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bija089 [108]
3 years ago
9

Iodine-123 is a radioactive substance used in medicine. It has a half-life of 13 hours. A nurse received a solution that initial

ly contained 48 grams of iodine-123. Now only 12 grams of the iodine-123 remain. How many hours have passed since the nurse received the solution?
Mathematics
1 answer:
zhuklara [117]3 years ago
5 0
Ok

A=P(1/2)^(t/h)
A=amount final
P=initial amount
t=time elapsed
h=time of half life

so

given h=13
and initial =48g
and final is 12g


12=48(1/2)^(t/13)
divide both sides by 48
1/4=(1/2)^(t/13)
what a coincidence
(1/2)^2=(1/2)^(t/13)
so
2=t/13
26=t

26 hours have passed
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Answer:

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You would have 2x

The combined total needs to be 6 hours or greater.

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This needs to be greater than or equal to 6.

I hope this helps you :)

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Lucy rented a movie. She started the movie at 9:48 a.m. and it was 3 hours 39 minutes long. When did the movie end
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The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assum
Vadim26 [7]

Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

Z = \frac{X - \mu}{s}

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

X = 25

Z = \frac{X - \mu}{s}

Z = \frac{25 - 26}{0.762}

Z = -1.31

Z = -1.31 has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

Z = \frac{X - \mu}{s}

-1.645 = \frac{X - 26}{0.762}

X - 26 = -1.645*0.762

X = 24.75

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

Z = \frac{X - \mu}{s}

1.645 = \frac{X - 26}{0.762}

X - 26 = 1.645*0.762

X = 27.25

Between $24.75 and $27.25.

3 0
3 years ago
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