Answer:
135%
Step-by-step explanation:
To represent a decimal to percentage, multiply by 100
1.35 x 100 = 135
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-Chetan K
9514 1404 393
Answer:
- late only: 15
- extra-late only: 24
- one type: 43
- total trucks: 105
Step-by-step explanation:
It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.
For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.
Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.
__
a) 15 trucks carried only late
b) 24 trucks carried only extra late
c) 4+15+24 = 43 trucks carried only one type
d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out
Answer
Find out the how much prize money did she win.
To proof
Let us assume that the total amount prize money did she win be u.
As given
Mrs.Gill won certain prize money in a cooking competition.
She spent half of the prize money on the clothes
![= u -\frac{u}{2}\\ = \frac{u}{2}](https://tex.z-dn.net/?f=%3D%20u%20-%5Cfrac%7Bu%7D%7B2%7D%5C%5C%20%3D%20%5Cfrac%7Bu%7D%7B2%7D)
one third on grocery
![= \frac{u}{2} -\frac{u}{3}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bu%7D%7B2%7D%20-%5Cfrac%7Bu%7D%7B3%7D)
(L.C.M of (2,3) =6 )
![= \frac{3u -2u}{6} \\=\frac{u}{6}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B3u%20-2u%7D%7B6%7D%20%5C%5C%3D%5Cfrac%7Bu%7D%7B6%7D)
gave away the remaining Rs.2000 to an orphanage
![2000 = \frac{u}{6}](https://tex.z-dn.net/?f=2000%20%3D%20%5Cfrac%7Bu%7D%7B6%7D)
u =Rs 12000
Rs 12000 prize money did she win.
Hence proved
Answer:
Step-by-step explanation:
J we be w e e e e d. D
![f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=f_X%28x%29%3D%5Cbegin%7Bcases%7D0.1e%5E%7B-0.1x%7D%26%5Ctext%7Bfor%20%7Dx%3E0%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
a. 9:00 AM is the 60 minute mark:
![f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248](https://tex.z-dn.net/?f=f_X%2860%29%3D0.1e%5E%7B-0.1%5Ccdot60%7D%5Capprox0.000248)
b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is
![\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B15%7D%5E%7B30%7Df_X%28x%29%5C%2C%5Cmathrm%20dx%5Capprox0.173)
c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is
![\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B40%7Df_X%28x%29%5C%2C%5Cmathrm%20dx%5Capprox0.982)
The probability of doing so for at least 2 of 5 days is
![\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D2%7D%5E5%5Cbinom5n%280.982%29%5En%281-0.982%29%5E%7B5-n%7D%5Capprox1)
i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.
d. Integrate the PDF to obtain the CDF:
![F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x](https://tex.z-dn.net/?f=F_X%28x%29%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5Exf_X%28t%29%5C%2C%5Cmathrm%20dt%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dx%3C0%5C%5C1-0.1e%5E%7B-0.1x%7D%26%5Ctext%7Bfor%20%7Dx%5Cge0%5Cend%7Bcases%7D)
Then the desired probability is
![F_X(30)-F_X(15)\approx0.950-0.777=0.173](https://tex.z-dn.net/?f=F_X%2830%29-F_X%2815%29%5Capprox0.950-0.777%3D0.173)