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sergeinik [125]
2 years ago
9

Carly placed 50 marbles in a bag. Fifteen of the marbles are pink, 8 marbles are black,

Mathematics
1 answer:
yuradex [85]2 years ago
4 0

When you go into this problem, you want to figure out your marble ammount to 50 so in this case we will say C for color and 50 for the total ammount of marbles.

We know 15 are pink, 8 are black, 2 are green, 18 are clear, and 7 are striped

15P/50

8B/50

2G/50

18C/50

7S/50 for a total of 50 marbles

Now we use the chart to decide our awnsers  

A. We know our propability of drawing a green and clear is 20/50 which if we simplify is a 2/5 ratio. If We put this in perspective 2/5 is rare and is unlikley to even.

B. We know a striped marble is 18/50 or 1.8/5 ratio which is mainly unlikely

C. We have 23/50 marbles that are black and pink, our propability is about 2.3/5 and gives us an even chance to get one of these

D. We know we have 33/50 marbles that are pink and clear and gives us a 3.3/5 chance of getting one of  these and gives us an even to likely chance of getting one of these.

E. If we have a total of 17 marbles in these 3 colors, we have a 1.7/5 chance of getting one of these and is probably impossible to unlikey.

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(b) The possible numbers are 8.45 to 8.49, inclusively

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In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc
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Answer:

Step-by-step explanation:

(a)

Consider the following:

A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°

Use sine rule,

\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}

Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
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Thus, the angle B is function of A is, B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Now find \frac{dB}{dA}

Differentiate implicitly the function \sin{B}=\sqrt{\frac{3}{2}}\sin{A} with respect to A to get,

\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}

And,

f'(A)=\frac{dB}{dA}=\sqrt{3} from part b

Therefore, the linear approximation at A=\frac{\pi}{4} is,

f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}

d)

Use part (c), when A=46°, B is approximately,

B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°

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Answer:

4/17

Step-by-step explanation:

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Thank you! This helped a lot!!! :)
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