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Naily [24]
3 years ago
11

Suppose a 99% confidence interval for the mean weight of high school girls in pounds is (102.3, 106.5). If we had measured the w

eights of each of the girls in kilograms (2.2 pounds = 1 kilogram) then the confidence interval for the mean weight of high school girls in kilograms would have been
Mathematics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

The confidence interval for the mean weight of high school girls in kilograms would have been would have been (46.5, 48.41).

Step-by-step explanation:

To find the interval, we can just make the conversion of the values from pounds to kilograms.

Each kilogram has 2.2 pounds. So, to realize the conversion from pounds to kilograms, we divide by 2.2.

Confidence interval: (102.3 pounds, 106.5 pounds).

102.3 pounds = 102.3/2.2 = 46.5 kilograms

106.5 pounds = 106.5/2.2 = 48.41 kilograms

The confidence interval for the mean weight of high school girls in kilograms would have been would have been (46.5, 48.41).

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One coin in a collection of 65 has two heads. The rest are fair. If a coin, chosen at random from the lot and then tossed, turns
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Answer:

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Researchers are monitoring two different radioactive substances. They have 300 grams of substance A which decays at a rate of 0.
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Answer:

231.59 years

Step-by-step explanation:

To model this situation we are going to use the exponential decay function:

f(t)= a (1-b)^t

where f(t) is the final amount remaining after t years of decay

a is the final amount

b is the decay rate in decimal form

t is the time in years

For Substance A:

Since  we have 300 grams of the substance, a=300. To convert the decay rate to decimal form, we are going to divide the rate by 100%:

r = 0.15/100 = 0.0015. Replacing the values in our function:

f(t) = a (1-b)^t

f(t) = 300 (1-0.0015)^t

f(t) = 300 (0.9985)^t equation (1)

For Substance B:

Since we have 500 grams of the substance, a= 500. To convert the decay rate to decimal form, we are going to divide the rate by 100%:

r=0.37/100= 0.0037. Replacing the values in our function:

f(t) = a (1-b)^t

f(t)= 500 (1-0.0037)^t

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Since they are trying to determine how many years it will be before the substances have an equal mass M, we can replace f(t) with M in both equations:

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M=500(0.9963)^t equation (2)

We can conclude that the system of equations that can be used to determine how long it will be before the substances have an equal mass, M, is :

{M=300(0.9985)^t

{M=500(0.9963)^t

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