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2 years ago

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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Arisa [49]

(a) One possibility is 8.45

Rounded to the nearest integer = 8

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3 0

3 years ago

Read 2 more answers

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

Troyanec [42]

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,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

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,

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

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

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

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

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