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Masteriza [31]
3 years ago
12

Tomb trees are standing side-by-side in the sunlight one is 50 feet tall and the other is 20 feet tall if the taller tree cast a

n 18 foot shadow what is the length in feet of the shadow of the shorter tree
Mathematics
1 answer:
8090 [49]3 years ago
7 0

Answer: Length of shorter tree = 7.2 feet .

Step-by-step explanation:

Given : Height of two trees as 50 feet and 20 feet.

At same time the angle made by sun rays would be same for both trees.

Both trees are vertical to the ground making right angle so triangles made by both triangles must be similar.

Corresponding sides of similar triangles are proportional.

So \dfrac{\text{length of the shadow of the shorter tree}}{\text{height of shorter tree}}=\dfrac{\text{length of the shadow of the bigger tree}}{\text{height of bigger tree}}\\\\=\dfrac{\text{length of the shadow of the shorter tree}}{20}=\dfrac{18}{50}\\\\\Rightarrow\ \text{length of the shadow of the shorter tree}=\dfrac{18}{50}\times20\\\\\Rightarrow\ \text{length of the shadow of the shorter tree} = 7.2

So, Length of shorter tree = 7.2 feet .

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Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav
Furkat [3]

Answer:

  • There is no significant evidence that p1 is different than p2 at 0.01 significance level.
  • 99% confidence interval for p1-p2 is  -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

H_{0}: p1-p2=0

H_{a}: p1-p2≠0

Test statistic can be found using the equation:

z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

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  • p2 is the sample proportion of the common attribute in population2 (\frac{1337}{1900} =0.704)
  • p is the pool proportion of p1 and p2 (\frac{16+1337}{30+1900}=0.701)
  • n1 is the sample size of the people from population1 (30)
  • n2 is the sample size of the people from population2 (1900)

Then z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}} ≈ 2.03

p-value of the test statistic is  0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

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Thus 99% confidence interval is

0.533-0.704±2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}} ≈ -0.171 ±0.237 that is (−0.408, 0.066)

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Simplest form of:
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