A total of 150 students voted for the class president.Bianca got 34 votes and Carlos recieved 38% of the votes. Which student re
1 answer:
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1) Mean: 6.5
2) Standard deviation: 3.2
Step-by-step explanation:
1)
A uniform random distribution is the distribution of a variable that can assume n different values, and each of these values is equally likely to occur.
This means that the probability of occurring of each of these values is the same:
For a uniform random distribution, the mean of the distribution is simply equal to the median value of the distribution, which is exactly at the center of the distribution itself. It can be calculated as
where
a is the lowest value of the distribution
b is the highest value of the distribution
Here the distribution consists of all values ranging from 1 to 12, so:
a = 1
b = 12
Therefore the mean value is
2)
For a uniform random distribution, it can be dimostrated that the variance of the distribution is
For the distribution in this problem, we have:
a = 1
b = 12
Therefore the variance is
The standard deviation is the square root of the variance, therefore for this distribution it is:
Learn more about mean and standard deviation:
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Answer:
The correct options are;
EFGH has 4 congruent sides
Diagonal FH bisects angles EFG and EHG
Angle FEH is congruent to angle FGH
Step-by-step explanation:
1) Given that for a reflection, we have;
The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection
Therefore;
The distance of the point E from the line HF = The distance of the point G from the line HF
Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)
We can show that from the length of a line given by the equation, that the length EH ≅ GH and EF ≅ GF
Therefore since we are given that EH = EF, we have;
EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides
2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;
ΔEHF ≅ ΔGHF
∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent
Similarly, ∠GFH ≅ ∠EFH
Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF
Therefore, diagonal FH bisects angles EFG and EHG
3) Given that ΔEHF ≅ ΔGHF, we have;
Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent