Answer:
0.289792.
Step-by-step explanation:
If we define a month with one or more accident as "success"; and
A month with no accident as "failure" .
- P(one or more accidents will occur during any given month)=0.60.
- P(no accident will occur during any given month)=1-0.60=0.40
We want to calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.
in this case, let k=number of failures before the fourth success
Therefore, 
k follows a negative binomial distribution.
Therefore, the probability is:
![P(k\geq 4)=1-P(k\leq 3)\\=1-\sum_{k=0}^{3}\left(\begin{array}{ccc}3+k\\k\end{array}\right)(0.60)^4(0.40)^k\\=1-(0.60)^4 \left[\left(\begin{array}{ccc}3\\0\end{array}\right)(0.40)^0+\left(\begin{array}{ccc}4\\1\end{array}\right)(0.40)^1+\left(\begin{array}{ccc}5\\2\end{array}\right)(0.40)^2+\left(\begin{array}{ccc}6\\3\end{array}\right)(0.40)^3\right]\\=1-(0.60)^4 \left[ 1+1.6+1.6+1.28\right]\\=1-(0.60)^4[5.48]\\=0.289792](https://tex.z-dn.net/?f=P%28k%5Cgeq%204%29%3D1-P%28k%5Cleq%203%29%5C%5C%3D1-%5Csum_%7Bk%3D0%7D%5E%7B3%7D%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D3%2Bk%5C%5Ck%5Cend%7Barray%7D%5Cright%29%280.60%29%5E4%280.40%29%5Ek%5C%5C%3D1-%280.60%29%5E4%20%5Cleft%5B%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C0%5Cend%7Barray%7D%5Cright%29%280.40%29%5E0%2B%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C1%5Cend%7Barray%7D%5Cright%29%280.40%29%5E1%2B%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C2%5Cend%7Barray%7D%5Cright%29%280.40%29%5E2%2B%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7D6%5C%5C3%5Cend%7Barray%7D%5Cright%29%280.40%29%5E3%5Cright%5D%5C%5C%3D1-%280.60%29%5E4%20%5Cleft%5B%201%2B1.6%2B1.6%2B1.28%5Cright%5D%5C%5C%3D1-%280.60%29%5E4%5B5.48%5D%5C%5C%3D0.289792)
The required probability is 0.289792.
Answer:
m would be 69!
Step-by-step explanation:
61 = m - 8
61 + 8 = m - 8 + 8
69 = m
Answer:
here?
Step-by-step explanation:
75÷600=0.125
if you meant
600÷75=8
Neither P, nor A are on the sketch
I guess P is the upper right corner of the rectangle
and A=(0,1)
P belongs to the line going through (1,0) and B(0,y)
<span>but we don't know the y-coordinate of B </span>
<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span>
now look at the right triangle BOA
<span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span>
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1
so the coordinates of B are (0,1)
the line going through (1,0) and (0,1) is L(x)=-x+1
P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1)
b) so if that's P, the height of the rectangle is -x+1 and the width=2x
<span>so its area A(x)=2x*(-x+1)= -2x²+2x
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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