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Black_prince [1.1K]
3 years ago
13

How do I put -(x-4)^2 -1 into standard form

Mathematics
1 answer:
Archy [21]3 years ago
4 0
Standard form is ax²+bx+c

expand then simlify
-(x-4)²=-(x-4)(x-4)=-(x²-8x+16)=-x²+8x-16
now we gots
-x²+8x-16-1
-x²+8x-17 is standard form
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Divide 1/7 divided by 6
ArbitrLikvidat [17]

Answer:

1/42

Step-by-step explanation:

1/7 ÷ 6/1

1 x 1 = 1

------------

7 x 6 = 42

1/42

already simplified to the fullest

hope that helps!

6 0
3 years ago
Which of the following is not a property of a chi-square distribution?
laiz [17]

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

\chi_k^2 = \matchal{N}_1^2 +  \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 +  \matchal{N}_k^2

Where N₁ , N₂m .... N_k are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

6 0
3 years ago
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Step-by-step explanation:

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