Answer: 10
Step-by-step explanation:
tony is 5 years younger than Marvin's age 1 year from now.
t = (m+1) - 5
t = m - 4
6 years ago tony was half as old as marvin was then.
t - 6 = 1/2(m - 6)
Get rid of the fraction, multiply both sides by 2
2(t-6) = m - 6
2t - 12 = m - 6
2t = m - 6 + 12
2t = m + 6
from the first statement, replace t with (m-4)
2(m-4) = m + 6
2m - 8 = m + 6
2m - m = 6 + 8
m = 14 is Marvin's present age
:
how old is tony now?
t = 14 - 4
t = 10 yrs is tony's age
Answer:
The value of this expression is -35.
Step-by-step explanation:
Let's use the order of operations (PEMDAS), to solve this problem.
In conclusion, the value of this expression is -35. Hope this helped :D
Answer: 2 meters.
Step-by-step explanation:
Let w = width of the cement path.
Dimensions of pool : Length = 15 meters , width = 9 meters
Area of pool = length x width = 15 x 9 = 135 square meters
Along width cement path, the length of region = 
width = 
Area of road with pool = 
Area of road = (Area of road with pool ) -(area of pool)
![\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2](https://tex.z-dn.net/?f=%5CRightarrow%5C%20112%20%3D4w%5E2%2B48w%2B135-%20135%5C%5C%5C%5C%5CRightarrow%5C%20112%3D%204w%5E2%2B48w%5C%5C%5C%5C%5CRightarrow%5C%204%20w%5E2%2B48w-112%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B12w-28%3D0%5C%20%5C%20%5C%20%5B%5Ctext%7BDivide%20both%20sides%20by%204%7D%5D%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B14w-2w-28%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%28w%2B14%29-2%28w%2B14%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%28w%2B14%29%28w-2%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%20w%3D-14%5C%20or%20%5C%20w%3D2)
width cannot be negative, so w=2 meters
Hence, the width of the road = 2 meters.
<span>Probability = 0.063
Fourth try = 0.0973
Let X be the number of failed attempts at passing the test before the student passes. This
is a negative binomial or geometric variable with x â {0, 1, 2, 3, . . .}, p = P(success) = 0.7
and the number of successes to to observe r = 1. Thus the pmf is nb(x; 1, p) = (1 â’ p)
xp.
The probability P that the student passes on the third try means that there were x = 2
failed attempts or P = nb(2, ; 1, .7) = (.3)2
(.7) = 0.063 . The probability that the student
passes before the third try is that there were two or fewer failed attmpts, so P = P(X ≤
2) = nb(0, ; 1, .7) + nb(1, ; 1, .7) + nb(2, ; 1, .7) = (.3)0
(.7) + (.3)1
(.7) + (.3)2
(.7) = 0.973 .</span>
