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2 years ago

What is the mode? 9 4 6 3 9 3 10 4 6 4

Mathematics

2 answers:

Brut [27]2 years ago

5 0

Answer:

Explanation:

The mode is what number occurs most often

Zepler [3.9K]2 years ago

3 0

The mode for this question is <u>4</u>

<u></u>

In a set of numbers, the mode is the number that occurs most often. In other words, it measures how frequently a particular number occurs within a set of numbers. It is a type of average, like its counterparts, the median and mean.

To find the mode of a set of numbers,

List the set of numbers in ascending or descending order, including any duplicates
Count the number of times each number in the set of numbers occurs. The number that occurs the highest number of times is the mode of the set of data

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Answer: The area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

Step-by-step explanation:

I will assume that the exercise says " $\frac{1}{3}$ times the base and height of Jerry’s rhombus".

The area of a rhombus can be calculated with the following formula:

$A=b*a$

Where "b" is the length of the base and "a" is the altitude or the height.

Then, you can calculate the area using the formula shown above.

Therefore, you get:

1. Jerry's rhombus:

$A_1=(15\ in)(9\ in)\\\\A_1=135\ in^2$

2. Charlene's rhombus:

$A_2=(\frac{1}{3}*15\ in)(\frac{1}{3}*9\ in)\\\\A_2=(5\ in)(3\ in)\\\\A_2=15\ in^2$

Dividing the area calculated, you get:

$\frac{135\ in^2}{15\ in^2}=9$

Therefore, you can conclude that the area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

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Then, our options are the following:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \therefore y=-\frac{2}{5}x-2 \\ \\ C) \ 2x-5y=-10 \therefore y=\frac{2}{5}x+2 \\ \\ D) \ y+4=-\frac{2}{5}(x-5) \therefore y=-\frac{2}{5}x-2 \\ \\ E) \ y-4=\frac{5}{2}(x + 5) \therefore y=\frac{5}{2}x+\frac{33}{2}$

For two perpendicular lines it is true that the product of its slopes is:

$m_{1}m_{2}=-1$

$m_{1}m_{2}=-1 \\ \\ m_{1} \ is \ the \ slope \ of \ y=\frac{5}{2}x+3, \ that \ is, \ m_{1}=\frac{5}{2} \\ \\ Then, \ the \ slope \ of \ a \ perpendicular \ line \ is: \\ \\ m_{2}=-\frac{2}{5}$

According to this, only A) B) and D) might be the perpendicular lines we are looking for. Notice that these lines are the same. The other condition is that the line must pass through the point (5, -4). By substituting this point in the equation, we have:

$y = -\frac{2}{5}x-2 \\ \\ -4=-\frac{2}{5}(5)-2 \\ \\ -4=-2-2 \\ \\ \boxed{-4=-4} \ True!$