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3 years ago

A coin is flipped 4 times. what is the probability of the coin landing on tails all 4 times

Mathematics

2 answers:

DaniilM [7]3 years ago

6 0

Half the chance of landing tail once(1/2)

Half the chance of land tails four times (1/2 of 1/2 of 1/2 of 1/2)

1/16

tankabanditka [31]3 years ago

3 0

the probablity of the coin landing on tails all four times is

1/4

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harkovskaia [24]

Step-by-step explanation:

i think the answer is 10600

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2 years ago

Write the linear equation from the given information in slope intercept form (-1,3); slope=-3

nadya68 [22]

Answer:

Step-by-step explanation:

The standard expression of equation of a line in slope-intercept form is expressed as;

y = mx+c

m is the slope

c is the intercept

Given

slope m = -3

Point (x, y) = (-1, 3)

x = -1 and y = 3

Get the intercept

To get the intercept c, we will substitute the given values into the equation above to have;

y = mx+c

3 = -3(-1)+c

3 = 3 + c

c = 3-3

c = 0

Substitute m = -3 and c = 0 into the equation above;

y = mx+c

y = -3x+0

y = -3x

Hence the required linear equation is y = -3x

4 0

2 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$