Answer:
y = 8 · 10ˣ
Step-by-step explanation:
y = abˣ
When x = 0, y = 8.
8 = ab⁰
8 = a
When x = 3, y = 8000.
8000 = 8b³
1000 = b³
10 = b
The function is:
y = 8 · 10ˣ
Answer:
The margin of error of u is of 3.8.
The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 28 - 1 = 27
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.7707
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error of u is of 3.8.
The lower end of the interval is the sample mean subtracted by M. So it is 31.2 - 3.8 = 27.4 minutes
The upper end of the interval is the sample mean added to M. So it is 31.2 + 3.8 = 35 minutes
The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.
Answer: see what you gotta do is/right/left/up/turn left/ down/up/down/down/ right up/ 2nd row turn left/ then do it all over again in you got it
Step-by-step explanation:
/right/left/up/turn left/ down/up/down/down/ right up/ ALSO CAN I BRAINLIEST IF I HELPED.
The equation to represent the number of people that can attend the event will be 18500 +/- 1200.
<h3>How to calculate the value?</h3>
Based on the information, equation to represent the number of people that can attend the event will be 18500 +/- 1200.
The maximum number will be:
= 18500 + 1200
= 19700
The minimum number will be:
= 18500 - 1200
= 17300
Learn more about equations on:
brainly.com/question/13763238
#SPJ1
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
![\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bthen%20we%20can%20say%20that%7D~%5Chfill%20%7D%7Bsin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%5Cimplies%20%5Ctheta%20%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Btherefore%20then%7D~%5Chfill%20%7D%7Bsin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D)
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B13%5E2-%28-5%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B144%7D%3Da%5Cimplies%20%5Cpm%2012%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B%5Cpm%2012%7D%7D%7B13%7D)
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.