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klemol [59]
3 years ago
12

The researcher trains a sample of 16 individuals in mindfulness and then collects data on how long they spend eating per day. Th

e sample mean for those trained in mindfulness equals 74; the SS = 150. What is the calculated t-value? Round your answer to two decimal places.
Mathematics
1 answer:
Mila [183]3 years ago
4 0

Answer:

The calculated t-value is t=0.47.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Americans typically eat quickly and do not spend time enjoying their meal. A researcher plans to test if those trained in mindfulness will take more time to enjoy their meal (that is, they will spend more time eating than the general population). The general population spends an average of 68 minutes per day combined eating their three meals. The researcher trains a sample of 16 individuals in mindfulness and then collects data on how long they spend eating per day. The sample mean for those trained in mindfulness equals 74; the SS = 150. What is the calculated t-value? Round your answer to two decimal places."</em>

The degrees of freedom are (n-1)=(16-1)=15

The standard deviation is:

s=\sqrt{\frac{SS}{n-1}}=\sqrt{\frac{150}{16-1}}=\sqrt{10}=3.16

The t-value can be calculated as:

t=\frac{M-\mu}{\sigma/\sqrt{n}} =\frac{74-68}{3.16/\sqrt{16}} \\\\t=\frac{6}{3.16*4}=\frac{6}{12.64}= 0.47

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Work out m and c for the line: y=3x
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Step-by-step explanation:

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A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

  • (a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

  • (b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

  • (c) The rate at which the angle formed with the wall is changing is approximately <u>0.286 rad/sec</u>.

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, \displaystyle \frac{dx}{dt} = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}

25² = x² + y²

y = √(25² - x²)

\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}

Which gives;

\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times \frac{dx}{dt} =  \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times2

\displaystyle \frac{dy}{dt} =  \mathbf{ \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times2}

When x = 15, we get;

\displaystyle \frac{dy}{dt} =   \frac{15 \times \sqrt{625-15^2}  }{15^2- 625}\times2 = \mathbf{-1.5}

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

\displaystyle \frac{dy}{dt} =   \frac{20 \times \sqrt{625-20^2}  }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6

The velocity of the top of the ladder = \underline{-2.\overline{6} \ m/s \ downwards}

When x = 24, we get;

\displaystyle \frac{dy}{dt} =   \frac{24 \times \sqrt{625-24^2}  }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}}  \approx -6.86

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) \displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}

Therefore;

\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}

\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}

\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}

Therefore;

\displaystyle \frac{dA}{dt} =  \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2

When the ladder is 24 feet from the wall, we have;

x = 24

\displaystyle \frac{dA}{dt} =  \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}

The rate the area formed by the ladder is changing, \displaystyle \frac{dA}{dt} ≈ <u>-75.29 ft.²/sec</u>

(c) From trigonometric ratios, we have;

\displaystyle sin(\theta) = \frac{x}{25}

\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}

\displaystyle \frac{d \theta}{dt}  = \frac{d \theta}{dx} \times \frac{dx}{dt}

\displaystyle\frac{d \theta}{dx}  = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}

Which gives;

\displaystyle \frac{d \theta}{dt}  =  -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}

When x = 24 feet, we have;

\displaystyle \frac{d \theta}{dt} =  -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is \displaystyle \frac{d \theta}{dt} ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

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