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sertanlavr [38]
3 years ago
5

The number of species of coastal dune plants in australia decreases as the latitude, in °s, increases. There are 34 species at 1

1°s and 26 species at 44°s.1 (a) find a formula for the number, n, of species of coastal dune plants in australia as a linear function of the latitude, l, in °s. Round the slope to four decimal places and the intercept to two decimal places.
Mathematics
1 answer:
Bas_tet [7]3 years ago
4 0

We have been given that the number of species of coastal dune plants in Australia decreases as the latitude, in °s, increases.

Further we know that there are 34 species at 11°s and 26 species at 44°s.

We can express the given information at two ordered pairs as shown below:

(n,l)=(11,34)\text{ and }(n,l)=(44,26)

Let us find slope of the line through these points:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{26-34}{44-11}=\frac{-8}{33}=-0.2424

Therefore, we can write the equation of line in slope intercept form as:

n=-0.2424l+b

Where b is the y intercept, and we can find its value using one of the two points.

34=-0.2424(11)+b\\34=-2.67+b\\b=34+2.67=36.67

Therefore, the required equation of the linear function is:

n=-0.2424l+36.67

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What will make 3/10 n= 4/5?
Paladinen [302]

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put 8/3

Step-by-step explanation:

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3 years ago
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

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a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

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".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
3 years ago
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= 46 units²

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2 years ago
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