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3 years ago

Answer #3 for 30 points FAST

Mathematics

2 answers:

meriva3 years ago

6 0

Answer:

first answer is (3, -8); second answer is (3, -5)

Step-by-step explanation:

Verizon [17]3 years ago

3 0

For substitution: -2x -2y = 10 y = 4x - 20
6x + 8y = -22 y = -5 (for both plug in the second equation as y in the first equation if that makes sense) (first answer is (3, -8); second answer is (3, -5)

for elimination: 3x + 4y = 2.5 5x - 4y = 25.5 (adding will get you to 8x = 28 bc it gives you 0y; answer is (7/2, -2))
3x + 4y = 2.84 3x - y = 1.79 (gives you 3y = 1.05 if you subtract; answer is (0.48, 0.35))

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Elden [556K]

Answer:

(a) The probability that the distance between two flaws is greater than 15 m is 0.2865.

(b) The probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(d) The standard deviation is 12.

(e) The 65th percentile of the distances is 12.61 m.

Step-by-step explanation:

The random variable X can be defined as the distance between flaws on a long cable.

The random variable X is exponentially distributed with mean, μ = 12 m.

The parameter of the exponential distribution is:

$\lambda=\frac{1}{\mu}=\frac{1}{12}=0.0833$

The probability density function of X is:

$f_{X}(x)=0.0833e^{-0.0833x};\ x\geq 0$

(a)

Compute the probability that the distance between two flaws is greater than 15 m as follows:

$P(X\geq15)=\int\limits^{\infty}_{15}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{\infty}_{15}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{\infty}_{15}\\=e^{0.0833\times 15}\\=0.2865$

Thus, the probability that the distance between two flaws is greater than 15 m is 0.2865.

(b)

Compute the probability that the distance between two flaws is between 8 and 20 m as follows:

$P(8\leq X\leq20)=\int\limits^{20}_{8}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{20}_{8}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{20}_{8}\\=e^{0.0833\times 8}-e^{0.0833\times 20}\\=0.51355-0.1890\\=0.32455\\\approx0.3246$

Thus, the probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(c)

The median of an Exponential distribution is given by:

$Median=\frac{\ln (2)}{\lambda}$

Compute the median as follows:

$Median=\frac{\ln (2)}{\lambda}$

$=\farc{0.69315}{0.08333}\\=8.322$

Thus, the median is 8.322.

(d)

The standard deviation of an Exponential distribution is given by:

$\sigma=\sqrt{\frac{1}{\lambda^{2}}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{\frac{1}{\lambda^{2}}}$

$=\sqrt{\frac{1}{0.0833^{2}}}\\=12.0048\\\approx 12$

Thus, the standard deviation is 12.

(e)

Let x be 65th percentile of the distances.

Then, P (X < x) = 0.65.

Compute the value of x as follows:

$\int\limits^{x}_{0}{0.0833e^{-0.0833x}}\, dx=0.65\\0.0833\times \int\limits^{x}_{0}{e^{-0.0833x}}\, dx=0.65\\0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{x}_{0}=0.65\\-e^{-0.0833x}+1=0.65\\-e^{-0.0833x}=-0.35\\-0.0833x=-1.05\\x=12.61$

Thus, the 65th percentile of the distances is 12.61 m.

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