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nignag [31]
3 years ago
11

30 random samples of high school students were asked if they played a sport for their high school. Each sample was 20 students.

The dot plot represents the 30 random sample proportions.
Choose ALL that are correct.


A) A sample proportion of 0.20 means that 4 of 20 high school students responded that they play a sport.

B) Based on the sample distribution, the best prediction for the proportion of all high school students that play a sport is 0.10.

C) The mean of the 30 sample proportions calculated to the nearest hundredth is 0.16.

D) The shape of the sample distribution is fairly symmetrical.

Mathematics
1 answer:
stiv31 [10]3 years ago
5 0

Answer: a. This option is correct

b. This will be true only if we take the mode as the best representative value

c. This is a not correct choice

d. This option is correct

Step-by-step explanation:  Took the test on USATest Prep

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(5+4) +10=5+(4+10)<br> A. True<br> B. False
Studentka2010 [4]

Answer:

A. True

Step-by-step explanation:

One of the ways to check this is solving each side of the equation.

(5+4)+10=5+(4+10)

Remember PEMDAS.

Parenthesis first.

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2 years ago
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A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha
Arlecino [84]

Answer:

A) F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }with a = \frac{2\Delta x}{t^2}

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha) (1)

where \alpha is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration a that can be calculated as a function of the given displacement and time interval as:

a = \frac{2\Delta x}{t^2} and the force equation is:

\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma

now substituting N from equation (1):

F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}

and putting the expression for the acceleration:

\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}} (2)

which is the requested expression

B) Substituting the values in (2) and using g=9.8\ \rm{ms^{-2}}:

F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}

The gravitational force acting on the box is:

F_g=mg=50\cdot9.8=490\ \rm{N}

F/F_g = (117.634/490)\cdot 100 = 24 \%

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.

8 0
4 years ago
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