Maths

The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation √13.44, then the standard deviation of the second sample is:

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Contributor-Level 10

1st sample: n? =100, x? =15, σ? =3. Σx? = 1500. Σx? ² = n? (σ? ²+x? ²) = 100 (9+225) = 23400.
Whole group: n=250, x? =15.6, σ²=13.44. Σx = 250*15.6 = 3900.
2nd sample: n? =150. Σy? = 3900 - 1500 = 2400. y? = 2400/150 = 16.
Σ (x+y)² = n (σ²+x? ²) = 250 (13.44+15.6²) = 250 (13.44+243.36) = 64200.
Σy? ² = 64200 - 23400 = 40800.
σ? ² = Σy? ²/n? - y? ² = 40800/150 - 16² = 272 - 256 = 16.
σ? = 4.

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New answer posted

a year ago

A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of any ellipse with eccentricity 1/3 and the distance of the nearer focus from this directrix is 8/√53, then the equation of the other directrix can be:

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Contributor-Level 10

Reflected point of (2,1) about y-axis is (-2,1).

Reflected ray passes through (-2,1) and (5,3).
Equation: (y-1)/ (x+2) = (3-1)/ (5+2) = 2/7 ⇒ 2x - 7y + 11 = 0.
This is one directrix. Let the other be 2x - 7y + α = 0.
Distance between directrices = 2a/e = |11-α|/√53.
Distance from focus to directrix = a/e - ae = 8/√53.
a/e (1-e²) = 8/√53.
e=1/3. a/e (8/9) = 8/√53 ⇒ a/e = 9/√53.
2a/e = 18/√53 = |11-α|/√53.
|11-α| = 18.
11-α = 18 ⇒ α = -7.
11-α = -18 ⇒ α = 29.
Other directrix: 2x-7y-7=0 or 2x-7y+29=0.

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New answer posted

a year ago

Consider functions f : A → B and g : B → C (A,B,C ⊂ R) such that (gof)⁻¹ exists, then:

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Maths Differential Equations

Contributor-Level 10

If (gof)? ¹ exist then gof is a bijective function. For gof to be bijective, f must be one-one and g must be onto.

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New answer posted

a year ago

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Contributor-Level 10

dy/dx = e^ (3x+4y) = e³? e?
e? dy = e³? dx
∫e? dy = ∫e³? dx
-e? /4 = e³? /3 + C
y (0)=0 ⇒ -1/4 = 1/3 + C ⇒ C = -7/12.
-e? /4 = e³? /3 - 7/12
e? = (7 - 4e³? )/3
y = (-1/4)ln (7-4e³? )/3)
x = -2/3 ln2 = ln (2? ²/³) = ln (1/4¹/³)
e³? = e^ (ln (1/4) = 1/4.
y = (-1/4)ln (7-1)/3) = (-1/4)ln2.
α = -1/4.

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New answer posted

a year ago

Let X be a random variable such that the probability function of a distribution is given by P(X=0) = 1/2, P(X = j) = (1/3)? (j = 1,2,3,.∞). Then the mean of the distribution and P(X is positive and even) respectively are:

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Contributor-Level 10

Mean = Σx? p? = 0 (1/2) + Σ? ∞ j (1/3)? = (1/3)/ (1-1/3)² = (1/3)/ (4/9) = 3/4
P (X is positive and even) = P (X=2) + P (X=4) + .
= (1/3)² + (1/3)? + . = (1/9)/ (1-1/9) = (1/9)/ (8/9) = 1/8

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New answer posted

a year ago

Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:

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Contributor-Level 10

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.

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New answer posted

a year ago

The value of cot(π/24) is:

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Contributor-Level 10

cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2

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New answer posted

a year ago

Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:

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Contributor-Level 10

S? : x² + y² - x - y - 1/2 = 0, C? : (1/2, 1/2), r? = √ (1/4)+ (1/4)+ (1/2) = 1.
S? : x² + y² - 4y + 7/4 = 0, C? : (0, 2), r? = √ (4 - 7/4) = 3/2.
S? : (x-2)² + (y-1)² ≤ r², C? : (2, 1).
A ∪ B ⊂ C means both circles S? and S? must be inside S?
Distance C? = √ (2-1/2)² + (1-1/2)²) = √ (9/4 + 1/4) = √10/2.
Condition: r ≥ C? + r? ⇒ r ≥ √10/2 + 1.
Distance C? = √ (2-0)² + (1-2)²) = √5.
Condition: r ≥ C? + r? ⇒ r ≥ √5 + 3/2.
√10/2 + 1 ≈ 1.58 + 1 = 2.58.
√5 + 3/2 ≈ 2.23 + 1.5 = 3.73.
So minimum r = (√10+2)/2 and (2√5+3)/2. We need the maximum of these two.
Let's recheck the question logic f

...more

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New answer posted

a year ago

The compound statement (P ∨ Q) ∧ (¬P) ⇒ Q is equivalent to:

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Maths Continuity and Differentiability

Contributor-Level 10

Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.

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New answer posted

a year ago

Let f: [-π/4, π/4] → R be defined as
f(x) = { (1 + |sin x|)³ᵃ/|sin x|, -π/4 < x < 0; b, x = 0; e^(cot 4x / cot 2x), 0 < x < /4 }
If f is continuous at x = 0, then the value of 6a + b² is equal to:

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