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KIM [24]
3 years ago
8

A doctor prescribes 150 milligrams of a therapeutic drug that decays by about 15% each hour. To the nearest hour, what is the ha

lf-life of the drug?
Mathematics
1 answer:
Nikolay [14]3 years ago
3 0

Answer:

4 hours

Step-by-step explanation:

The amount remaining at the end of each hour is 0.85 of the amount at the start of the hour. You want to find the exponent such that ...

... 0.85^x = 0.5 . . . . . x = the number of hours until half is left

... x·log(0.85) = log(0.5) . . . . . take the log to turn it to a linear equation

... x = log(0.5)/log(0.85) ≈ 4.265024

Rounded to the nearest  hour, x = 4.

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Andrews [41]

You and your 2 nieces bakes 27 cupcakes. This means that there were 27 cupcakes to be distributed evenly between the 3 of you. So divide 27 by 3 to get your answer.

27 ÷ 3 =  9.

This means that each of you would get 9 miniture cupcakes to take home.

8 0
3 years ago
A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w
melamori03 [73]

Answer:

The hypothesis is:

<em>H₀</em>: p_{X}-p_{Y}=0.

<em>Hₐ</em>: p_{X}-p_{Y}.

Step-by-step explanation:

Let <em>X</em> = number of men who exercise regularly and <em>Y</em> = number of women who exercise regularly.

The information provided is:

n_{X}=150\\X=88\\n_{Y}=200\\Y=130

Compute the sample proportion of men and women who exercise regularly as follows:

\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587

\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and \hat p_{X}=0.587.

The random variable <em>Y</em> also follows a Binomial distribution with parameters <em>n</em> = 200 and \hat p_{Y}=0.65.

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

\hat p=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion <em>z</em>-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

The hypothesis for this test cab be defined as:

<em>H₀</em>: The proportion of women is same as men who exercise regularly, i.e. p_{X}-p_{Y}=0.

<em>Hₐ</em>: The proportion of women is more than men who exercise regularly, i.e. p_{X}-p_{Y}.

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If you add it up it will be 6 2/3
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