Complete options are;
A) Paired t-test for means
B) Paired Z-test for means
C) Z-test for proportion
D) t-test for means
E) Z-test for means
F) t-test for proportion
Answer:
B) Paired Z-test for means
Step-by-step explanation:
We are told that the distances travelled are normally distributed.
We are also trying to find out if the average distance travelled by the Titleist Pro V1 golf balls is farther than that of the Callaway Chrome Soft golf balls.
Also, the standard deviation of both cases are known.
Thus, this is where Paired Z-test for means is used.
You can turn 9/6 into a mixed number by doing it 2 ways:
9/6=1 3/6 or 1 1/2
<span>Or you can simplify it to 3/2 to make it easier to form into a mixed number. 1 1/2
Hope this helped :)</span>
recall your d = rt, distance = rate * time.
let's say we have two trains, A and B, A is going at 85 mph and B at 65 mph.
they are 210 miles apart and moving toward each other, at some point they will meet, when that happens, the faster train A has covered say d miles, and the slower B has covered then the slack from 210 and d, namely 210 - d.
When both trains meet, A has covered more miles than B because A is faster, however the time both have been moving, is the same, say t hours.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{Train A}&d&85&t\\ \textit{Train B}&210-d&65&t \end{array}\\\\ \dotfill\\\\ \begin{cases} \boxed{d}=85t\\ 210-d=65t\\[-0.5em] \hrulefill\\ 210-\boxed{85t}=65t \end{cases} \\\\\\ 210=150t\implies \cfrac{210}{150}=t\implies \cfrac{7}{5}=t\implies \stackrel{\textit{one hour and 24 minutes}}{1\frac{2}{5}=t}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7BTrain%20A%7D%26d%2685%26t%5C%5C%20%5Ctextit%7BTrain%20B%7D%26210-d%2665%26t%20%5Cend%7Barray%7D%5C%5C%5C%5C%20%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D85t%5C%5C%20210-d%3D65t%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20210-%5Cboxed%7B85t%7D%3D65t%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20210%3D150t%5Cimplies%20%5Ccfrac%7B210%7D%7B150%7D%3Dt%5Cimplies%20%5Ccfrac%7B7%7D%7B5%7D%3Dt%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20hour%20and%2024%20minutes%7D%7D%7B1%5Cfrac%7B2%7D%7B5%7D%3Dt%7D%20)
Answer:
The picture is 4.25 inches from the side of the paper
Step-by-step explanation:
- Taylor wants to center a 3.5 inch picture on a piece of paper that is
12 inches wide
- Lets think about that he want to put the picture in the center of the
paper, then divide the length of the paper into two equal parts and the
picture into two equal part
∵ The width of the paper is 12 inches
∵ 12 ÷ 2 = 6 inches
∵ The width of the picture is 3.5 inches
∵ 3.5 ÷ 2 = 1.75
- Now lets subtract from 6 inches (half paper) 1.75 inches (half picture)
to find the distance between the side of the paper and the picture
∵ 6 - 1.75 = 4.25
∴ The distance from the side of the paper to the picture is 4.25 inches
* <em>The picture is 4.25 inches from the side of the paper</em>
* Look to attached figure for more understand