A=-.25+6.7t, s=.75+4.5t when Amir catches up a=s so:
-.25+6.7t=.75+4.5t add .25 to both sides
6.7t=1+4.5t subtract 4.5t from both sides
2.2t=1 divide both sides by 2.2
t=10/22 hr
t≈0.45 hr (to nearest hundredth)
Answer:
![P(Sum = 9) = \frac{1}{9}](https://tex.z-dn.net/?f=P%28Sum%20%3D%209%29%20%3D%20%5Cfrac%7B1%7D%7B9%7D)
Step-by-step explanation:
Given
Outcomes of rolling two dice
Required
P(Sum = 9)
From the given outcomes, we have:
-- the sample size
![Sum = 9:=\{(3,6),(4,5),(5,4),(6,3)\}](https://tex.z-dn.net/?f=Sum%20%3D%209%3A%3D%5C%7B%283%2C6%29%2C%284%2C5%29%2C%285%2C4%29%2C%286%2C3%29%5C%7D)
So.
![n(Sum = 9) = 4](https://tex.z-dn.net/?f=n%28Sum%20%3D%209%29%20%3D%204)
The probability is:
![P(Sum = 9) = \frac{n(Sum = 9)}{n(S)}](https://tex.z-dn.net/?f=P%28Sum%20%3D%209%29%20%3D%20%5Cfrac%7Bn%28Sum%20%3D%209%29%7D%7Bn%28S%29%7D)
![P(Sum = 9) = \frac{4}{36}](https://tex.z-dn.net/?f=P%28Sum%20%3D%209%29%20%3D%20%5Cfrac%7B4%7D%7B36%7D)
Simplify
![P(Sum = 9) = \frac{1}{9}](https://tex.z-dn.net/?f=P%28Sum%20%3D%209%29%20%3D%20%5Cfrac%7B1%7D%7B9%7D)
See the attached figure to better understand the problem
we know that
AM=MD
so
triangle AMD is an isosceles right triangle
therefore
its height is half its width.
Then
AB = (1/2)AD----------Equation 1
Perimeter=2*[AB+AD]=34 in ---------> AB+AD=17-------> Equation 2
I substitute 1 in 2
(1/2)AD +AD = 17
(3/2)AD=17
AD=17*2/3----------> AD=34/3-------> 11 1/3 in
AB=(1/2)AD--------> AB=(1/2)*34/3--------> AB=17/3-------> AB=5 2/3 in
the answers are
AD=11 1/3 inAB = 5 2/3 in
Answer:
uuhh sorry
Step-by-step explanation: