All categories

2 years ago

The most commonly used descriptive statistics for a set of numerical scores are the mean and the ____.

Mathematics

1 answer:

nasty-shy [4]2 years ago

8 0

Answer: Standard Deviation.

Step-by-step explanation:

DescriptionIn insights, the standard deviation is a proportion of the measure of variety or scattering of a lot of qualities. A low standard deviation demonstrates that the qualities will in general be near the mean of the set, while an elevated expectation deviation shows that the qualities are spread out over a more extensive territory.

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Can you help me with these questions please?

Marianna [84]

The answers are following

sol1 y = x-1

sol2 y= x+ 7

sol4 y = x+4

Step-by-step explanation:

Solution 1

Here we are going put the point in the given equation ( given in the options)

If the each point satisfies any equation, that will be the relationship between x and y.

Given points are (-3,-4) , (2,1) , (5,4)

Let us take the option C as y = x -1

-4 = -3-1

1 = 2-1

4=5-1

Putting each value in the given equation we get LHS and RHS are same for all points. so, C is the answer.

Similarly, option B is the answer of Solution 2

Each point satisfies the equation y = x +7

Solution 4

y = x +4

You are requested to check the options of Q3 once.

5 0

3 years ago

Find the LCM (Least Common Factor) of 12 and 27 using prime factorizations

Aneli [31]

The LCM (least common factor) of 12 and 27 is the number: 3

5 0

3 years ago

Assignment The operation teeter is defined on the set T= { 2, 3, 5, 7 } by x teeter y=[x+y+xy) mod.8 a - Construct mod.8 table f

rosijanka [135]

Answer:

Case A) tau_net = -243.36 N m, case B) tau_net = 783.36 N / m, tau_net = -63.36 N m, case C) tau _net = - 963.36 N m,

Explanation:

For this exercise we use Newton's relation for rotation

Σ τ = I α

In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N

we will assume that the counterclockwise turns are positive.

case a

tau_net = m g x - F x2

tau_nett = -28.8 9.8 1.5 + 180 1

tau_net = -243.36 N m

in this case the man's force is downward and the system rotates clockwise

case b

2 force clockwise, the direction of

the force is up

tau_nett = -28.8 9.8 1.5 - 180 2

tau_net = 783.36 N / m

in case the force is applied upwards

3) counterclockwise

tau_nett = -28.8 9.8 1.5 + 180 2

tau_net = -63.36 N m

system rotates clockwise

case c

2 schedule

tau_nett = -28.8 9.8 1.5 - 180 3

tau _net = - 963.36 N m

3 counterclockwise

tau_nett = -28.8 9.8 1.5 + 180 3

tau_net = 116.64 Nm

the sitam rotated counterclockwise

4 0

1 year ago

A building that is 100 ft tall casts a shadow that makes a 30 degree angle with the ground. Approximately how long, in feet, is

Naddik [55]

The awnser is 130 ft.

8 0

3 years ago

Read 2 more answers

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

3 years ago