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Given :

$X=$the amount of time (in minutes)

The probability distribution of $X$can be modeled by a uniform density curve on the interval from: data-custom-editor="chemistry" $0\mathrm{minutes}\mathrm{to}8$minutes.

Random variable :$V=60X$

We know that $X$denotes the time (in minutes) that Sally will have to wait for the bus on a random day.

As a result, $V=60X$.

We also know that a minute is divided into role="math" localid="1654005209290" $60$seconds.

If we increase the time in minutes by $60$, we get the time in seconds.

As a result, $V$denotes the amount of time (in seconds) that Sally will have to wait for the bus on a random day.

Given :

$X=$the amount of time (in minutes)

The probability distribution of $X$can be modeled by a uniform density curve on the interval from: data-custom-editor="chemistry" $0\mathrm{minutes}\mathrm{to}8\mathrm{minutes}$.

Random variable :$V=60X$

$X$exhibits a uniform density curve on the interval data-custom-editor="chemistry" $0\mathrm{to}8$minutes for probability distribution.

We know that $X$denotes the time (in minutes) that Sally will have to wait for the bus on a random day.

In addition, $X$has a consistent density curve from $0\mathrm{to}8$minutes.

Because $V=60x$, The constant$60$is multiplied by every data value in the $X$distribution.

If every data value is multiplied by the same constant, the form of the distribution remains intact.

The density curve of $X$is uniform.

$V,$on the other hand, has a homogeneous density curve.

Zero minutes and zero seconds are the same for a uniform density curve.

Similarly, $8$minutes is the same as $480$seconds.

As a result, $V$exhibits a consistent density curve spanning the time interval $0\mathrm{to}480$seconds.