Answer:
The sum of the probabilities is greater than 100%; and the distribution is too uniform to be a normal distribution.
Step-by-step explanation:
The sum of the probabilities of a distribution should be 100%. When you add the probabilities of this distribution together, you have
22+24+21+26+28 = 46+21+26+28 = 67+26+28 = 93+28 = 121
This is more than 100%, which is a flaw with the results.
A normal distribution is a bell-shaped distribution. Graphing the probabilities for this distribution, we would have a bar up to 22; a bar to 24; a bar to 21; a bar to 26; and bar to 28.
The bars would not create a bell-shaped curve; thus this is not a normal distribution.
I mean it would be 14.0 because it’s a whole number unless you are getting into 14 tenths or 14 hundredths then it’s like 0.14 or 0.014.
First you would make the mixed fractions into improper fraction by multiplying the denominator and the whole number, then adding the numerator. Then you would flip the second fraction and multiply.
Ex. 1 1/2 ÷1 1/2
1st. 3/2÷3/2
2nd: 3/2×2/3
3rd 3/2x2/3=1
<h3>
Answer: False</h3>
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Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
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Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
Answer:
It is in simplest form
Step-by-step explanation:
