Step-by-step explanation:
every triangle inscribed into a circle with the baseline being a diameter of the circle must be a right-angled triangle.
therefore,
angle XVY = 90°.
for VY we can use Pythagoras
c² = a² + b²
with c being the Hypotenuse (the line opposite of the 90° angle, in our case XY).
XY = 2×ZY = 2×17 = 34.
so,
34² = 30² + VY²
VY² = 34² - 30² = 1156 - 900 = 256
VY = 16
Equation of a circle in standard form:
(x-h)²+(y-k)²=r²
r=radius
(h,k) is the center.
In this case:
(h,k) =(0,0) (h=0,k=0 )
r=4
Therefore:
(x-0)²+(y-0)²=4²
x²+y²=16
Answer: x²+y²=16
Interior angles on parallel lines cut by a traversal are supplementary (they add up to 180°). These can be identified as "c" angles, due to their shape. Knowing this, we can figure out the value of x:
7x+(2x+36)=180
7x+2x+36=180
Simplify the equation
9x+36=180
Collect like terms
9x=144
Divide by 9 on both sides to isolate x
x=16
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>
Answer:
4xy
Step-by-step explanation:
2 x 2 = 4, then just plop the x and y down, and there's your answer
