Question 5 help please

Amy earns $300 per week working at a shoe store. She also earns $4 for each pair of shoes she sells. This week, she wants to ear

Anna007 [38]

Answer:

B

Step-by-step explanation:

3 0

3 years ago

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This is a warmup for your first project. The drug valium is eliminated from the bloodstream through a decay process with a half-

frosja888 [35]

Step-by-step explanation:

a. We have v as the amount of drug(mg) in the blood at time t (hrs)

b. Flow rate in = k litre/he

For a circulatory system we have the flow rate in and flow rate out to be the same

Therefore,

Flow rate out = k litre/hr

concentration of drug going in = 0mg/litre

Concentration of drug going out = v mg/litre

V(0) = 20mg since 20mg was taken at midnight

V(0) = 20

Half life t1/2 = 36 hours

V(36) = v(0)/2

= 20/2

= 10mg

C. $ivp = \frac{dv}{dt}$ $= -kv$

$v(0) = 20mg\\v(36) = 10mg$

d. solution

$\frac{dv}{dt} = -kt\\ln(v) = ln(c) - kt$

$\frac{v}{c} = e^{-kt} = v=ce^{-kt}$

v(0) = 20

$ce^{-k(0)} =20$

c = 20

so

$v(t) = 20e^{-kt}$

e.

$t\frac{1}{2}=36hours\\ v(36)= 10$

$10 = 20e^{-36k}$

$\frac{1}{2} =e^{-36k}$

we take log

$k=\frac{ln(2)}{36}$

please check attachment for answer f

8 0

3 years ago

Fill in the blank and which goes where, drag and drop.. 80% 95% 33 1/3% 15%

scoundrel [369]

1. = 80 2 = 95 3 = 33 4 = 15

4 0

2 years ago

380102.804 word form

MrMuchimi

Three hundred eighty thousand, one hundred two and eight hundred four thousandths

7 0

3 years ago

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A manufacturer of nails claims that only 4% of its nails are defective. A random sample of 20 nails is selected, and it is found

irakobra [83]

Answer:

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

Step-by-step explanation:

A manufacturer of nails claims that only 4% of its nails are defective.

At the null hypothesis, we test if the proportion is of 4%, that is:

$H_0: p = 0.04$

At the alternative hypothesis, we test if the proportion is more than 4%, that is:

$H_a: p > 0.04$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

4% is tested at the null hypothesis

This means that $\mu = 0.04, \sigma = \sqrt{0.04*0.96}$

A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective.

This means that $n = 20, X = 0.1$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.1 - 0.04}{\frac{\sqrt{0.04*0.96}}{\sqrt{20}}}$

$z = 1.37$

P-value of the test and decision:

Considering an standard significance level of 0.05.

The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.

Looking at the z-table, z = 1.37 has a p-value of 0.9147

1 - 0.9147 = 0.0853

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

7 0

2 years ago

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