Answer:
Figure attached
We can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.
Explanation:
We can use the following R code to obtain the data for wdiff:
source("http://www.openintro.org/stat/data/cdc.R") #obtain the info
nrow(cdc) # number of elements
names(cdc) # obtain the name for the variable
[1] "genhlth" "exerany" "hlthplan" "smoke100" "height" "weight" "wtdesire" "age"
[9] "gender"
wdiff represent the difference between desired weight (wtdesire) and current weight (weight) and we can obtain the data with the following code:
wdiff <- (cdc$weight-cdc$wtdesire)
And now we can create the histogram with this code
hist(wdiff,xlim =c(-100,150))
> mean(wdiff)
[1] 14.5891
> median(wdiff)
[1] 10
And the result is on the figure attached.
And we can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.
Answer:
option B
Explanation:
given,
Force exerted by the hydraulic jack piston = F₁ = 250 N
diameter of piston, d₁ = 0.02 m
r₁ = 0.01 m
diameter of second piston, d₂ = 0.15 m
r₂ = 0.075 m
mass of the jack to lift = ?
now,
F₂ = 14062.5 N
F = m g
m = 1435 Kg
hence, the correct answer is option B
Answer:
4 (please see the attached file)
Explanation:
While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.
So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.
When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.
This acceleration is related with the tangential acceleration, by this expression:
at = α*r
This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
Answer: It would increase.
Explanation:
The equation for determining the force of the gravitational pull between any two objects is:
Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.
Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.
Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.