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goldenfox [79]
3 years ago
15

መመሪያ 5 ፡- ለሚከተሉት ቁጥሮች ትንሹ የጋራ ብዜታቸውን ፈልጉ፡፡ ሀ. 20 እና 16 ሊ 34 እና 60 ሐ 48 20 እና 8​

Mathematics
1 answer:
strojnjashka [21]3 years ago
7 0

what is this, i don't get it

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Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t
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Answer:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

df = n-1= 22-1=21

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

Step-by-step explanation:

Data given

\bar X=33.9 represent the sample mean

s=4.1 represent the sample standard deviation

n=22 sample size  

\mu_o =26 represent the value that we want to test

\alpha=0.025 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:  

Null hypothesis:\mu \geq 36  

Alternative hypothesis:\mu < 36  

The statistic is given by:

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

And replacing we got:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

Now we can calculate the critical value but first we need to find the degreed of freedom:

df = n-1= 22-1=21

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

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3 years ago
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P³ = 1/8 please help me if anybody can .
ivanzaharov [21]

Answer:

p= \dfrac{1}{2}

Step-by-step explanation:

Given equation:

p^3=\dfrac{1}{8}

Cube root both sides:

\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}

\implies p= \sqrt[3]{\dfrac{1}{8}}

\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:

\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}

\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:

\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad 1^a=1:

\implies p= \dfrac{1}{8^{\frac{1}{3}}}

Rewrite 8 as 2³:

\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}

Simplify:

\implies p= \dfrac{1}{2^{\frac{3}{3}}}

\implies p= \dfrac{1}{2^{1}}

\implies p= \dfrac{1}{2}

3 0
1 year ago
Read 2 more answers
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