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Mila [183]
2 years ago
9

I need help with this I don't understand this math

Mathematics
1 answer:
Dmitriy789 [7]2 years ago
7 0

The exponential regression equation is y = 172.21(0.99)ˣ and at x = 60 minutes y = 94.22 degree Fahrenheit and at x = 240 minutes y = 15.43 degree Fahrenheit.

<h3>What is correlation?</h3>

It is defined as the relation between two variables which is a quantitative type and gives an idea about the direction of these two variables.

\rm r = \dfrac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{{[n\sum x^2- (\sum x)^2]}}\sqrt{[n\sum y^2- (\sum y)^2]}}

We have data shown in the table:

We can find the exponential regression equation using the data shown.

\rm y = ab^x

After calculating from the data:

a = 172.21 and b = 0.99

\rm y = 172.21(0.99)^x

The correlation coefficient will be:

r = -0.99

The range will be all real numbers.

The domain will be y > 0

As we can see in the graph x ∈R and y>0

After an hour,

Plug x = 60 minutes in the exponential regression equation:

\rm y = 172.21(0.99)^{60}

y = 94.22 degree Fahrenheit

After 4 hours, plug x = 240

\rm y = 172.21(0.99)^{240}

y = 15.43 degree Fahrenheit

 

Similarly, we can find any data from the exponential regression equation.

Thus, the exponential regression equation is y = 172.21(0.99)ˣ and at x = 60 minutes y = 94.22 degree Fahrenheit and at x = 240 minutes y = 15.43 degree Fahrenheit.

Learn more about the correlation here:

brainly.com/question/11705632

#SPJ1

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Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

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The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

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