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Evgesh-ka [11]
1 year ago
15

Deondra is 1.35 meters tall. At 10 a.m., she measures the length of a tree's shadow to

Mathematics
1 answer:
pashok25 [27]1 year ago
6 0

Answer:

If the length of a tree's shadow is 35.25 meters. The height of the tree to the nearest hundredth of a meter will be : 11.74m

Given:

Denora height=1.35 meters

Length =35.25 meters

Width =31.2 meters

Height of the tree=x

Proportion:

1.35 : 35.25 :: x : 31.2

Now let's determine the height of the tree:

35.25 - 31.2 / 1.35 = 35.25 / x

4.05 / 1.35 = 35.25 / x

Cross multiply

4.05x = 35.25 × 1.35

4.05x = 47.58

Divide both sides

x = 47.58 / 4.05

<u>x = 11.74</u>

In conclusion, if the length of a tree's shadow is 35.25 meters. The height of the tree to the nearest hundredth of a meter will be: 11.74m

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The length and breadth of a rectangular field are 312m and 186m respectively; correct to the nearest metres. Between what limits
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P \geq 1000 \; meters

Step-by-step explanation:

<u>Given the following data;</u>

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Breadth = 186 meters

To find the perimeter of the rectangle;

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Perimeter = 2(L + W)

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The following are the last 7 days adjusted closing stock prices (in $) for Citigroup: 22, 25, 20, 15, 40, 55, 28 A) Compute the
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(A) The 25th percentile is 20.

(B) The mean of the data set is 20 and the median of the data set is 25. The data set is left-skewed.

Step-by-step explanation:

The data set arranged in ascending order is:

S = {15, 20, 22, 25, 28, 40, 55}

(A)

Compute the 25th percentile as follows:

25^{th} \text{Percentile}=[\frac{n+1}{4}]^{th}obs.\\\\=[\frac{7+1}{4}]^{th}obs.\\\\=2^{nd}obs.\\\\=20

Thus, the 25th percentile is 20. This value implies that exactly 25% of the observations are less than 20.

(B)

Compute the sample mean as follows:

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Compute the median as follows:

For a set with odd number of observations, the median value if the middle value.

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Thus, the mean of the data set is 20 and the median of the data set is 25.

The Mean < Median.

This implies that the data set is left-skewed.

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