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Natali5045456 [20]

3 years ago

6

There is a spinner with 15 equal areas, numbered 1 through 15. If the spinner is spun one time, what is the probability that the

result is a multiple of 5 and a multiple of 2?

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

6.67% probability that the result is a multiple of 5 and a multiple of 2

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

Multiples of 2 AND 5

Between 1 and 15, the multiples of 2 are: 2,4,6,8,10,12,14

Between 1 and 15, the multiples of 5 are: 5,10,15

So only 10 is a multiply of both 2 and 5, so only one desired outcome, which means that $D = 1$

Total outcomes:

Any number between 1 and 15, there are 15, so $T = 15$

Probability:

$p = \frac{D}{T} = \frac{1}{15} = 0.0667$

6.67% probability that the result is a multiple of 5 and a multiple of 2

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3 years ago

How do you find the limit?

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Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

<u>Differentiation</u> (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

<u>Differential</u> (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

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Answer:

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The area of a trapezoid is 243cm2. The height is 18cm and the length of one of the parallel sides is 10cm. Find the length of th

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Answer:

$Side_2 = 17\ cm$

Step-by-step explanation:

Given

Shape: Trapezoid

$Area = 243cm^2$

$Height = 18cm$

$Side_1 = 10cm$

Required

Determine the length of the second parallel side

The area of a trapezoid is:

$Area = \frac{1}{2}(Side_1 + Side_2) * Height$

Substitute values for Area, Height and Side1

$243 = \frac{1}{2}(10 + Side_2) * 18$

Multiply both sides by 2

$2 * 243 = 2 * \frac{1}{2}(10 + Side_2) * 18$

$486 = (10 + Side_2) * 18$

Divide both sides by 18

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$Side_2 = 17\ cm$

Hence;

<em>The length of the second parallel side is 17cm</em>

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