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yarga [219]
3 years ago
13

The equation below represents the function p.

Mathematics
1 answer:
Inessa [10]3 years ago
5 0
Answers:

1)Tthe first answer is that as x increases the value of  p(x) approaches a number that is greater than  q (x).


2) the y-intercept of the function p is greater than the y-intercept of the function q.


Explanation:


1) Value of the functions as x increases.


Function p:


p(x)= [ 2/5 ] ˣ - 3


As x increases, the value of the function is the limit when x → ∞.


Since [2/5] is less than 1, the limit of [2/5]ˣ when x → ∞ is 0, and the limit of p(x) is 0 - 3 =  -3.


While in the graph you see that the function q has a horizontal asymptote that shows that the limit of q (x) when x → ∞ is - 4.


Then, the first answer is that as x increases the value of  p(x) approaches a number that is greater than  q (x).


2) y - intercepts.


i) To determine the y-intercept of the function p(x), just replace x = 0 in the equation:


p(x) = [ 2 / 5]⁰ - 3 = 1 - 3 = - 2

ii) The y-intercept of q(x) is read in the graph. It is - 3.


Then the answer is that the y-intercept of the function p is greater than the y-intercept of the function q.
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Suppose the weights of tight ends in a football league are normally distributed such that σ2=400. A sample of 11 tight ends was
AlladinOne [14]

Answer:

217.636-1.96\frac{20}{\sqrt{11}}=205.817    

217.636+1.96\frac{20}{\sqrt{11}}=229.455    

So on this case the 95% confidence interval would be given by (205.82;229.46)    

Step-by-step explanation:

Assuming the following data: Weight 150 169 170 196 200 218 219 262 269 270 271

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma= \sqrt{400}= 20 represent the population standard deviation

n=11 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean we can use the following formula:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

The mean calculated for this case is \bar X=217.636

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

217.636-1.96\frac{20}{\sqrt{11}}=205.817    

217.636+1.96\frac{20}{\sqrt{11}}=229.455    

So on this case the 95% confidence interval would be given by (205.82;229.46)    

8 0
4 years ago
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