Empirical probability is the ratio of the number of outcomes with the actual number of trials. This means that it relies on experimental data and not on the theoretical probability.

The only options that makes use of actual data or the trials done is
A. Darren threw a football to Ted 15 times, and Ted caught it 13 times. The probability Ted will catch a throw is .

7 0

3 years ago

GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial vel

zhuklara [117]

<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>

<em><u>Solution:</u></em>

Given that,

<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

$h(t) = -16t^2 + 60t$

<em><u>At what times will the object be at a height of 56 feet</u></em>

<em><u>Substitute h = 56</u></em>

$56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0$

Solve the above equation by quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75$

Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

5 0

4 years ago

Check all that apply

Pavlova-9 [17]

Get her done ...........................

5 0

3 years ago