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scZoUnD [109]
3 years ago
9

The space shuttle flight control system called PASS (Primary Avionics Software Set) uses four independent computers working in p

arallel. At each critical step, the computers "vote" to determine the appropriate step. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is 0.003. Let X denote the number of computers that vote for a left roll when a right roll is appropriate. What is the probability mass function of X
Mathematics
1 answer:
tigry1 [53]3 years ago
6 0

Answer:

f_X(k)=\binom{4}{k}0.003^k0.997^{4-k}

Step-by-step explanation:

Recall that a <em>probability mass function</em> defined on a discrete random variable X is just a function that gives the probability that the random variable equals a certain value k

f_X(k)=P(X=k)

In this case we have the event  

“The computer will ask for a roll to the left when a roll to the right is appropriate” with a probability of 0.003.

Then we have 2 possible events, either the computer is right or not.

Since we have 4 computers in parallel, the situation could be modeled with a binomial distribution and the probability mass function

f_X(k)=\binom{4}{k}0.003^k0.997^{4-k}

This gives the probability that k computers are wrong at the same time.

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Answer:

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(x - (-2))^2 + (y - (3))^2 = (4)^2

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This corresponds to answer choice C.

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Step-by-step explanation:

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If Sanjay wants to make 24 but the recipe makes 8, then do the recipe 3 times to get 24 (8 x 3 = 24)

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

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\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

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We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

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\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

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Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

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