The answer would be 56.10
Since, each side of the fence should have the 1 3/4 feet wide decorative piece, the width of the fence should be subtracted by 2 pieces (1 3/4 feet)
= 26.5 feet
Then, divide the remaining width by 4.5 feet and the answer is 5.88. Thus, approximately 5 pieces of the 4 1/2 feet should be installed.
Answer:
It compresses up 2
Step-by-step explanation:
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<u><em>Answer:</em></u>sin (C)
<u><em>Explanation:</em></u><u>In a right-angled triangle, special trig functions can be applied. These functions are as follows:</u>
sin (theta) = </span>

<span>
cos (theta) = </span>

<span>
tan (theta) = </span>

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<u>Now, let's check the triangle we have:</u>
<u>We have two options:</u>
<u>First option:</u>5 is the hypotenuse of the triangle
4 is the side adjacent to angle B
Therefore, we can apply the <u>cos function</u>:
cos (B) = </span>

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<u>Second option:</u>5 is the hypotenuse of the triangle
4 is the side opposite to angle C
Therefore, we can apply the <u>sin function</u>:
sin (C) = </span>

<span>
Among the two options, the second one is the one found in the choices. Therefore, it will be the correct one.
Hope this helps :)
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Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:

The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.
X = 82:


Z = 1
Z = 1 has a p-value of 0.84.
X = 72:


Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at brainly.com/question/24663213
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