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solmaris [256]
3 years ago
13

The sum of the square of a positive number and the square of 44 more than the number is 8080. what is the​ number?

Mathematics
2 answers:
Elodia [21]3 years ago
8 0
<span>Let the unknown number be x
   x^2 + (x + 44)^2 = 8080
 x^2 +x^2 +88x +44^2 = 8080
 2x^2 +88x + 1936-8080=0
 2x^2 +88x - 6144 = 0
   divide through by 2
 x^2 + 44x -3072 =0
   Solving the quadratic equation, we have
 x= 37.65 or - 81.65</span>
mash [69]3 years ago
5 0
Let the number be x, therefore, its square is x², and 44 more than the number is x+44 whose square is (x+4)².
Thus the sum will be x² + (x+44)²² = 8080
                                  x² + x² + 88x + 1936= 8080 Combining the right terms 
                                  2x² + 8x - 6144 = 0 dividing by 2
                                     x² + 44x - 3072 =0 solving for x
                          x = 37.6 or -81.6
Therefore, the positive number is 37.6
                                
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t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

p_v =P(t_{(49)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation  

\bar X=25.02 represent the sample mean

s=4.83 represent the sample standard deviation

n=50 sample size  

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

\alpha 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is at least 26 mpg, the system of hypothesis would be:  

Null hypothesis:\mu \geq 26  

Alternative hypothesis:\mu < 26  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=50-1=49  

Since is a one sided test the p value would be:  

p_v =P(t_{(49)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

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