Most everyday situations involving chance and likelihood are examples of ______. simple probability permutations conditional pro
1 answer:
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Answer:
(x, y) = (0, 1/2) or (1, 3)
Step-by-step explanation:
The first equation factors as ...
x(3x -y) = 0
This has solutions x=0 and y=3x.
__
<u>x = 0</u>
Using this in the second equation gives ...
2y -0 = 1
y = 1/2
(x, y) = (0, 1/2) is a solution
__
<u>y = 3x</u>
Using the expression for y in the second equation, we get ...
2(3x) -5x = 1
x = 1 . . . . . . . . . simplify
y = 3x = 3 . . . . using x=1 in the first equation
(x, y) = (1, 3) is a solution
_____
Interestingly, the (red line) graph of 3x^2 -xy = 0 produced by this graphing calculator has a "hole" at x=0, It says that point is (0, undefined). In a sense, y is undefined, in that it can be <em>anything</em>. A more appropriate graph would graph that equation as the two lines x=0 and y=3x.
Answer:
We conclude that the sample is from a population of songs with a mean greater than 210 seconds.
Step-by-step explanation:
We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.
Assume that the standard deviation of song lengths is 54.5 sec.
Let = <u><em>population mean length of the songs</em></u>
So, Null Hypothesis, :
210 seconds {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}
Alternate Hypothesis, :
> 210 seconds {means that the sample is from a population of songs with a mean greater than 210 seconds}
The test statistics that will be used here is <u>One-sample z-test statistics </u>because we know about population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean length of songs = 252.5 seconds
= population standard deviation = 54.5 seconds
n = sample of current hit songs = 40
So, <u><em>the test statistics</em></u> =
= 4.932
The value of z-test statistics is 4.932.
Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.
Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.
Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.