Answer:
2.7
Step-by-step explanation:
Answer: the second one
4x^2+76+32
Step-by-step explanation:
i used photomath lol
Answer:
t is less than or equal to 4
Step-by-step explanation:
Answer:
{-2, -1 , 3}
Step-by-step explanation:
When we have a set like:
{x₁, x₂, x₃}
The mode is the value that appears the most, so if there is no mode, then each value appears just one time.
The median is the middle value, here we know that the median is -1, then we can rewrite the set as:
{x₁, -1 , x₃}
The mean is computed as:
Mean = (x₁ + x₂ + x₃)/3
in this case we know that the mean is 0, then:
0 = (x₁ + x₂ + x₃)/3
then the numerator must be zero, so:
0 = (x₁ + x₂ + x₃)
replacing the value of x₂ = -1 we get:
0 = (x₁ - 1 + x₃)
where:
-5 < x₁ < -1 < x₃ ≤ 3
Now we can select the values of x₁ and x₃ such that the sum is equal to zero, and it meets the wanted restrictions.
here we can choose x₃ to be equal to 3 (the maximum allowed value), I do this because I noticed that the other values that are larger than -1 will not work (just with quick math).
then:
0 = x₁ - 1 + 3
Now we can solve this for x₁
0 = x₁ + 2
-2 = x₁
Then the set is:
{-2, -1 , 3}
Answer:
Step-by-step explanation:
Given that X is a normal random variable with parameters µ = 10 and σ 2 = 36,
X is N(10, 6)
Or z = 
is N(0,1)
a) P(X > 5),
=
(b) P(4 < X < 16),
=
(c) P(X < 8),
=
(d) P(X < 20),
=
(e) P(X > 16).
=P(Z>-0.6667)
= 0.2524
