How can you solve it if you don't know the equations? :P Just kidding...
n+d=22, so we can say that d=22-n
5n+10d=120, and using d found above in this equation gives you:
5n+10(22-n)=120
5n+220-10n=120
-5n=-100
n=20, and since d=22-n, d=2
So there are two dimes and twenty nickels...
check...
20(5)+2(10)=100+20=120 cents which is $1.20
n+d=22, 20+2=22, 22=22
Answer: -5/21.
Step-by-step explanation:
![\frac{2}{5}*[\frac{-3}{7} +(\frac{-1}{6})]=\frac{2}{5} *(-\frac{3}{7} -\frac{1}{6}) =\frac{2}{5}*(-\frac{3*6+1*7)}{7*6} )=\frac{2}{5}*(-\frac{18+7}{42})=\frac{2}{5}*(-\frac{25}{42})=-\frac{5}{21} .](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D%2A%5B%5Cfrac%7B-3%7D%7B7%7D%20%2B%28%5Cfrac%7B-1%7D%7B6%7D%29%5D%3D%5Cfrac%7B2%7D%7B5%7D%20%2A%28-%5Cfrac%7B3%7D%7B7%7D%20-%5Cfrac%7B1%7D%7B6%7D%29%20%3D%5Cfrac%7B2%7D%7B5%7D%2A%28-%5Cfrac%7B3%2A6%2B1%2A7%29%7D%7B7%2A6%7D%20%20%29%3D%5Cfrac%7B2%7D%7B5%7D%2A%28-%5Cfrac%7B18%2B7%7D%7B42%7D%29%3D%5Cfrac%7B2%7D%7B5%7D%2A%28-%5Cfrac%7B25%7D%7B42%7D%29%3D-%5Cfrac%7B5%7D%7B21%7D%20.)
Good luck an' have a nice day!
To solve this problem you must apply the proccedure shown below:
1. You have the following equation of a parabola, given in the problem above:
x<span>=1/16y^2
2. Then, based on the graph attached, you have:
p=y^2/4x
p=8^2/(4)(4)
p=64/16
p=4
3. The directrix is:
directrix=h-p
directrix=0-4
directrix=-4
The answer is:-4</span>
Answer:
The song is called "The Upside Down"
Step-by-step explanation:
Answer:
a) <u>0.4647</u>
b) <u>24.6 secs</u>
Step-by-step explanation:
Let T be interval between two successive barges
t(t) = λe^λt where t > 0
The mean of the exponential
E(T) = 1/λ
E(T) = 8
1/λ = 8
λ = 1/8
∴ t(t) = 1/8×e^-t/8 [ t > 0]
Now the probability we need
p[T<5] = ₀∫⁵ t(t) dt
=₀∫⁵ 1/8×e^-t/8 dt
= 1/8 ₀∫⁵ e^-t/8 dt
= 1/8 [ (e^-t/8) / -1/8 ]₀⁵
= - [ e^-t/8]₀⁵
= - [ e^-5/8 - 1 ]
= 1 - e^-5/8 = <u>0.4647</u>
Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>
<u></u>
b)
Now we find t such that;
p[T>t] = 0.95
so
t_∫¹⁰ t(x) dx = 0.95
t_∫¹⁰ 1/8×e^-x/8 = 0.95
1/8 t_∫¹⁰ e^-x/8 dx = 0.95
1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t = 0.95
- [ e^-x/8]¹⁰_t = 0.96
- [ 0 - e^-t/8 ] = 0.95
e^-t/8 = 0.95
take log of both sides
log (e^-t/8) = log (0.95)
-t/8 = In(0.95)
-t/8 = -0.0513
t = 8 × 0.0513
t = 0.4104 (min)
so we convert to seconds
t = 0.4104 × 60
t = <u>24.6 secs</u>
Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>
