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stiv31 [10]
3 years ago
14

After a history test with 20 questions, the teacher asks his students to write down the number of the questions that was the mos

t difficult. He then asks a couple of his students to organize this data and inform the class on the results. Here are their results: then the students calculate the median and find it to be 4. Which of the following sentences best describes their teachers response to this report

Mathematics
2 answers:
Elis [28]3 years ago
6 0
Yup. D
Median does not tell which one was the hardest. That would be mode
asambeis [7]3 years ago
4 0

Answer:

i think it would be d

Step-by-step explanation:

the median is just the middle of the numbers, it doesnt say which is the hardest question though

hope this helps.

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2 years ago
These triangles are similar. find Y and X
CaHeK987 [17]

It these triangles are similar, then the sides of the triangles are in proportion:

\dfrac{y}{10}=\dfrac{y+4}{15}      <em>cross multiply</em>

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15y=10y+40       <em>subtract 10y from both sides</em>

5y=40           <em>divide both sides by 5</em>

\boxed{y=8}


\dfrac{x}{4}=\dfrac{6}{8}            <em>cross multiply</em>

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3 0
3 years ago
Find the values of the sine, cosine, and tangent for ZA C A 36ft B <br> 24ft
Reptile [31]
<h2>Question:</h2>

Find the values of the sine, cosine, and tangent for ∠A

a. sin A = \frac{\sqrt{13} }{2},  cos A = \frac{\sqrt{13} }{3},  tan A = \frac{2 }{3}

b. sin A = 3\frac{\sqrt{13} }{13},  cos A = 2\frac{\sqrt{13} }{13},  tan A = \frac{3}{2}

c. sin A = \frac{\sqrt{13} }{3},  cos A = \frac{\sqrt{13} }{2},  tan A = \frac{3}{2}

d. sin A = 2\frac{\sqrt{13} }{13},  cos A = 3\frac{\sqrt{13} }{13},  tan A = \frac{2 }{3}

<h2>Answer:</h2>

d. sin A = 2\frac{\sqrt{13} }{13},  cos A = 3\frac{\sqrt{13} }{13},  tan A = \frac{2 }{3}

<h2>Step-by-step explanation:</h2>

The triangle for the question has been attached to this response.

As shown in the triangle;

AC = 36ft

BC = 24ft

ACB = 90°

To calculate the values of the sine, cosine, and tangent of ∠A;

<em>i. First calculate the value of the missing side AB.</em>

<em>Using Pythagoras' theorem;</em>

⇒ (AB)² = (AC)² + (BC)²

<em>Substitute the values of AC and BC</em>

⇒ (AB)² = (36)² + (24)²

<em>Solve for AB</em>

⇒ (AB)² = 1296 + 576

⇒ (AB)² = 1872

⇒ AB = \sqrt{1872}

⇒ AB = 12\sqrt{13} ft

From the values of the sides, it can be noted that the side AB is the hypotenuse of the triangle since that is the longest side with a value of 12\sqrt{13} ft (43.27ft).

<em>ii. Calculate the sine of ∠A (i.e sin A)</em>

The sine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the hypotenuse side of the triangle. i.e

sin Ф = \frac{opposite}{hypotenuse}             -------------(i)

<em>In this case,</em>

Ф = A

opposite = 24ft (This is the opposite side to angle A)

hypotenuse = 12\sqrt{13} ft (This is the longest side of the triangle)

<em>Substitute these values into equation (i) as follows;</em>

sin A = \frac{24}{12\sqrt{13} }

sin A = \frac{2}{\sqrt{13}}

<em>Rationalize the result by multiplying both the numerator and denominator by </em>\sqrt{13}<em />

sin A = \frac{2}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }

sin A = \frac{2\sqrt{13} }{13}

<em>iii. Calculate the cosine of ∠A (i.e cos A)</em>

The cosine of an angle (Ф) in a triangle is given by the ratio of the adjacent side to that angle to the hypotenuse side of the triangle. i.e

cos Ф = \frac{adjacent}{hypotenuse}             -------------(ii)

<em>In this case,</em>

Ф = A

adjacent = 36ft (This is the adjecent side to angle A)

hypotenuse = 12\sqrt{13} ft (This is the longest side of the triangle)

<em>Substitute these values into equation (ii) as follows;</em>

cos A = \frac{36}{12\sqrt{13} }

cos A = \frac{3}{\sqrt{13}}

<em>Rationalize the result by multiplying both the numerator and denominator by </em>\sqrt{13}<em />

cos A = \frac{3}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }

cos A = \frac{3\sqrt{13} }{13}

<em>iii. Calculate the tangent of ∠A (i.e tan A)</em>

The cosine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the adjacent side of the triangle. i.e

tan Ф = \frac{opposite}{adjacent}             -------------(iii)

<em>In this case,</em>

Ф = A

opposite = 24 ft (This is the opposite side to angle A)

adjacent = 36 ft (This is the adjacent side to angle A)

<em>Substitute these values into equation (iii) as follows;</em>

tan A = \frac{24}{36}

tan A = \frac{2}{3}

6 0
3 years ago
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