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Novosadov [1.4K]
3 years ago
7

15 Points!

Mathematics
1 answer:
Virty [35]3 years ago
6 0
5; -3; -1 -2i
<span>In polynomials we have not of an imaginary unit (i), therefore we have the fourth zeros -1-2i.</span>

f(x)=(x-5)(x+3)\left[x-(-1-2i)\right]\left[x-(-1+2i)\right]\\\\=(x^2+3x-5x-15)[x^2-x(-1+2i)-x(-1-2i)+(-1)^2-(2i)^2]\\\\=(x^2-2x-15)(x^2+x-2i+x+2i+1+4)\\\\=(x^2-2x-15)(x^2+2x+5)\\\\=x^4+2x^3+5x^2-2x^3-4x^2-10x-15x^2-30x-75\\\\=x^4-14x^2-40x-75

Answer: B)


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Consider an ANOVA procedure to compare five population means. In this case, the total number of pairs to be considered in multip
Oliga [24]

Answer:

The right response is "0.0085". A further explanation is given below.

Step-by-step explanation:

Let,

P (across each pair, this same null hypothesis would be wrongly rejected) be "a".

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3 years ago
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Tomtit [17]
To solve the question we use the following analogy:
Given that the odds of winning is a/b, then the probability is b/(a+b)
hence to solve the question we proceed as follows:

a]
P(E)=11/36
odds against E=(36-11)/11
simplifying this gives us:
25/11

b] P(E)=31/36
odds against E=(36-31)/31
simplifying this we get:
5/31
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c]P(E)=32/36
odds in favor of E will be:
36/(36-32)
simplifying the above we get:
36/4
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d]
P(E)=30/36
odds against E will be:
(36-30)/36
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6 0
3 years ago
Solve the logarithmic equation. Show steps
Morgarella [4.7K]

\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \underset{\stackrel{\uparrow }{\textit{let's use this one}}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] ~\dotfill

\log_4(x+10)-\log_4(x-2)=\log_4(x)\implies \log_4\left( \cfrac{x+10}{x-2} \right)=\log_4(x) \\\\\\ \stackrel{\textit{exponentializing both sides}}{4^{\log_4\left( \frac{x+10}{x-2} \right)}=4^{\log_4(x)}}\implies \cfrac{x+10}{x-2}=x\implies x+10=x^2-2x \\\\\\ 10=x^2-3x\implies 0=x^2-3x-10 \\\\\\ 0=(x-5)(x+2)\implies x= \begin{cases} 5~~\checkmark\\ -2 \end{cases}

notice, -2 is a valid value for the quadratic, however, the argument value for a logarithm can never 0 or less, it has to be always greater than 0, so for the logarithmic expression with (x-2), using x = -2 will give us a negative value, so -2 is no dice.

4 0
2 years ago
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