Answer:
The Normal distribution is a continuous probability distribution with possible values all the reals. Some properties of this distribution are:
Is symmetrical and bell shaped no matter the parameters used. Usually if X is a random variable normally distributed we write this like that:

The two parameters are:
who represent the mean and is on the center of the distribution
who represent the standard deviation
One particular case is the normal standard distribution denoted by:

Example: Usually this distribution is used to model almost all the practical things in the life one of the examples is when we can model the scores of a test. Usually the distribution for this variable is normally distributed and we can find quantiles and probabilities associated
Step-by-step explanation:
The Normal distribution is a continuous probability distribution with possible values all the reals. Some properties of this distribution are:
Is symmetrical and bell shaped no matter the parameters used. Usually if X is a random variable normally distributed we write this like that:

The two parameters are:
who represent the mean and is on the center of the distribution
who represent the standard deviation
One particular case is the normal standard distribution denoted by:

Example: Usually this distribution is used to model almost all the practical things in the life one of the examples is when we can model the scores of a test. Usually the distribution for this variable is normally distributed and we can find quantiles and probabilities associated
6 positive tiles because there’s 6 negative so 6 positive will cancel it out
Answer:
Step-by-step explanation:
√((5-1)²+ (4 - (-6))²
√(4)² + (10)²
√(16+100)
√116
2√(29)
Answer:
x=12
Step-by-step explanation:
(X-8)/7=2
X-8=14
X=22
The function you seek to minimize is
()=3‾√4(3)2+(13−4)2
f
(
x
)
=
3
4
(
x
3
)
2
+
(
13
−
x
4
)
2
Then
′()=3‾√18−13−8=(3‾√18+18)−138
f
′
(
x
)
=
3
x
18
−
13
−
x
8
=
(
3
18
+
1
8
)
x
−
13
8
Note that ″()>0
f
″
(
x
)
>
0
so that the critical point at ′()=0
f
′
(
x
)
=
0
will be a minimum. The critical point is at
=1179+43‾√≈7.345m
x
=
117
9
+
4
3
≈
7.345
m
So that the amount used for the square will be 13−
13
−
x
, or
13−=524+33‾√≈5.655m