Answer:
P(0) = 0.6825
P(1) = 0.2607
Step-by-step explanation:
From the given information, the number of homicide is 107 and the total number of homicides per city –year is 280.
Let us denote the number of homicides per city-year as X.
The mean value, X is calculated as:
\begin{array}{c}\\{\rm{Mean}} = \frac{{107}}{{280}}\\\\ = 0.382\\\end{array}
Mean= 107/280 = 0.382
The mean number of homicides per city- year \left( {\lambda = \mu } \right)(λ=μ) is 0.382.
a. The probability that zero homicides is obtained is as below:
\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}
P(X=0) = e −0.382 (0.382)⁰/1
= (0.6825)(1)
/1
P(X=0) = 0.6825
Thus, the probability that zero homicides P(0) is 0.6825.
b. The probability that one homicides is obtained is as below:
\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}
P(X=1) = e −0.382 (0.382)¹/1
= (0.6825)(0.382)/1
P(X=1) = 0.2607
Thus, the probability that zero homicides P(0) is 0.2607.
Assuming that am supposed to calculate the the mean absolute deviation of
1,2,3,4,5,6,7
mean absolute deviation is calculated by first calculating the mean;
(1+2+3+4+5+6+7) =28/7
= 4
Then, find the absolute value (the difference between each value in the data and the mean) (x-μ)
= -3,-2,-1,0,1,2,3
Absolute values= 3,2,1,0,1,2,3
Then we find the mean of the values,
3+2+1+0+1+2+3 = 12
= 12/7
= 1.7143
Therefore, the mean is 4, and the Mean absolute deviation is 1.7143
Answer:
1) x³-y³ = (x-y)(x²-xy+y²)
2)x²-y² =(x-y)(x+y)
3)x-y = (x-y)
hcf = (x-y)
lcm = (x-y)(x²-xy+y²)(x+y)= (x³-y³)(x+y)
Step-by-step explanation:
it will take Lisa 11 months to have paid more than Kelly
