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Question

The mean of a set of observation is $stack x with ̄ on top$ . If each observation is divided by a, a¹ 0 and then is increased by 10, then the mean of the new set is

$\frac{\bar{x}}{α}$
$\frac{\bar{x} + 10}{α}$
$\frac{\bar{x} + 10 α}{α}$
$α \bar{x} + 10$

hint Hint:

The correct answer is: $fraction numerator stack x with ̄ on top plus 10 alpha over denominator alpha end fraction$

To find the mean of the new data.
Let $x_{1}, x_{2} \dots \dots, x_{n}$ be n observations.
$stack x with ̄ on top$
$y subscript i$

$N e w m e a n =$ $fraction numerator sum y subscript i over denominator n end fraction$
$fraction numerator x with bar on top over denominator alpha end fraction$ $fraction numerator stack x with ̄ on top plus 10 alpha over denominator alpha end fraction$

Therefore, new mean = $fraction numerator stack x with ̄ on top plus 10 alpha over denominator alpha end fraction$

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The median of the distribution is[AMU 1994]</span

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Matrix M_r is defined as M_r= $open square brackets table row r cell r – 1 end cell row cell r – 1 end cell r end table close square brackets$ , r $element of$ N. Then det. (M₁) + det.(M₂) + ........... + det.(M₂₀₀₉) = ?

maths-General

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The mean of a set of observation is . If each observation is divided by a, a¹ 0 and then is increased by 10, then the mean of the new set is

The correct answer is:

To find the mean of the new data.Let x1,x2……,xn be n observations.Then, = let = +10New mean =New mean = +10 =

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The following data gives the distribution of height of studentsThe median of the distribution is[AMU 1994]</span

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The mean of a set of observation is $stack x with ̄ on top$ . If each observation is divided by a, a¹ 0 and then is increased by 10, then the mean of the new set is

The correct answer is: $fraction numerator stack x with ̄ on top plus 10 alpha over denominator alpha end fraction$

The following data gives the distribution of height of students

The median of the distribution is[AMU 1994]</span

The following data gives the distribution of height of students

The median of the distribution is[AMU 1994]</span

The order of [x y z] $open square brackets table row a h g row h b f row g f c end table close square brackets open square brackets table row x row y row z end table close square brackets$ is

The order of [x y z] $open square brackets table row a h g row h b f row g f c end table close square brackets open square brackets table row x row y row z end table close square brackets$ is

If the matrix $open square brackets table row 1 3 cell alpha plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then $alpha$ =

If the matrix $open square brackets table row 1 3 cell alpha plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then $alpha$ =

If the matrix $open square brackets table row 1 3 cell lambda plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then the value of  is :

If the matrix $open square brackets table row 1 3 cell lambda plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then the value of  is :

Let A = $open square brackets table row cell negative 1 end cell 2 cell negative 3 end cell row cell negative 2 end cell 0 3 row 3 cell negative 3 end cell 1 end table close square brackets$ be a matrix, then(determinant of A ) × (adjoint of inverse of A)is equal to-

Let A = $open square brackets table row cell negative 1 end cell 2 cell negative 3 end cell row cell negative 2 end cell 0 3 row 3 cell negative 3 end cell 1 end table close square brackets$ be a matrix, then(determinant of A ) × (adjoint of inverse of A)is equal to-

If k . $open square brackets table row cell negative 1 end cell 2 2 row 2 cell negative 1 end cell 2 row 2 2 cell negative 1 end cell end table close square brackets$ is an orthogonal matrix, then k is equal to-

If k . $open square brackets table row cell negative 1 end cell 2 2 row 2 cell negative 1 end cell 2 row 2 2 cell negative 1 end cell end table close square brackets$ is an orthogonal matrix, then k is equal to-

If A is square matrix such that A² = A, then |A| equals-

If A is square matrix such that A² = A, then |A| equals-

If the matrix A = $open square brackets table row 8 cell negative 6 end cell 2 row cell negative 6 end cell 7 cell negative 4 end cell row 2 cell negative 4 end cell lambda end table close square brackets$ is singular, then $lambda$ equal to:

If the matrix A = $open square brackets table row 8 cell negative 6 end cell 2 row cell negative 6 end cell 7 cell negative 4 end cell row 2 cell negative 4 end cell lambda end table close square brackets$ is singular, then $lambda$ equal to:

If A² = 8A + kI where A = $open square brackets table row 1 0 row cell negative 1 end cell 7 end table close square brackets$ then k is :

If A² = 8A + kI where A = $open square brackets table row 1 0 row cell negative 1 end cell 7 end table close square brackets$ then k is :

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ and B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ such that A² = B the $alpha$ n is :

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ and B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ such that A² = B the $alpha$ n is :

Matrix M_r is defined as M_r= $open square brackets table row r cell r – 1 end cell row cell r – 1 end cell r end table close square brackets$ , r $element of$ N. Then det. (M₁) + det.(M₂) + ........... + det.(M₂₀₀₉) = ?

Matrix M_r is defined as M_r= $open square brackets table row r cell r – 1 end cell row cell r – 1 end cell r end table close square brackets$ , r $element of$ N. Then det. (M₁) + det.(M₂) + ........... + det.(M₂₀₀₉) = ?