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3 years ago

Find the three geometric means between 15 and 1215.

Mathematics

1 answer:

irina [24]3 years ago

6 0

Hello,

g_m(15,1215)=√(5*1215)=9√15≈77,942286340....
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15. A rectangle has a length of 2x - 1 and a width of 3x + 7. Find the perimeter of this rectangle.

andrew11 [14]

Answer:

10x + 12 and 6x² + 11x - 7

Step-by-step explanation:

The perimeter (P) of a rectangle is calculated as

P = 2l + 2w ( l is length and w is width )

= 2(2x - 1) + 2(3x + 7) ← distribute parenthesis

= 4x - 2 + 6x + 14 ← collect like terms

= 10x + 12

--------------------------------------------------

The area (A) of a rectangle is calculated as

A = lw

= (2x - 1)(3x + 7) ← expand using FOIL

= 6x² + 14x - 3x - 7 ← collect like terms

= 6x² + 11x - 7

7 0

3 years ago

Can somebody help me on this problem please!!

harkovskaia [24]

It is Rong the answer is 2

8 0

3 years ago

Read 2 more answers

Three quarters of the batch of twenty cookies burned when you forgot to take them out of the oven how many cookies burned

Natali5045456 [20]

15 cookies burned because if you Turn $\frac{3}{4}$ into a decimal then multiply it by 20 you get 15 and thats your answer.

7 0

2 years ago

G verify that the divergence theorem is true for the vector field f on the region

Alenkasestr [34]

$\mathbf f(x,y,z)=\langle z,y,x\rangle\implies\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial x}{\partial z}=0+1+0=1$

Converting to spherical coordinates, we have

$\displaystyle\iiint_E\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=6}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=288\pi$

On the other hand, we can parameterize the boundary of $E$ by

$\mathbf s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\pi$ . Now, consider the surface element

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\dfrac{\mathbf s_v\times\mathbf s_u}{\|\mathbf s_v\times\mathbf s_u\|}\|\mathbf s_v\times\mathbf s_u\|\,\mathrm du\,\mathrm dv$
$\mathrm d\mathbf S=\mathbf s_v\times\mathbf s_u\,\mathrm du\,\mathrm dv$
$\mathrm d\mathbf S=36\langle\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v\rangle\,\mathrm du\,\mathrm dv$

So we have the surface integral - which the divergence theorem says the above triple integral is equal to -

$\displaystyle\iint_{\partial E}\mathbf f\cdot\mathrm d\mathbf S=36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv$
$=\displaystyle36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}(12\cos u\cos v\sin^2v+6\sin^2u\sin^3v)\,\mathrm du\,\mathrm dv=288\pi$

as required.

3 0

3 years ago

(a) Let {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it app

Nataliya [291]

Answer:

0.625

Step-by-step explanation:

Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.

Since A1 and A2 are mutually exclusive and exhaustive, we can say

b) P(B) = P(A1B)+P(A2B)

Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads. $P(B/A1) = 0.50 \\P(B/A2) = 0.75\\P(A1B) = 0.5(0.5) = 0.25 \\P(A2B) = 0.75(0.5) = 0.375\\P(B) = 0.625$

c) Using Bayes theorem

conditional probability that it wasthe biased coin

= $\frac{0.375}{0.625} \\=\frac{3}{5}$

d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin= $\frac{0.25*0.5}{0.25*0.5+0.5*0.5} \\=\frac{1}{3}$

7 0

3 years ago