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3 years ago

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Airline Fatalities One study showed that in a certain year, airline fatalities occur at the rate of 0.011 deaths per 100 million

miles. Find the probability that, during the next 100 million miles of flight, there will be

1 answer:

Lelu [443]3 years ago

8 0

Answer:

$P(X = 0) = 0.9891$

Step-by-step explanation:

Given

$\lambda = 0.011$

Required [This completes the question]

The probability of exactly 0 deaths

This probability follows a Poisson distribution, and it is given by:

$P(X = x) = \frac{e^{-\lambda}\lambda^{x}}{x!}$

For 0 deaths;

$x = 0$

So, the expression becomes

$P(X = 0) = \frac{e^{-\lambda}\lambda^{0}}{0!}$

$P(X = 0) = \frac{e^{-\lambda}\lambda^{0}}{1}$

$P(X = 0) = \frac{e^{-\lambda}*1}{1}$

$P(X = 0) = e^{-\lambda}$

Substitute 0.011 for $\lambda$

$P(X = 0) = e^{-0.011}$

$P(X = 0) = 0.9891$

<em>The probability of having exactly death is 0.9891</em>

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Estimate the quotient for 317÷42 a.3 b.6 c.7 d.8

lesya [120]

42×3=126
42×6=252
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Nearest is Product of 7
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3 years ago

1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is

sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by $d(t)=\frac{t^{2}}{6}+4t$ where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

$V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}$

so let's take the first interval:

(a) [1.95, 2.05]

$V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}$

we get that:

$d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125$

$d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167$

so:

$V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s$

(b) [1.995, 2.005]

$V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}$

we get that:

$d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831$

$d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835$

so:

$V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s$

(c) [1.9995, 2.0005]

$V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}$

we get that:

$d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358$

$d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358$

so:

$V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s$

(d) [2, 2.00001]

$V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}$

we get that:

$d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333$

$d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333$

so:

$V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s$

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

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3 years ago

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g100num [7]

x = 136/2187 = 0.062

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3^4*x-5-(1/27)=0

<span>Step by step solution :
</span><span>Step 1 :</span> 1 Simplify —— 27 <span>Equation at the end of step 1 :</span><span> 1 (((34) • x) - 5) - —— = 0 27 </span><span>Step 2 :</span>

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<span> 3.1 </span> Subtracting a fraction from a whole

Rewrite the whole as a fraction using <span> 27 </span> as the denominator :

81x - 5 (81x - 5) • 27 81x - 5 = ——————— = —————————————— 1 27

<span>Equivalent fraction : </span>The fraction thus generated looks different but has the same value as the whole

<span>Common denominator : </span>The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

<span> 3.2 </span> Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

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Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the <span>denominator, </span>Tiger multiplys both sides of the equation by the denominator.

Here's how:

2187x-136 ————————— • 27 = 0 • 27 27

Now, on the left hand side, the <span> 27 </span> cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
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Solving a Single Variable Equation :

<span> 4.2 </span>     Solve  :    2187x-136 = 0<span>

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One solution was found : <span> x = 136/2187 = 0.062</span>

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