To find the area of a trapezoid, multiply the sum of the bases (the parallel sides) by the height (the perpendicular distance between the bases), and then divide by 2.

Step-by-step explanation:

3 0

3 years ago

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 200 m above the groun

kirill115 [55]

Step-by-step answer:

Given:

mean, mu = 200 m

standard deviation, sigma = 30 m

sample size, N = 5

Maximum deviation for no damage, D = 100 m

Solution:

Z-score for maximum deviation

= (D-mu)/sigma

= (100-200)/30

= -10/3

From normal distribution tables, the probability of right tail with

Z= - 10/3

is 0.9995709, which represents the probability that the parachute will open at 100m or more.

Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.

P(all five safe) = 0.9995709^5 = 0.9978565

So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m

6 0

3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

3 0

2 years ago

Can someone please please help me. ?????

stira [4]

Answer:

x intercept is -5

y intercept is -5

y=8x

Step-by-step explanation:

6 0

2 years ago