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3 years ago

Select all of the complex numbers in the given table

Mathematics

2 answers:

S_A_V [24]3 years ago

7 0

Answer:

2,3,5,6 are complex numbers

kap26 [50]3 years ago

7 0

Answer:

options B, C, E and F

Step-by-step explanation:

Complex number is a number that has a decimal under the square root.

Option A has a negative outside the square root . So it is not a complex

Option B has a square root (-3) so it is a complex number

Option C has a square root (-16) so it is a complex number

Option D has a negative outside the square root . So it is not a complex

Option E has a square root (-9) so it is a complex number

Option F has a negative fraction under the square root so it is a complex number

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3 years ago

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A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 436.0 gram setting. It

Vilka [71]

Answer:

We conclude that the machine is under filling the bags.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 436.0 gram

Sample mean, $\bar{x}$ = 429.0 grams

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 436.0\text{ grams}\\H_A: \mu < 436.0\text{ grams}$

We use one-tailed(left) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{429 - 436}{\frac{23}{\sqrt{40}} } = -1.92$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Since,

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that the machine is under filling the bags.

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3 years ago

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

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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