Answer:
9
The middle numbers are 8 and 10. 9 is the number in between them which means it is the median.
Answer:
Step-by-step explanation:
Hope this helps :D
Answer:
3 no solution
Step-by-step explanation:
I'm a smart man I would know trust me
I have my masters degree
What is the given matrix?
you didn't show a matrix.
the components of the matrix are the elements.
and the elements of the inverse matrix are the elements of the inverse matrix
1)
so if matrix A is 1 4 1
2 5 2
3 6 3
2)
14114
25225
36336
3)
15 24 12
15 12 24
4)
[A]=(15+24+12=51)-(15+12+24=51)
[A]=0
5)
52 23 25
63 23 36
41 11 14
63 33 36
41 12 14
52 12 25
6)
15-12 6-6 12-15
12-6 3-3 6-12
8-5 2-2 5-8
7)
3 0 -3
6 0 -6
3 0 -3
8)
(+)3 (-)0 (+)-3
(-)6 (+)0 (-)-6
(+)3 (-)0 (+)-3
=
3 0 -3
-6 0 6
3 0 -3
9)
3 -6 3
N= 0 0 0
-3 6 -3
10) the inverse of [A]= 1/|A|*[N]
which is in this case ⁻A= 0
because |A|=0
So the waiting time for a bus has density f(t)=λe−λtf(t)=λe−λt, where λλ is the rate. To understand the rate, you know that f(t)dtf(t)dt is a probability, so λλ has units of 1/[t]1/[t]. Thus if your bus arrives rr times per hour, the rate would be λ=rλ=r. Since the expectation of an exponential distribution is 1/λ1/λ, the higher your rate, the quicker you'll see a bus, which makes sense.
So define <span><span>X=min(<span>B1</span>,<span>B2</span>)</span><span>X=min(<span>B1</span>,<span>B2</span>)</span></span>, where <span><span>B1</span><span>B1</span></span> is exponential with rate <span>33</span> and <span><span>B2</span><span>B2</span></span> has rate <span>44</span>. It's easy to show the minimum of two independent exponentials is another exponential with rate <span><span><span>λ1</span>+<span>λ2</span></span><span><span>λ1</span>+<span>λ2</span></span></span>. So you want:
<span><span>P(X>20 minutes)=P(X>1/3)=1−F(1/3),</span><span>P(X>20 minutes)=P(X>1/3)=1−F(1/3),</span></span>
where <span><span>F(t)=1−<span>e<span>−t(<span>λ1</span>+<span>λ2</span>)</span></span></span><span>F(t)=1−<span>e<span>−t(<span>λ1</span>+<span>λ2</span>)</span></span></span></span>.
