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3 years ago

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An experiment has four equally likely outcomes: E1, E2, Es, and E.Round your answers to three decimal places. a. What is the pro

bability that E occurs? .25 b. What is the probability that any two of the outcomes occur (e.g., E1 or Es)? 375 c. What is the probability that any three of the outcomes occur (e.g., E1 or E2 or E4)?

1 answer:

V125BC [204]3 years ago

6 0

Answer:

a. 0.25

b. 0.50

c. 0.75

Step-by-step explanation:

P(E1) = 0.25

P(E2) = 0.25

P(Es) = 0.25

P(E) = 0.25

a. Since there are 4 outcomes with the same probability of occurring, the probability of E occurring is:

$P(E) = \frac{1}{4}=0.25$

b. The probability of any two outcomes occurring is twice the probability of a single outcome occurring

$P(two\ outcomes) = 2*P(one\ outcome) = \frac{2}{4}=0.50$

c. The probability of any three outcomes occurring can be written as 100% minus the probability of the remaining outcome occurring

$P(three\ outcomes) = 1-P(one\ outcome) = 1-\frac{1}{4}=0.75$

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Step-by-step explanation:

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2 years ago

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2 years ago

Read 2 more answers

You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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