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Ann [662]
3 years ago
9

A pet store has three fish tanks, each holding a different volume of water and a different number of fish. Tank AAA holds 40\tex

t{ L}40 L40, start text, space, L, end text and 555 fish. Tank BBB holds 100\text{ L}100 L100, start text, space, L, end text and 121212 fish. Tank CCC holds 180\text{ L}180 L180, start text, space, L, end text and 232323 fish.
Mathematics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

The tanks by volume per fish from least to greatest:

Tank C, Tank A, Tank B

Step-by-step explanation:

A pet store has three fish tanks, each holding a different volume of water and a different number of fish.

Order the tanks by volume per fish from least to greatest.

Tank A holds 40 L and 5 fish.

= 40L/5 fishes

= 8 volume per fish

Tank B holds 100 L and 12 fish

= 100L/12 fishes

= 8.3333333333 volume per fish

Tank C holds 180 L and 23 fish.

= 180L /23 fish

= 7.8260869565 volume per fish

The tanks by volume per fish from least to greatest

Tank C, Tank A, Tank B

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In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study
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z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055    

The p value can be calculated from the alternative hypothesis with this probability:

p_v =2*P(Z>2.055)=0.0399    

And the best option for this case would be:

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Step-by-step explanation:

Information provided

X_{1}=1642 represent the number of smokers from the sample in 1995

X_{2}=1415 represent the number of smokers from the sample in 2010

n_{1}=4276 sample from 1995

n_{2}=3908 sample from 2010  

p_{1}=\frac{1642}{4276}=0.384 represent the proportion of smokers from the sample in 1995

p_{2}=\frac{1415}{3908}=0.362 represent the proportion of smokers from the sample in 2010

\hat p represent the pooled estimate of p

z would represent the statistic    

p_v represent the value for the pvalue

System of hypothesis

We want to test the equality of the proportion of smokers and the system of hypothesis are:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

The statistic is given by:

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1642+1415}{4276+3908}=0.374  

Replacing the info given we got:

z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055    

The p value can be calculated from the alternative hypothesis with this probability:

p_v =2*P(Z>2.055)=0.0399    

And the best option for this case would be:

C. between 0.01 and 0.05.

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