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Afina-wow [57]
2 years ago
12

The half-life of cesium-137 is 30 years. Suppose we have a 170 mg sample.(a) Find the mass that remains after t years.(b) How mu

ch of the sample remains after 60 years? (Round your answer to two decimal places.)(c) After how long will only 1 mg remain? (Round your answer to one decimal place.)
Mathematics
1 answer:
saul85 [17]2 years ago
5 0

Answer:

212.9 years

Step-by-step explanation:

Given that the half-life of cesium-137 is 30 years. Suppose we have a 170 mg sample

P0 = 175

P(30) = 87.5

So we can write equation as

P(t) = 170(\frac{1}{2} )^{\frac{t}{30} }

b) After 60 years t = 30

In 30 years it becomes half and hence in 60 years it would become 1/4

i.e. P(60) = 137(1/2^4) = 34.25 mg

c) If P(t) =1, let us find t

137(1/2)^{\frac{t}{30} } =1\\\\n = 7.098 years\\n = 7.1*30 = 212.9 years

212.9

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Two random samples are taken from private and public universities
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Answer:

Step-by-step explanation:

For private Institutions,

n = 20

Mean, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

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Summation(x - mean)² = (43120 - 34623.05)^2+ (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

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For public Institutions,

n = 20

Mean, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

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This is a test of 2 independent groups. Let μ1 be the mean out-of-state tuition for private institutions and μ2 be the mean out-of-state tuition for public institutions.

The random variable is μ1 - μ2 = difference in the mean out-of-state tuition for private institutions and the mean out-of-state tuition for public institutions.

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-B. H0: μ1 = μ2 ; H1: μ1 > μ2

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(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

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df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

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p value = 0.000065

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