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1 year ago

How much of the initial mass remains after 80 years? (Round your answer to three decimal places.)

Mathematics

1 answer:

aleksandr82 [10.1K]1 year ago

4 0

Given

$y=11(\frac{1}{2})^{\frac{t}{30}}$

When t = 80

We substitute for t

$y=11(\frac{1}{2})\frac{}{}^{\frac{80}{30}}$

Simplify

$y=11(\frac{1}{2})^{\frac{8}{3}}$

Simplify further

$y=1.732g$

The final answer

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Explanation A random sample of 100 observations from a normally distributed population possesses a mean equal to 83.2 and a stan

Olenka [21]

Answer:

The 95% confidence interval would be given by (81.933;84.467)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=83.2$ represent the sample mean

$\mu$ population mean (variable of interest)

s=6.4 represent the sample standard deviation

n=100 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=100-1=99$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that $t_{\alpha/2}=1.98$

Now we have everything in order to replace into formula (1):

$83.2-1.98\frac{6.4}{\sqrt{100}}=81.933$

$83.2+1.98\frac{6.4}{\sqrt{100}}=84.467$

So on this case the 95% confidence interval would be given by (81.933;84.467)

7 0

3 years ago

What is the equation of the line that passes through the point (- 2, - 1) and has a slope of 5/2

Phoenix [80]

Answer:

$equation \: of \: a \: line \\ y - y1 = m(x - x1) \\ y - ( - 1) = \frac{5}{2}(x - ( - 2)) \\ 2(y + 1 )= 5(x + 2) \\ 2y + 2 = 5x + 10 \\ 2y - 5x - 8 = 0$

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2 years ago

The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an

Genrish500 [490]

Answer:

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,

Standard deviation $(\sigma)$ = 7.1

Mean $\overline{(y)}$ = 110

$\alpha =$ 1 - confidence interval = 1 - .98 = .02

$z_{\frac{\alpha}{2}}$ = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine = $\overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}$