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Arte-miy333 [17]
2 years ago
14

Find two numbers if their ratio is 9:11 and their difference is 6.

Mathematics
2 answers:
erastova [34]2 years ago
8 0

Answer:

Step-by-step explanation:

27,33

9*3 is 27

11*3 is 33

33-27 is 6

i only have one, sorry!

svlad2 [7]2 years ago
3 0

Answer:

Step-by-step explanation:

Let the larger number = y

Let the smaller number = x

x/y = 9/11

y - x = 6

Add x to both sides of the second equation

y = 6 + x

Substitute into the first equation

x/(x + 6) = 9/11

Cross multiply

11x = 9(x + 6)

Remove the brackets

11x = 9x + 54

Subtract 9x from both sides.

2x = 54

x = 54/2

x = 27

y = 27 + 6

y = 33

(27,33)

The way I've done it, there isn't a second pair.

I suppose you could do it this way.

9/11 = -27/-33

-27 - - 33 = 6

So the second pair could be

(-27, - 33)

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Answer:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

df=n-1=44-1=43  

p_v =P(t_{(43)}>1.436)=0.079  

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

\bar X=13.88 represent the sample mean

s=3.14 represent the sample standard deviation

n=44 sample size  

\mu_o =13.2 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:  

Null hypothesis:\mu \leq 13.2  

Alternative hypothesis:\mu > 13.2  

The statistic for this case is:

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

Calculating the statistic

Replacing the info given we got:

t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436    

P-value

The degrees of freedom are:

df=n-1=44-1=43  

Since is a right tailed test the p value is:  

p_v =P(t_{(43)}>1.436)=0.079  

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

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3 years ago
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