 8^2} = 6\text{ cm}.
3
Calculate the area of the triangle
Area = 48 cm248\text{ cm}^2
Multiply half the base by the perpendicular height: 12×16×6=48 cm2\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2.

Anahtar Kavram

Perimeter and Area of Isosceles Triangles using Pythagorean Theorem
Tahmini Süre:1m 30s
Soru 7520Soru

A binary operation \star defined on the set of real numbers R\mathbb{R} is given by ab=2a3b+aba \star b = 2a - 3b + ab. If (2x)3=16(2 \star x) \star 3 = 16, what is the value of xx?

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Cevap: 1-1

Cevap

1-1
Applying the binary operation rule ab=2a3b+aba \star b = 2a - 3b + ab step-by-step gives 2x=4x2 \star x = 4 - x. Operating on this with 33 yields (4x)3=2(4x)3(3)+3(4x)=115x(4 - x) \star 3 = 2(4 - x) - 3(3) + 3(4 - x) = 11 - 5x. Setting 115x=1611 - 5x = 16 gives x=1x = -1.

Adım Adım Çözüm

1
Evaluate the inner expression 2x2 \star x using the given definition ab=2a3b+aba \star b = 2a - 3b + ab.
2x=2(2)3(x)+(2)(x)=43x+2x=4x2 \star x = 2(2) - 3(x) + (2)(x) = 4 - 3x + 2x = 4 - x
The expression inside the parentheses must be simplified first.
2
Substitute 4x4 - x into the outer operation (4x)3(4 - x) \star 3.
(4x)3=2(4x)3(3)+(4x)(3)=82x9+123x=115x(4 - x) \star 3 = 2(4 - x) - 3(3) + (4 - x)(3) = 8 - 2x - 9 + 12 - 3x = 11 - 5x
Apply the operation rule with first element a=4xa = 4 - x and second element b=3b = 3.
3
Equate the resulting expression to 16 and solve for xx.
115x=16    5x=5    x=111 - 5x = 16 \implies -5x = 5 \implies x = -1
Solve the linear equation to determine the value of xx.

Anahtar Kavram

Non-commutative nested binary operation evaluation
Tahmini Süre:2m 0s
ÖncekiSayfa 376 / 697Sonraki
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