PLEASE HELP, MARKING BRAINLIEST!!!! If 30/t = 18/r, then t/r =
2 answers:
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Answer:
t/r = 5/3
Step-by-step explanation:
Every proportion can be written 4 ways. I like to think of them as "upside down and sideways." They are ...
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Knowing this, you can simply rewrite your proportion as ...
t/r = 30/18
t/r = 5/3
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More formally, you could multiply both sides by t/18:
(30/t)(t/18) = (18/r)(t/18)
30/18 = t/r . . . . as above
Reducing the fraction gives ...
t/r = 5/3
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<em>Additional comments</em>
The usually-recommended procedure for solving something like this is to "cross-multiply" as a first step. That will give ...
30r = 18t
Of course, that is equivalent to multiplying both sides by rt, the product of the denominators.
From here, to get t/r, you need to divide both sides by 18r.
(30r)/(18r) = (18t)/(18r)
30/18 = t/r
The combination of the two steps, multiply by rt, divide by 18r, is equivalent to multiplying by (rt)/(18r) = t/18, as we did above. That is, you can save some work by doing both steps at once--or by simply rewriting the proportion.
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Answer:
We conclude that the standard deviation of the diameter of a golf ball is less than 0.005 inch.
Step-by-step explanation:
We are given that a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.005 inch.
Assume that the population is normally distributed: 1.678, 1.681, 1.676, 1.684, 1.676, 1.679, 1.681, 1.681, 1.677, 1.676, 1.681, 1.683.
Let = <u><em>population standard deviation of the diameter of a golf ball.</em></u>
SO, Null Hypothesis, :
0.005 inch {means that the standard deviation of the diameter of a golf ball is more than or equal to 0.005 inch}
Alternate Hypothesis, :
< 0.005 inch {means that the standard deviation of the diameter of a golf ball is less than 0.005 inch}
The test statistics that would be used here <u>One-sample Chi-square test statistics</u>;
T.S. = ~
where, s = sample standard deviation = = 0.00281 inch
n = sample size = 12
So, <u><em>the test statistics</em></u> = ~
= 3.47
The value of chi-square test statistics is 3.47.
Also, the P-value of test statistics is given by the following formula;
P-value = P( < 3.47) = 0.0182
Since, the P-value of the test statistics is less than the level of significance as 0.0182 < 0.05, so we reject our null hypothesis.
<u>Now, at 0.05 significance level the chi-square table gives critical value of 4.575 at 11 degree of freedom for left-tailed test.</u>
Since our test statistic is less than the critical value of chi-square as 3.47 < 4.575, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.
Therefore, we conclude that the standard deviation of the diameter of a golf ball is less than 0.005 inch.