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2 years ago

If you toss three coins, what are the odds in favor of getting exactly two tails and one head?

Mathematics

1 answer:

Leni [432]2 years ago

3 0

$\displaystyle
|\Omega|=2^3=8\\
|A|=3\\\\
P(A)=\dfrac{3}{8}\Rightarrow \text{A}$

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Can someone help plssss which of the equations below could be the equation of this parabola?

RideAnS [48]

Answer:

Step-by-step explanation:

Both of the y= make upward parabolas

It is also going into the positive direction so it is $x = \frac{1}{4}y^2$

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2 years ago

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Could someone please help me

slava [35]

78
because 5 times 2 equals 10 so then u times 39 times 2 and get 78

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3 years ago

Convert 4 months to fraction form assuming 360-day year

PIT_PIT [208]

I converted the four months into a fraction, and got 1/3. There's 12 months in a year, and divided by 4, I got 3.

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3 years ago

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What is -4d 3 + 28d 2 - 4d

Svetlanka [38]

Answer: 20d + 5

Step-by-step explanation:

First we need to combine like terms.

In this equation we can see that they're are 3 numbers containing the letter D we have -4, -4, 28.

We're going to combine -4 and -4 to get -8 and then combine -8 to 28 getting 20d

Afterwards we're going to combine 3, 2 equaling 5.

7 0

3 years ago

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

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3 years ago