گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!
  فرم معتبر نیست.

The distance travelled, in kilometres, by a Grippo brake pad before it needs to be replaced is modelled by $10000X$, where $X$ is a random variable having the probability density function

$f\left( x \right) = \left\{ \begin{gathered}   - k\left( {{x^2} - 5x + 6} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,2 \leqslant x \leqslant 3, \hfill \\  0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise. \hfill \\ \end{gathered}  \right.$

The graph of $y = f\left( x \right)$ is shown in the diagram.

a) Show that $k = 6$.

b) State the value of $E\left( X \right)$ and find $Var\left( X \right)$.

c) Sami fits four new Grippo brake pads on his car. Find the probability that at least one of these brake pads will need to be replaced after travelling less than $22000{\text{ }}km$.

پاسخ تشریحی :
نمایش پاسخ

a) $ - k\int_2^3 {\left( {{x^2} - 5x + 6} \right)dx}  = 1$

$( - k(\frac{{{3^3}}}{3} - 5 \times \frac{{{3^2}}}{2} + 6 \times 3 - [\frac{{{2^3}}}{3} - 5 \times \frac{{{2^2}}}{2} + 6 \times 2]) = 1)$

$ - k \times ( - \frac{1}{6}) = 1$ or $k \times \frac{1}{6} = 1$

$\left( {k = 6{\text{ }}AG} \right)$

b) $E\left( X \right) = 2.5$

$ - 6\int_2^3 {({x^4} - 5{x^3} + 6{x^2})\,dx} \,\,\,\,\,\left( { =  - 6 \times \left( { - 1.05} \right)} \right)$

$ - ''2.5'{'^2}$

$ = 0.05$

c) $ - 6\int_2^{2.2} {({x^2} - 5x + 6)dx} \,\,\,\,\,\,\left( { = 0.104} \right)$

$1 - {\left( {1 - ''0.104''} \right)^4}$

$ = 0.355/0.356$

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!