PLEASE HELP ITS URGENT
12. Carla is lying on the ground between two trees in her backyard. She is 67 feet from the tree to her left. The angle of elevation from her eyes to the top of the tree on the left is 50°. Carla is 75 feet from the tree to her right and the angle of elevation from her eyes to the top of that tree is 36°. a. Determine the height of each tree to the nearest foot. Show your work. b. State which tree is taller, and by how much.

PLEASE HELP ITS URGENT12. Carla Is Lying On The Ground Between Two Trees In Her Backyard. She Is 67 Feet

Answers

Answer 1

Answer:

the left tree is taller

Step-by-step explanation:

shown in work above :)

PLEASE HELP ITS URGENT12. Carla Is Lying On The Ground Between Two Trees In Her Backyard. She Is 67 Feet
Answer 2

The first tree is taller than the second tree by 25.3 ft.

We have to use the trigonometric ratios and angle of elevation to obtain the height of each tree as follows;

Let the height of the first tree be h1 and the height of the second tree be h2.

For the first tree;

tan 50° = h1/67

h1 = 67 tan 50° = 79.8 ft

For the second tree;

tan 36° = h2/75

h2 = 75tan 36°

h2 = 54.5 ft

The first tree is taller than the second tree by 25.3 ft.

Learn more about angle of elevation: https://brainly.com/question/12483071


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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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DONT TRUST ME ON  THIS

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Answers

Answer:

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.

This means that [tex]\mu = 74, \sigma = 2.7[/tex]

What proportion of childrenin this age range between 70 lbs and 85 lbs.

This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So

X = 85

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{85 - 74}{2.7}[/tex]

[tex]Z = 4.07[/tex]

[tex]Z = 4.07[/tex] has a pvalue of 1

X = 70

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 74}{2.7}[/tex]

[tex]Z = -1.48[/tex]

[tex]Z = -1.48[/tex] has a pvalue of 0.0694

1 - 0.0694 = 0.9306

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

The question is in the jpg.

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

Answer:

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Have a great day!

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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