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musickatia [10]

3 years ago

14

Counting the number of times a third-grade student leaves his seat without permission during a 30-minute observation period is a

n example of using what modality of measurement?

1 answer:

notka56 [123]3 years ago

6 0

Answer:

Behavioral modality of measurement

Step-by-step explanation:

Observing a 3rd grade student for 30 minutes as to how many time he leaves his seat without permission is a clear example of using behavioral modality of measurement.

In this modality of measurement the behavior of the object or system is observed when it is put into action.

This example shows as to what will happen or what is supposed to take place when a 3rd grade student is is seated and observed for 30 minutes. It counts as to how many times he leaves his seat in reaction to his stimulus from the environment.

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Answer the question in the pic below (math)

Marta_Voda [28]

Answer:

46.95 in.²

Step-by-step explanation:

Surface area of the kaleidoscope = surface area of a triangular prism = 2(BA) + P*H

Base area (B) = 2(½*bh) = 2(½*1.5*1.3) = 1.95 in.²

Perimeter of base (P) = 1.5 + 1.5 + 1.5 = 4.5 in.

Height of prism (H) = 10 in.

Plug in the values

Surface area = 1.95 + 4.5*10 = 46.95 in.²

7 0

3 years ago

Please I need help!!!

Natalija [7]

Answer:

58

Step-by-step explanation:

-14 + x = 44

x = 44 + 14

x = 58

3 0

2 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

2 years ago

marysya [2.9K]

12.5%. there is a 1 in 8 chance that all three times it will land on heads.

5 0

3 years ago

Read 2 more answers

show that the sum of the x- and y- intercepts of any tangent line to the curve sqrt(x) sqrt(y) = sqrt(c) is equal to c.

expeople1 [14]

The answer is <span>√x + √y = √c </span>
<span>=> 1/(2√x) + 1/(2√y) dy/dx = 0 </span>
<span>=> dy/dx = - √y/√x </span>

<span>Let (x', y') be any point on the curve </span>
<span>=> equation of the tangent at that point is </span>
<span>y - y' = - (√y'/√x') (x - x') </span>

<span>x-intercept of this tangent is obtained by plugging y = 0 </span>
<span>=> 0 - y' = - (√y'/√x') (x - x') </span>
<span>=> x = √(x'y') + x' </span>

<span>y-intercept of the tangent is obtained by plugging x = 0 </span>
<span>=> y - y' = - (√y'/√x') (0 - x') </span>
<span>=> y = y' + √(x'y') </span>

<span>Sum of the x and y intercepts </span>
<span>= √(x'y') + x' + y' + √(x'y') </span>
<span>= (√x' + √y')^2 </span>
<span>= (√c)^2 (because (x', y') is on the curve => √x' + √y' = √c) </span>
<span>= c. hope this helps :D</span>

5 0

3 years ago