Need help need help hellllllllllllppppppppppppp

Brenda studies frequently for her class in hopes of earning an A in the class. She is trying to keep her grade point average as

OleMash [197]

The correct answer is extrinsic motivation.

Extrinsic motivation refers to behavior that is driven by external rewards. In other words, the motivation to engage in a behavior arises from the outside sources because it is naturally satisfying to them. Practically, the behavior is driven by a reward given by someone or something else if the particular thing is done correctly. In this example Brenda is motivated to study more because she wants to earn an A, thus a reward for her effort of studying.

5 0

3 years ago

Read 2 more answers

How do effective parents discipline their children?

rodikova [14]

Answer:

Show and tell. Teach children right from wrong with calm words and actions. ...

Set limits.

Give consequences. .

Hear them out.

Give them your attention.

Catch them being good.

Know when not to respond.

Be prepared for trouble.

Explanation:

5 0

3 years ago

An ancient Greek physician believed performing lobotomies helped release inner spirits that caused mental illness. T or f

777dan777 [17]

Answer:

It is false on edge. Good luck with the test

5 0

3 years ago

Read 2 more answers

Use midpoints to approxiamte the area under the curve y=f(x) 5 sin(xx)!+ 2.5 cos(4xx) on the interval [0,1] using 10 equal subdi

alisha [4.7K]

Subdividing [0, 1] into 10 equally spaced intervals of length $\Delta x=\frac{1-0}{10}=\frac1{10}$ gives the partition

$[0,1] = \left[0,\dfrac1{10}\right] \cup \left[\dfrac1{10},\dfrac2{10}\right] \cup \cdots \cup \left[\dfrac9{10},1\right]$

The $i$ -th subinterval has left and right endpoints, respectively, given by

$\ell_i = \dfrac{i-1}{10} \text{ and } r_i = \dfrac i{10}$

where $i\in\{1,2,3,\ldots,10\}$ .

The midpoint of the $i$ -th interval is the average of these,

$m_i = \dfrac{\ell_i+r_i}2 = \dfrac{2i-1}{20} \in \left\{\dfrac1{20},\dfrac3{20},\dfrac5{20},\ldots,\dfrac{19}{20}\right\}$

We approximate the area under $f(x)$ over [0, 1] by the Riemann sum,

$\displaystyle \int_0^1 f(x) \, dx \approx \sum_{i=1}^{10} f(m_i) \Delta x \\\\ ~~~~~~~~ = \frac1{10} \sum_{i=1}^{10} \bigg(5\sin(\pi m_i) + 2.5 \cos(4\pi m_i)\bigg) \\\\ ~~~~~~~~ = \frac{\sin\left(\frac\pi{20}\right) + \sin\left(\frac{3\pi}{20}\right) + \cdots + \sin\left(\frac{19\pi}{20}\right)}2 \\\\ ~~~~~~~~~~~~ + \dfrac{\cos\left(\frac\pi5\right) + \cos\left(\frac{3\pi}5\right) + \cdots + \cos\left(\frac{19\pi}5\right)}4 \\\\ ~~~~~~~~ \approx 3.19623$

(D)

6 0

2 years ago

The average sale price (online) for four year old Harley Davidson touring motorcycles is approximately normally distributed with

grandymaker [24]

Answer:

A) Molly can buy 60% of the available motorcycles.

B) The 30th percentile is $11920

C) $25,000 is considered an outlier because is greater than: $24792

Explanation:

Data:

mean, μ = $14,000

standard deviation, σ = $4,000.

A) Using standard normal distribution table (see first figure attached) we have to compute Z (the standard variable) for an x = $15,000 (x is the actual variable) as follows:

Z = (x - μ)/σ

Z = (15000 - 14000)/4000

Z = 0.25

From the table, that represent 60%, that is, Molly can buy 60% of the available motorcycles.

B) From the second figure attached, we can see that 30% of the results (here, the motorcycles prices) correspond to a value of Z = -0.52. That correspond to a price of:

x = μ + Z*σ

x = 14000 + (-0.52)*4000

x = $11920

C) The 1.5 x IQR rule states that any value bigger than 1.5 times the interquartile range (IQR) added to the third quartile is an outlier.

The third quartile is located at Z = 0.6745, in terms of the actual variable:

x = μ + Z*σ

x = 14000 + 0.6745*4000

x = $16698

The first quartile is located at Z = -0.6745, in terms of the actual variable:

x = μ + Z*σ

x = 14000 - 0.6745*4000

x = $11302

Then:

IQR = 16698 - 11302 = $5396

The third quartile value added to 1.5 x IQR is equal to 16698 + 1.5*5396 = $24792

6 0

3 years ago