Ask question

Ask question

All categories

3 years ago

8

The average distance between the variable scores and the mean in a set of data is the __________. A. range B. standard deviation

C. mean D. median

2 answers:

ololo11 [35]3 years ago

6 0

The average distance between the variable scores and the mean in a set of data is the standard deviation.

Yuri [45]3 years ago

5 0

Answer: B. standard deviation

Explanation:

In statistics , Standard deviation is a term which is used to to represent the measure of dispersion of data from the mean-value.

It is used to determine how closely the data values are with the mean of the entire data.

It is the average distance from each data value to the mean.

Hence, the average distance between the variable scores and the mean in a set of data is the <u>standard deviation</u>.

You might be interested in

car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

<em>Length of the road, L = 320 km</em>
<em>Distance covered = 240 km at 75 km/h</em>
<em>time spent refueling, t₂ = 0.6 hr</em>
<em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

$t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours$

The time spent by the car for the remaining journey;

$t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr$

The total time of the journey is calculated as follows;

$t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours$

The average velocity of the car for the whole journey is calculated as follows;

$v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h$

Learn more about average velocity here: brainly.com/question/6504879

6 0

3 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

Stephanie serves a volleyball from a height of 0.80 m and gives it an initial velocity of +7.2 m/s straight up. how high will th

Papessa [141]

<span>3.78 m Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes. 7.2 m/s / 9.81 m/s^2 = 0.77945 s The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2 d = 4.905 m/s^2 0.607542 s^2 d = 2.979995 m So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height. d = 2.979995 m + 0.8 m = 3.779995 m Rounding to 2 decimal places gives us 3.78 m</span>

7 0

3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago

A particle, of mass 6 kg, is in equilibrium on a rough horizontal plane under a force o-f magnitude T N, which acts at an angle

Helga [31]

Answer:

T is less than or equal to 19 N

Explanation:

3 0

3 years ago