Need help with this triangle problem

Solve to x to the nearest foot

Arisa [49]

Answer:

x= 1325 ft (nearest foot)

Step-by-step explanation:

Please see the attached pictures for the full solution.

3 0

2 years ago

A random sample is made of 256 middle managers concerning their annual incomes. The sample mean is computed to be $35,420. If th

ziro4ka [17]

Answer: a. $128.125

Step-by-step explanation:

The standard error of the mean is given by :

$SE=\dfrac{\sigma}{\sqrt{n}}$ , where $\sigma$ = population standard deviation , n= sample size.

Given: $\sigma=\$2,050$

n= 256

Then, the standard error of the mean:-

$SE=\dfrac{2050}{\sqrt{256}}\\\\=\dfrac{2050}{16}=\$128.125$

Hence, option a. is correct.

3 0

3 years ago

Which of the following would have the same graphic representation as the function f(x) = 125 (5)x ? Select all that apply.

Sveta_85 [38]

F(x) = 34x is the correct answer

4 0

2 years ago

Read 2 more answers

Suppose that hot dogs come in packs of 9 and buns come in packs of 7. Jim bought the same number of hot dogs as buns. It was the

Ilia_Sergeevich [38]

List the multiples of each qty:

7, 14, 21, 28,35,42,49,56,63

9, 18, 27, 36, 45, 54, 63

The least common multiple is 63

He bought 63 hotdogs and buns

3 0

3 years ago

Read 2 more answers

Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).

Elan Coil [88]

Answer: $\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV$ = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

$a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$

Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

Then, triple integral is:

$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

Calculating:

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

$\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx$

$\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx$

$\frac{1}{45} \int\limits^5_0 {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4} } \, dx$

$\frac{1}{45} [30375-60750+118462.5-39150]$

$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$ = 1087.5

The volume of the tetrahedon is 1087.5 cubic units.

3 0

3 years ago