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hram777 [196]
2 years ago
12

At the end of a snow storm, Amelia saw there was a lot of snow on her front lawn. The temperature increased and the snow began t

o melt at a steady rate. After the storm, the snow started melting at a rate of 1.5 inches per hour and it is known that 4 hours after the storm ended, the depth of snow was down to 12 inches. Write an equation for S, in terms of t, representing the depth of snow on Amelia's lawn, in inches, t hours after the snow stopped falling.
Mathematics
1 answer:
Andrei [34K]2 years ago
5 0

Answer:

The equation is;

S = 18 - 1.5t  

Step-by-step explanation:

From the question, we are made to know that the depth of the snow is S and the time taken to melt in hours is t

After 4 hours of melting, the snow depth is 12 inches  

Now, recall that the rate of melting is 1.5 inches/ hour  

So in the 4 hours, the amount of snow melted will be 1.5 * 4 = 6 inches  

So the total depth of the snow initially will be 6 + 12 = 18 inches  

So let’s write the equation;

S = 18 - 1.5t  

Where S is the depth of the snow after a particular number of hours t

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Step-by-step explanation:

Data given and notation

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p_o=0.35 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

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

p_v =P(z>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

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