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3 years ago

What is the completely simplified equivalent of 2/(5+i)?

1 answer:

4 0

namely, let's rationalize the denominator in the fraction, for which case we'll be using the <u>conjugate</u> of that denominator, so we'll multiply top and bottom by its <u>conjugate</u>.

so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.

$\bf \cfrac{2}{5+i}\cdot \cfrac{5-i}{5-i}\implies \cfrac{2(5-i)}{\stackrel{\textit{difference of squares}}{(5+i)(5-i)}}\implies \cfrac{2(5-i)}{\stackrel{\textit{recall }i^2=-1}{5^2-i^2}}\implies \cfrac{2(5-i)}{25-(-1)} \\\\\\ \cfrac{2(5-i)}{25+1}\implies \cfrac{2(5-i)}{26}\implies \cfrac{5-i}{13}$

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C.) 1 5/b boxes

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3 years ago

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Step-by-step explanation:

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sergejj [24]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$

c) $p_v =P(t_{34}>2.958)=0.0028$

d) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

Step-by-step explanation:

Data given and notation

The sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=17$ represent the sample mean

$s=4$ represent the sample standard deviation

$n=35$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less or equal than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$

Part c P-value

The degrees of freedom are given by:

$df = n-1= 35-1=34$

Since is a right tailed test the p value would be:

$p_v =P(t_{34}>2.958)=0.0028$

Part d Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

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3 years ago

(1 2/7-3/14)/8/15trying to remember how to do this

SOVA2 [1]

We will write it as a fraction in ordet to solve, that is:

$\frac{(1\frac{2}{7}-\frac{3}{14})}{(\frac{8}{15})}$

We then operate as follows:

$\frac{((\frac{7}{7}+\frac{2}{7})-\frac{3}{14})}{(\frac{8}{15})}\Rightarrow\frac{(\frac{9}{7}-\frac{3}{14})}{(\frac{8}{15})}\Rightarrow\frac{(\frac{15}{14})}{(\frac{8}{15})}\Rightarrow\frac{15\cdot15}{14\cdot8}\Rightarrow\frac{225}{112}$

We have this, since 1 integer will be equal as a numerator divided by a denominator with equal values. Examples 1 = 2/2, 1 = 45/45, ...

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1 year ago