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Marianna [84]
3 years ago
11

Line AB contains points A(4, 5) and B(9, 7). What is the slope of Aß?

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
3 0

<em>Let the common root is ‘x’</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1⇒ x = 1</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1⇒ x = 1⇒ {1 + a + b = 0} (From equation (1))</em>

<em>Let the common root is ‘x’x2 + ax + b = 0 ……(1)x2 + bx + a = 0 ……(2)Subtract equation (2) from (1), we get(a – b)x + (b –a) = 0⇒ x = (a-b)/(a-b) = 1⇒ x = 1⇒ {1 + a + b = 0} (From equation (1))⇒ a + b = –1</em>

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{:}\longrightarrow\sf y=-x+50

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Exercise 5.2. Suppose that X has moment generating function
soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

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or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Step-by-step explanation:

Given:

moment generating function  of X as:

MX(t) = \frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}

a)  Now

Mean, E(X) = M_{X}'(t=0)

Thus,

M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}

or

M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}

also,

E(X^{2})=M_{X}''(t=0)

Thus,

M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}

or

M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}

Therefore,

Mean, E(X) = M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}

or

Mean, E(X) = - 0.5

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E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}

or

E(X^{2}) = 9.5

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Variance(X) = E(X²) - E(X)²

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b) Now,

Let f(x) be the PMF of X

Thus,

M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Therefore,

at x = 0, P(x) = \frac{1}{2}

at x= - 4 ,P(x) = \frac{1}{3}

at x = 5, P(x) = \frac{1}{6}

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E(X) =\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})

or

E(X) = - 0.5

also,E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})

E(X^{2})  = 9.5

Hence,

Var(X) = E(X²) - E(X)²

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= 9.25

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