“Number of teeth each child has in second grade class.”<br><br> Low or high variability?

data set A is {30, 45, 32, 50, 33, 40, 44, 32}. Data set B is {28, 43, 30, 48, 35, 42, 46, 34}. which statement best compares th

3241004551 [841]

Answer:

the mean in set B is equal to the mean in set A (option C)

Question:

The question is incomplete as the answer choices were not given.Let's consider the following question:

Data set A is {30, 45, 32, 50, 33, 40, 44, 32}. Data set B is {28, 43, 30, 48, 35, 42, 46, 34}. which statement best compares the two data sets?

a) median for set A is equal to the median for set B

b) Range for set A is greater than range for set B

c) The mean in set B is equal to the mean in set A

Step by step explanation:

We can describe a data set using four ways:

Center, spread, shape and unusual features.

Let's consider the center and spread.

Center: This is the median of the distribution.

Spread: This is the variation of the data set. If the range is wide, the spread is larger and If the range is small, the spread is smaller.

Rearranging the data set:

A = {30, 32, 32, 33, 40, 44, 45, 50}

B = {28, 30, 34, 35, 42, 43, 46, 48}

From the data:

The median for set A = (33+40)/2 = 73/2= 36.5

The median for set B = (35+42)/2 = 77/2= 38.5

Range = highest value - lowest value

The data ranges from 30 to 50 (range = 20) for A

The data ranges from 28 to 48 (range = 20) for B

Mean for set A = (30+32+32+33+40+4445+50)/8 = 306/8 = 38.25

Mean for set B = (28+30+34+35+42+43+46+48)/8= 306/8 = 38.25

In both data set the mean is equal to 38.25.

Therefore the statement that best compares the two data sets is the mean in set B is equal to the mean in set A (C)

8 0

3 years ago

HELP PLEASE ANYONE STEP BY STEP TIMED

UNO [17]

Answer:

$\huge\boxed{\text{B) 4}}$

Step-by-step explanation:

We can use the properties of some of these circles to solve and find the value of x.

FIrst off - we know that to find the side of a right triangle we can use the Pythagorean Theorem, which states that $a^2 + b^2 = c^2$ (a and b being legs, c being the hypotenuse.)

<h2><u>Finding the hypotenuse</u></h2>

We also know that this triangle is in a circle. Part of the hypotenuse and the altitude of the triangle are in the circle.

Point W is the center of the circle. Therefore, any points stretching to the edge from it will be equal (since this is a circle).

With this knowledge - UW and WZ are equal.

Therefore, since UW is $x-1$ , WZ will be too.

We also know the length of VZ. We can add this to WZ to find the length of the whole hypotenuse.

$\displaystyle x-1+2 \\ x+1$

So the length of the hypotenuse, WV, is $x+1$ .

<h2><u>Finding the value of x</u></h2>

Now that we know the length of the hypotenuse, we can use the Pythagorean Theorem like we stated before to find the value of x.

We know the two legs are $x-1$ and $2x-4$ , while the hypotenuse is

$\displaystyle (x-1)^2 + (2x-4)^2 = (x+1)^2$
$(x^2 - 2x + 1) + (4x^2-16x+16) = (x^2 + 2x + 1)$ (use FOIL to find the value of each expression squared)
$5x^2 - 18x + 17 = x^2 + 2x + 1$ (simplify the left side)
$4x^2 -20x + 16 = 0$ (subtract both sides by $x^2 +2x + 1$ )
$x = \frac{{ -b \pm \sqrt {b^2 - 4ac} }}{{2a}}$ (Quadratic formula)
$\frac{{ -(-20) \pm \sqrt {(-20)^2 - 4 \cdot 4 \cdot 16} }}{{2 \cdot 4}}$ (plug in values from equation)
$\frac{{ 20 \pm \sqrt {400 - 256} }}{8}}$ (simplify)
$\frac{{ 20 \pm \sqrt {144} }}{8}}$ (simplify)
$\frac{{ 20 \pm 12 }}{8}}$ (simplify)
$\displaystyle x = \frac{20+12}{8} \ \text{and} \ \frac{20-12}{8}$ (use plus/minus to find two roots)
$\displaystyle x = \frac{32}{8} \ \text{and} \ \frac{8}{8}$ (simplify)
$\displaystyle x = 4 \ \text{and} \ 1$