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3 years ago

7

An archer is able to hit the bull's-eye 53% of the time. If she shoots 10 arrows, what is the probability that she gets exactly

4 bull's-eyes?

1 answer:

Vesnalui [34]3 years ago

6 0

Answer:

P(x=4)=.2458

hope it helps!

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Convert the polar equation r = 2 secØ to a Cartesian equation.y = 2x = 2x^2 = 2

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Polar and cartesian equation<h2>Initial explanation</h2>

Let's analyze the relation between r and x and y:

We have that between the indicated value of r (of the polar coordinates) and x and y (of the cartesian coordinates) there is a relation because they form a triangle. If r changes, then the value of x and y will change.

<h2>STEP 1: given equation</h2>

Using the given equation

r = 2 secØ

we have that

$\begin{gathered} r=2secØ \\ \downarrow \\ \frac{r}{2}=secØ \end{gathered}$ <h2>STEP 2: secØ equation</h2>

Observing the image of the initial explanation we have a right triangle, we know that the equation of

secØ for any right triangle is given by:

$\sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}$

In this case,

hypotenuse = r

adjacent side = x

then,

$\begin{gathered} \sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{r}{x} \\ \sec Ø=\frac{r}{x} \end{gathered}$ <h2>STEP 3: comparison between given equation and secØ equation</h2>

Then, we have that:

$\begin{gathered} \sec Ø=\frac{r}{x} \\ \sec Ø=\frac{r}{2} \end{gathered}$

This means that:

$\begin{gathered} \frac{r}{x}=\sec Ø=\frac{r}{2} \\ \downarrow \\ \frac{r}{x}=\frac{r}{2} \end{gathered}$

Then,

x = 2

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1 year ago

Element X decays radioactively with a half life of 12 minutes. If there are 160

Salsk061 [2.6K]

Answer:

It would take 75.8 minutes for the element to decay to 2 grams.

Step-by-step explanation:

The number of grams of element x, after t minutes, is given by the following equation:

$X(t) = X(0)(1-r)^{t}$

In which X(0) is the initial amount and r is the decay rate.

There are 160 grams of Element X

This means that X(0) = 160.

So

$X(t) = X(0)(1-r)^{t}$

$X(t) = 160(1-r)^{t}$

Half life of 12 minutes.

This means that X(12) = 0.5*X(0) = 0.5*160 = 80. So

$X(t) = 160(1-r)^{t}$

$80 = 160(1-r)^{12}$

$(1 - r)^{12} = 0.5$

$\sqrt[12]{(1 - r)^{12}} = \sqrt[12]{0.5}$

$1 - r = 0.9438$

So

$X(t) = 160(1-r)^{t}$

$X(t) = 160(0.9438)^{t}$

How long would it take for the element to decay to 2 grams?

This is t for which X(t) = 2. So

$X(t) = 160(0.9438)^{t}$

$2 = 160(0.9438)^{t}$

$(0.9438)^{t} = \frac{2}{160}$

$\log{(0.9438)^{t}} = \log{\frac{2}{160}}$

$t\log{0.9438} = \log{\frac{2}{160}}$

$t = \frac{\log{\frac{2}{160}}}{\log{0.9438}}$

$t = 75.8$

It would take 75.8 minutes for the element to decay to 2 grams.

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