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blagie [28]
3 years ago
10

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
3 + 12 + 48 + 192 + ...<br> Whats the answer
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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.

8 0
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

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

4 0
3 years ago
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