Answer:
f(x) = 30 • 0.989x
Step-by-step explanation:
Given the data :
10 26.8
20 23.9
30 21.3
40 19
50 16.9
60 15.1
Using technology, the exponential model equation obtained by plotting the data is :
y = 30.068(0.989)^x
Based on the general exponential formula :
y = ab^x
y = predicted value
Initial value, a = 30.068
Rate = b = 0.989
The most appropriate model equation from the options given is :
f(x) = 30 • 0.989^x
PEMDAS
100 times 7 is 700
678 times 56 is 37968
700 +8 - 37968
-37260
The key features of the above given functions are correctly matched to their corresponding definition.
<h3>Definition of terms</h3>
- Negative sections of the graph: They are the parts where the graph is below the x-axis. That is option C.
- End behaviour: This is what happens to the graph on the far left or far right. That is option E.
- Positive sections of the graph: This is the parts where the graph is above the x-axis. That is option D.
- Intercepts: This is the points where the graph crosses an axis. That is option B
- Relative extrema: This is the points of relative minimum or maximum in a graph. That is option A.
Learn more about graphs here:
brainly.com/question/25799000
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Answer:
The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
The information provided is:
<em>σ</em> = $60
<em>MOE</em> = $2
The critical value of <em>z</em> for 95% confidence level is:
Compute the sample size as follows:
![n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Csigma%20%7D%7BMOE%7D%5D%5E%7B2%7D)
![=[\frac{1.96\times 60}{2}]^{2}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B1.96%5Ctimes%2060%7D%7B2%7D%5D%5E%7B2%7D)
Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
If you add the standard deviation to the average, it is 14.84. Because Marcus's score was lower than that, he will not receive a certificate.
I think this should be right for the reason and the calculation.
