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mario62 [17]
1 year ago
6

Alladin went grocery shopping for candy at Costco in Rego Park. He wants to buy the package of candy with the greatest percentag

e of Hershey's Cookie Crème Candy Bar. Package A has 13 Cookie Crème Candy out of a total of 25. Package B has 11 Cookie Crème Candy out of a total of 20. Which package should Alladin buy?
Mathematics
1 answer:
Pavlova-9 [17]1 year ago
3 0

We need to find what percentage represents the number of Hershey's Cookie Crème Candy Bars in each package, and then compare which one is better.

For package A.

It has 13 Hershey's Cookie Crème Candy Bars out of 25. To find which percentage is 13 out of 25 we do the following:

• Divide 100 by the total amount of candy 25, and multiply the result by 13:

\frac{100}{25}\times13

we get the following percentage:

\frac{100}{25}\times13=4\times13=52

Package A has 52% of Hershey's Cookie Crème Candy Bars.

For package B.

It has 11 Hershey's Cookie Crème Candy Bars out of 20. we do the same as before to find the percentage, only that this time, instead of dividing by 25, we divide 100 by 20, and intead of multiplying by 12, we multiply by 11:

\frac{100}{20}\times11

The result is:

\frac{100}{20}\times11=5\times11=55

Package B has 55% of Hershey's Cookie Crème Candy Bars.

Which package should Alladin buy? He should buy Package B because it has a greater percentage.

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<h3>Mean or Average</h3>

Mean is defined as the sum of data to the total number of samples.

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Using the formula:

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Hence the ratio that represents Kyle's batting average at the end of the season is h:10

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Answer:

\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}

Let a=\sin^{-1}(\frac{2}{3}).

With some restriction on a this means:

\sin(a)=\frac{2}{3}

We need to find \tan(a).

\sin^2(a)+\cos^2(a)=1 is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

(\frac{2}{3})^2+\cos^2(a)=1

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Subtract 4/9 on both sides:

\cos^2(a)=\frac{5}{9}

Take the square root of both sides:

\cos(a)=\pm \sqrt{\frac{5}{9}}

\cos(a)=\pm \frac{\sqrt{5}}{3}

The cosine value is positive because a is a number between -\frac{\pi}{2} and \frac{\pi}{2} because that is the restriction on sine inverse.

So we have \cos(a)=\frac{\sqrt{5}}{3}.

This means that \tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}.

Multiplying numerator and denominator by 3 gives us:

\tan(a)=\frac{2}{\sqrt{5}}

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

\tan(a)=\frac{2\sqrt{5}}{5}

Let's continue on to letting b=\cos^{-1}(\frac{1}{7}).

Let's go ahead and say what the restrictions on b are.

b is a number in between 0 and \pi.

So anyways b=\cos^{-1}(\frac{1}{7}) implies \cos(b)=\frac{1}{7}.

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of b.

\cos^2(b)+\sin^2(b)=1

(\frac{1}{7})^2+\sin^2(b)=1

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Subtract 1/49 on both sides:

\sin^2(b)=\frac{48}{49}

Take the square root of both sides:

\sin(b)=\pm \sqrt{\frac{48}{49}

\sin(b)=\pm \frac{\sqrt{48}}{7}

\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}

\sin(b)=\pm \frac{4\sqrt{3}}{7}

So since b is a number between 0 and \pi, then sine of this value is positive.

This implies:

\sin(b)=\frac{4\sqrt{3}}{7}

So \tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}.

Multiplying both top and bottom by 7 gives:

\frac{4\sqrt{3}}{1}= 4\sqrt{3}.

Let's put everything back into the first mentioned identity.

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}

\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}

\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}

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