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gtnhenbr [62]
3 years ago
7

A study finds that there is a positive correlation between height and math test scores. The researcher concludes that taller peo

ple are better at math. However, another researcher points out that height is also correlated with age, and age is correlated with greater knowledge of math. What issue does this example represent?
Physics
1 answer:
Citrus2011 [14]3 years ago
8 0

Answer:

Spurious relationship

Explanation:

In statistics a spurious relationship is a correlated mathematical  relationship of covariation between two variables or more which by themselves have no cause and effect relationship as a result of a coincidence or the existence of a third or confounding factor

The variables have covariation due to their correlation but they don't have a cause and effect relationship

An example of a spurious relationship is the increase in ice cream sales when there is an observed increase in the rate of drowning.

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A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above
babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

h(t) = h_0 + ut - \frac{1}{2}gt^2 (1)

where

h_0 is the height at time t = 0

u is the initial vertical velocity

g=32 ft/s^2 is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

f(t)=-16t^2+44t+88 (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

h_0 = 88 ft

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

f(t)=0

which means when

0=-16t^2+44t+88

First of all, we can simplify the equation by dividing each term by 4:

0=-4t^2+11t+22

This is a second-order equation, so we solve it using the usual formula and we find:

t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

v=u-gt

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

v=0

which means

0=u-gt

Here we have

u=+44 ft/s (initial velocity)

g=32 ft/s^2 (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

t=\frac{u}{g}=\frac{44}{32}=1.38 s

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

f(t)=-16t^2+44t+88

Substituting t = 1.38 s, we find:

f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m

So, the maximum height reached by the rock during its motion is

h_{max}=118.2 m

Which means 118.2 m above the water.

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Explanation:

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