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3 years ago

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Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 18% each week. The following function represents

the weekly weed growth: f(x) = 86(1.18)x. Rewrite the function to show how quickly the weeds grow each day.

1 answer:

8090 [49]3 years ago

6 0

Answer:

f(z) = 86.(0.169)z

Step-by-step explanation:

Let's call

x = number of weeks

z = number of days

We have the growing function f(x) = 86.(1.18)x. If we want to write f in terms of z, we need to consider that x = z/7, since there are 7 days in each week. Then,

86.(1.18)x = 86.(1.18)z/7 = 86.(0.169)z

f(z) = 86.(0.169)z

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Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

$\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$

From the above plot,

$n = 83$

The median is calculated as:

$Median = \frac{n+1}{2}th$

$Median = \frac{83+1}{2}th$

$Median = \frac{84}{2}th$

$Median = 42nd$

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf .4

So the median is:

$Median = 90.4$

The lower quartile (Q) is calculated as:

$Q_1 = \frac{n+1}{4}th$

$Q_1 = \frac{83+1}{4}th$

$Q_1 = \frac{84}{4}th$

$Q_1 = 21st$

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

$Q_1 = 88.6$

The upper quartile (Q3) is calculated as:

$Q_3 = 3 * \frac{n+1}{4}th$

$Q_3 = 3 * \frac{83+1}{4}th$

$Q_3 = 3 * \frac{84}{4}th$

$Q_3 = 3 * 21th$

$Q_3 = 63rd$

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

$Q_3 = 92.2$

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3 years ago