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dusya [7]
3 years ago
6

Need help ASAP. The diagram shows 5ft student standing near a tree. The shadow of the student and the shadow of the tree at endp

oint A. What is the height of the tree. Show all your work, and make sure to include units in your final answer.

Mathematics
1 answer:
inna [77]3 years ago
8 0

Answer:

The height of the tree = 25 feet

 Step-by-step explanation:  

From the given diagram : EF = 5 feet, FA = 8 feet, CA = 40 feet

∠AFE = 90° and ∠ACB = 90°

To find : CB, the height of the tree.  

Solution : In ΔAEF and ΔAB  C

∠AFE = ∠ACB = 90°  

∠A is common angle.  

So, By AA postulate of similarity of triangle, ΔAEF ~ ΔABC  

Now, sides of similar triangles are proportional to each other

\implies\frac{FA}{CA}=\frac{EF}{CB}\\\\\implies \frac{8}{40}=\frac{5}{CB}\\\\\implies CB=\frac{40\times 5}{8}\\\\\implies CB=25

Hence, The height of the tree = 25 feet

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Answer: The correct answer is NF_3

Explanation:

In NF_3 molecule, there are three bond pairs and one lone pair present in the molecule. In a molecule, lone pair present on the nitrogen atom will have a dipole moment in upward direction where as total dipole moment formed by three fluorine atoms will be more in magnitude and it will be directed in downward direction which means that the net dipole moment will be in downward direction. As shown in image attached.

Hence, covalent bond present between nitrogen and fluorine will become polar in nature.

So, the correct answer is NF_3.

7 0
3 years ago
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The solution of the equation 7x+2=23 is ?
ikadub [295]
7x+2=23
Move 2 to the other side
7x=23-2
7x=21 

To isolate the x divide both sides by 7
x=3 

To check plug the value of x back in 

7(3)+2=23
21+2=23
23=23
3 0
3 years ago
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In need of help PLEASE!!!!​
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Answer:

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Step-by-step explanation:

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Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{1048}{5}}

Standard deviation = \sqrt{209.6}

Standard deviation = 14.48

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{14.48}{142}*100

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

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