David is 38 years old and his son Frank is 8. In how many years will David be twice as old as Frank?

A freshly prepared sample of curium-243 undergoes 3312 disintegrations per second. after 6.00 yr, the activity of the sample dec

Lerok [7]

I took the liberty of finding for the complete question. And here I believe that the problem asks for the half life of Curium. Assuming that the radioactive decay of Curium is of 1st order, therefore the rate equation is in the form of:

A = Ao e^(-kt)

where,

A = amount after t years = 2755

Ao = initial amount = 3312

k = rate constant

t = number of years passed = 6

Therefore the rate constant is:

2755/3312 = e^(-6k)

-6k = ln (2755/3312)

k = 0.0307/yr

The half life, t’, can be calculated using the formula:

t’ = ln 2 / k

Substituting the value of k:

t’ = ln 2 / 0.0307

t’ = 22.586 years

or

t’ = 22.6 years

4 0

4 years ago

Find the slope of the graph

OverLord2011 [107]

Answer: The slope of the graph is -2

Step-by-step explanation:

Take any two points from the graph, and do rise/run

For example, we take (1,3) and (2,1)

Apply rise/run... rise: 2, run: -1

2/-1 = -2

m = -2

For the next question, the slope is 4

And the last question, the slope is 1 1/2

Hope this helped!

6 0

3 years ago

You are going to take a 10-question True/False test. How many ways can this test be answered if leaving questions blank is a via

FinnZ [79.3K]

Answer:

2^10 = 1024

Step-by-step explanation:

The test most likely can be answered in two different ways. the different ways that 10 questions can be answered will be:

2^10 = 1024

6 0

3 years ago

Mislabeled seafood In 2013 the environmental group Oceana (usa.oceana.org) analyzed 1215 samples of seafood purchased across the

Step2247 [10]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

<h3>z-distribution interval:</h3>

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

For this problem:

1215 samples, hence $n = 1215$ .

33% was mislabeled or misidentified, hence $p = 0.33$ .

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

<h3>The lower limit of this interval is:</h3>

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 - 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3036$

<h3>The upper limit of this interval is:</h3>

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 + 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3564$

The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

You can learn more about the use of the z-distribution to build a confidence interval at brainly.com/question/25730047

4 0

2 years ago

there are 3 liters of apple juice at a school party. 10 students want to drink all the apple juice, Andi all want to get exactly

pochemuha

Answer:

0.3 L

Step-by-step explanation:

3 L / 10 students = 3/10 = 0.3 Liters per student

4 0

3 years ago