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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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Terry and Callie do word processing. For a certain prospectus Callie can prepare it two hours faster than Terry can. If they wor
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Time taken by jerry alone is 10.1 hours

Time taken by callie alone is 8.1 hours

<u>Solution:</u>

Given:- For a certain prospectus Callie can prepare it two hours faster than Terry can

Let the time taken by Terry be "a" hours

So, the time taken by Callie will be (a-2) hours

Hence, the efficiency of Callie and Terry per hour is \frac{1}{a-2} \text { and } \frac{1}{a} \text { respectively }

If they work together they can do the entire prospectus in five hours

\text {So, } \frac{1}{a-2}+\frac{1}{a}=\frac{1}{5}

On cross-multiplication we get,

\frac{a+(a-2)}{(a-2) \times a}=\frac{1}{5}

\frac{2 a-2}{(a-2) \times a}=\frac{1}{5}

On cross multiplication ,we get

\begin{array}{l}{5 \times(2 a-2)=a \times(a-2)} \\\\ {10 a-10=a^{2}-2 a} \\\\ {a^{2}-2 a-10 a+10=0} \\\\ {a^{2}-12 a+10=0}\end{array}

<em><u>using quadratic formula:-</u></em>

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

x=\frac{12 \pm \sqrt{144-40}}{2}

\begin{array}{l}{x=\frac{12 \pm \sqrt{144-40}}{2}} \\\\ {x=\frac{12 \pm \sqrt{104}}{2}} \\\\ {x=\frac{12 \pm 2 \sqrt{26}}{2}} \\\\ {x=6 \pm \sqrt{26}=6 \pm 5.1} \\\\ {x=10.1 \text { or } x=0.9}\end{array}

If we take a = 0.9, then while calculating time taken by callie = a - 2 we will end up in negative value

Let us take a = 10.1

So time taken by jerry alone = a = 10.1 hours

Time taken by callie alone = a - 2 = 10.1 - 2 = 8.1 hours

3 0
3 years ago
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