Drake drew a 120 degree angle on his paper. He drew a line

What is the value of x if the average of 10, 11, 12 and x is 9?

frosja888 [35]

Answer: The value of x is 3.

Step-by-step explanation:

Since we have given that

10,11, 12 and x

Average of above numbers = 9

As we know that

Average is given by

$\dfrac{\text{Sum of observation}}{\text{Number of observation}}\\\\\\\dfrac{10+11+12+x}{4}=9\\\\10+11+12+x=9\times 4\\\\33+x=36\\\\x=36-33\\\\x=3$

Hence, the value of x is 3.

6 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$