Find the mean,median,range of 6 6,8,5,4,6,4,3,8,4

2 answers:

8 0

The numbers in order are
3, 4, 4, 4, 5, 6, 6, 6, 8

The median is the number in the middle, which is 5.
The range is the largest number minus the smallest number, 8 - 3 = 5
The mean is the sum of all the numbers divided by how many number are the in the list, 3 + 4 + 4 + 4 + 5 + 6 + 6 + 6 + 8 = 46, 46/9 = 5.11

Natasha_Volkova [10]3 years ago

4 0

Mean is 5.4
You find this by adding all the numbers together and dividing by how many numbers there are.

Median is 5.5
This is found by putting the numbers in order from least to greatest. After that, you can cross out a number from the left and then the right, repeating the process until you get tot eh middle number. If there are two numbers in the middle, add them together and divide by two.

Range is 5
To find this you subtract the biggest number by the smallest number.

If you would like me to specify on anything, just ask and I will respond when I can.

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2 years ago

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In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.

Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

• the altitude to the hypotenuse is denoted AN,

• AB = 2√5 in,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

$h^2=a^2+b^2\text{.}$

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

• a = ,NC = 1,,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

• a = BN,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

$\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}$

Equalling the last two equations, we have:

$\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}$

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

• a = AC,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have: