Drag each tile to the correct cell in the table.

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luda_lava [24]

A = area of triangle + area of rectangular + half of the circle area

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triangle area = 1/2 × base × height

triangle area = 1/2 × 3 × 5 = 15/2 = 7.50

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rectangular area = length × width

rectangular area = 6 × 5 = 30

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circle half area = 1/2 × pi × radius^2

circle half area = 1/2 × 3.14 × 3^2

circle half area = 1.57 × 9

circle half area = 14.13

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A = 7.50 + 30 + 14.13

A = 51.63 m^2

7 0

3 years ago

The distance between flaws on a long cable is exponentially distributed with mean 12 m.

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.

(c) The median is 8.322.

(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.

4 0

3 years ago

Ron is tiling a countertop. He needs to place 54 square tiles on each of 8 rows to cover the counter. He wants to randomly place

Eduardwww [97]

22 white and 32 blue since 8 times 4 is 32 and 54 minus 32 is 22.

5 0

3 years ago

What is the coefficient of the term of degree 5 in the polynomial below

Olegator [25]

I think the answer is a)4.

5 0

3 years ago

Read 2 more answers

The equation of line CD is (y−3) = − 2 (x − 4). What is the slope of a line perpendicular to line CD?. . .

Allisa [31]

The given equation is in the point-slope form. The slope can be obtained by simple identification of the parts of the equation. This is shown below:

Given equation:

(y - 3) = -2 (x - 4)
y - y1 = m(x - x1)

Where slope, m = -2

The slope of a line perpendicular to the line CD is the negative reciprocal of the slope of the line CD. Therefore:

Slope = 1/2

4 0

3 years ago