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3 years ago

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A bag contains 5 red marbles , 4 blue marbles, 3 green marbles , and no other marbles. if a marble is chosen at random from the

bag, what is the probability that marble choosen will not be red?

1 answer:

OlgaM077 [116]3 years ago

6 0

That will be any color out of 12 that is the probaility.

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Beth had 1/4 box of candy. she ate 1/5 of the candy. If she decides to refill the box , what fraction of the box would need to b

Colt1911 [192]

19/20 of the box would need to be refilled.

6 0

3 years ago

Find the area of the surface. The surface with parametric equations x = u2, y = uv, z = 1 2 v2, 0 ≤ u ≤ 2, 0 ≤ v ≤ 1.

ehidna [41]

Answer:

Area = 32/3

Step-by-step explanation:

x = u ²

y = uv

z = 12v ²

0 ≤ u ≤ 2

0 ≤ v ≤ 1

Since ru = <2u, v, 0> and rv = <0, u, 24v>, we have

Where ru is the differentiation of x, y, z with respect to u and rv is the differentiation of x, y, z wit respect to v.

we find the cross product of ru and rv

ru × rv = 24v²i - 48uv²j + 2u²k

absolute value of ru × rv = 2u² + 24v²

We can now find the area

∫₀² du ∫₀¹ dv (2u² + 24v²) = ∫₀² du [2u²v + 8v³]₀¹ = 32/3

Detailed description can be found in the attachment

5 0

3 years ago

Read 2 more answers

A mad scientist develops a new memory pill and wishes to test it on a group of 11th grade students. He has a large number of sug

Arisa [49]

The adjective that best describes the experiment listed in detail above would be double-blind. This is because both the mad scientist and the manufacturer are both blind to each other's experiments.<span />

6 0

3 years ago

Read 2 more answers

Please prove this........

Crazy boy [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

4 years ago

Inverse Function In Exercise,analytically show that the functions are inverse functions.Then use the graphing utility to show th

Anna [14]

Step-by-step explanation:

We need to show whether

$f^{-1}(x) = g(x)$

or

$g^{-1}(x) = f(x)$

so we'll do either one of them,

we'll convert f(x) to f^-1(x) and lets see if it looks like g(x).

$f(x) = e^x - 1$

we can also write it as:

$y = e^x - 1$

now all we have to do is to make x the subject of the equation.

$y+1 = e^x$

$\ln{(y+1)} = x$

$x=\ln{(y+1)}$

now we'll interchange the variables

$y=\ln{(x+1)}$

this is the inverse of f(x)

$f^{-1}(x)=\ln{(x+1)}$

and it does equal to g(x)

$g(x)=\ln{(x+1)}$

Hence, both functions are inverse of each other!

This can be shown graphically too:

we can see that both functions are reflections of each other about the line y=x.

4 0

4 years ago