3 years ago

A single numerical value used as an estimate of a population parameter is known as

1 answer:

4 0

A point estimate, I believe

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deff fn [24]

D since the other options increase at a constant rate. For A, each increase in x is +4. For B, each increase in x is +1. For C, each increase is 13

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3 years ago

aliina [53]

Angle bisector would be the answer

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3 years ago

WITCHER [35]

X=7 cause -5(7) = -35

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3 years ago

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

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3 years ago

Virty [35]

Answer:

Step-by-step explanation:

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3 years ago