well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
![\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}](https://tex.z-dn.net/?f=%5Ccfrac%7B300~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B2%5Cpi%20~rad%7D%7B~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B60secs%7D%5Cimplies%20%5Ccfrac%7B%28300%29%282%5Cpi%20%29rad%7D%7B60secs%7D%5Cimplies%2010%5Cpi%20~%5Cfrac%7Brad%7D%7Bsecs%7D%5Capprox%2031.42~%5Cfrac%7Brad%7D%7Bsecs%7D)
Answer:
No
Step-by-step explanation:
Explanation 1)A vertical line can be represented by the equation x=a, where a can be any number. If you think about it, there is no y in the function, this making it not a function.
Explanation 2) For a line to be a function it has to pass the vertical line test, so you will draw a vertical line through the function and if it touches more than one point then it is not a function. In this case if you draw a vertical line through a vertical line there are an infinite number of places where the lines intercept, so it is not a function.
Answer:
t=5.5080( to 3 d.p)
Step-by-step explanation:
From the data given,
n =20
Deviation= 34/20= 1.7
Standard deviation (sd)= 1.3803(√Deviation)
Standard Error = sd/√n
= 1.3803/V20 = 0.3086
Test statistic is:
t = deviation /SE
= 1.7/0.3086 = 5.5080
ndf = 20 - 1 = 19
alpha = 0.01
One Tailed - Right Side Test
From Table, critical value of t =2.5395
Since the calculated value of t = 5.5080 is greater than critical value of t = 2.5395, the difference is significant. Reject null hypothesis.
t score = 5.5080
ndf = 19
One Tail - Right side Test
By Technology, p - value = 0.000
Since p - value is less than alpha , reject null hypothesis.
Conclusion:
From the result obtained it can be concluded that ,the data support the claim that the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10.
I believe the Answer is a/22 ( that’s a fraction )
