Evaluate StartFraction n exclamation mark Over left parenthesis n minus x right parenthesis exclamation mark x...
Evaluate StartFraction n exclamation mark Over left parenthesis n minus x right parenthesis exclamation mark x exclamation mark EndFraction p Superscript x Baseline times left parenthesis 1 minus p right parenthesis Superscript n minus xn!(n−x)!x!px•(1−p)n−x for nequals=1818, pequals=one fifth15, xequals=22. The answer is ?
Homework Answers
Solution
Given that ,
Using binomial probability formula ,
P(X = x) = (n C x) * px * (1 - p)n - x
P(X = 2) = (18 C 2) * 0.22 * (0.8)16
= 0.1723
Probability = 0.1723
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Evaluate StartFraction n exclamation mark Over left parenthesis
n minus x right parenthesis exclamation mark x...
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Over v EndFractionx=u2+v2 and y=uv.
Find
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at
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1. Suppose $26002600 is invested annually into an annuity that earns 55% interest. If P dollars are invested annually at an interest rate of r, the value of the annuity after n years is given by the following equation. Upper W equals Upper P left bracket StartFraction left parenthesis 1 plus r right parenthesis Superscript n Baseline minus 1 Over r EndFraction right bracketW=P(1+r)n−1r How much is the annuity worth after 5 years? 2. Suppose that $90,000 is invested at...
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Let
z equals f left parenthesis x comma y right parenthesis
commaz=f(x,y) ,
where
x equals u squared plus v squared and y equals StartFraction u
Over v EndFractionx=u2+v2 and y=uv.
Find
StartFraction partial derivative z Over partial derivative u
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at
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parenthesis negative 6 comma negative 6 right
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,
given that :
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