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lara31 [8.8K]
4 years ago
15

A 6 ft vertical pole casts a 14 in shadow at the same time a nearby cell phone tower casts a 119 ft shadow how tall is the cell

phone tower
Mathematics
1 answer:
o-na [289]4 years ago
7 0
The tower is 612 feet. to solve this you have to convert the 6ft from into inches which is 72in then divide youll get 5.14285714 then times it by 119ft. because its about ratio.
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The following data represent the weights in pounds of a sample of 25 police officers:
DochEvi [55]

Given:

The data values are:

164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, and 171.

To find;

a. Lower quartile.

b. Upper quartile.

c. Interquartile range.

Solution:

We have,

164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, 171.

Arrange the data values in ascending order.

134, 137, 140, 145, 146, 148, 151, 152, 154, 156, 156, 157, 162, 163, 164, 165, 168, 168, 169, 171, 172, 173, 174, 174, 177.

Divide the data values in two equal parts.

(134, 137, 140, 145, 146, 148, 151, 152, 154, 156, 156, 157), 162, (163, 164, 165, 168, 168, 169, 171, 172, 173, 174, 174, 177)

Divide each parentheses in two equal parts.

(134, 137, 140, 145, 146, 148), (151, 152, 154, 156, 156, 157), 162, (163, 164, 165, 168, 168, 169), (171, 172, 173, 174, 174, 177)

a. Location of lower quartile is:

Q_1=\dfrac{1}{4}(n+1)\text{th term}

Q_1=\dfrac{1}{4}(25+1)\text{th term}

Q_1=\dfrac{26}{4}\text{th term}

Q_1=6.5\text{th term}

The lower quartile of the weights is:

Q_1=\dfrac{148+151}{2}

Q_1=\dfrac{299}{2}

Q_1=149.5

Therefore, the location of the lower quartile of the weights is between 6th term and the 7th term. The value of the lower quartile is 149.5.

b. Location of upper quartile is:

Q_3=\dfrac{3}{4}(n+1)\text{th term}

Q_3=\dfrac{3}{4}(25+1)\text{th term}

Q_3=\dfrac{3\cdot 26}{4}\text{th term}

Q_3=19.5\text{th term}

The upper quartile of the weights is:

Q_3=\dfrac{169+171}{2}

Q_3=\dfrac{340}{2}

Q_3=170

Therefore, the location of the upper quartile of the weights is between 19th term and the 20th term. The value of the upper quartile is 170.

c. The interquartile range of the given data set is:

IQR=Q_3-Q_1

IQR=170-149.5

IQR=20.5

Therefore, the interquartile range of the weights is 20.5.

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3 years ago
the length of a rectangular parking lot at the airport is 2/3 mile.if the area is 1/2 square mile, what is the width of the park
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<h3>Answer:</h3>

3/4 mile

<h3>Explanation:</h3>

Area is the product of length and width. Put the given values into that formula and solve for width.

... A = L·W

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... (1/2 mi²)/(2/3 mi) = W = (1/2)·(3/2) mi

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Answer:

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If lines are ||, corresponding angles are equal : m∠1 = m∠5

Substitution : m∠1 + m∠7 = 180°

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4 years ago
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The mean of their scores is their sum divided by the number of scores.
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How many numbers facts can you find using this set of numbers. 4, 5, 8, 10, 40
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Only one number is odd,
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