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3 years ago

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A newspaper conducted a survey to find out how many high school students play video games.The two ways table below displays the

data from the survey. Based on these data in the table, which statement is true? A) There were 2,451 boys surveyed, and about 29% of them play video games. B) There were 2,996 girls surveyed, and about 45% of them play video games. C) There were 5,447 students surveyed, and about 54% of them do not play video games. D) There were 2,493 students surveyed, and about 34% of them are girls who do not play veideo games.

1 answer:

zepelin [54]3 years ago

7 0

Answer:

Step-by-step explanation:

Given that

We are given the following:

1. Boys who do play video games = 1593

2. Boys who do not play video games = 858

3. Total boys = 2451

4. Girls who do play video games = 1361

5. Boys who do not play video games = 1635

6. Total Girls = 2996

7. Total number of candidates = Total boys + Total girls

=(2451 + 2996) = 5447

8. Total number of students, who play video games = 2954

9. Total number of students, who do not play video games = 2493

Percentage of boys who play video games:

$\dfrac{1593}{2451} \times 100 = 64.99\%$

So, option A is not correct.

Percentage of girls who play video games:

$\dfrac{1361}{2996} \times 100 = 45.43\%$

so option B is correct. The percentage is near about 45%,

Percentage of Total number of students, who do not play video games:

$\dfrac{2493}{5447} \times 100 = 45.77\%$

So, option C is not correct.

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<u>The problem is as following:</u>

The graph shows the relationship between the total cost and the number of gift cards that Raj bought for raffle prizes. What would be the cost for 5 of the gift cards? $80, $90, $100 Or $110

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<u>Solution</u>:

The points of the figure represents a straight line

The general form of the straight line is ⇒ y = mx + c

where m is the slope and c is constant

let x ⇒ The number of gifts cards , y ⇒ Total cost (in $)

we will find m and c using the points (1,20) , (2,40)

The slope = m = $\frac{y_{2} - y_{1}}{x_{2}-x_{1}} =\frac{40-20}{2-1} =\frac{20}{1}=20$

∴ y = 20x + c

Substitute with the point (1,20) ⇒ y = 20 at x = 1

∴ 20 = 20 * 1 + c

∴ 20 = 20 + c

∴ c = 0

∴ The equation of the line ⇒⇒⇒ y = 20x

To find the cost for 5 of the gift cards substitute with x = 5

∴ y = 20 * 5 = 100

So, the cost for 5 of the gift cards = $100

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first find the slope of the line using the two points

$slope \: m = \frac{y2 - y1}{x2 - x1} = \frac{23 - 5}{6 - ( - 3)} \\ = \frac{18}{9} \\ slope \: m = 2$

substitute 2 for m and use one of the points for x and y and solve for b

$5 = 2( - 3) + b \\ 5 = - 6 + b \\ 5 + 6 = b \\ b = 11$

Now that you found b, the y-intercept, you can substitute that into the equation for your line

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How do you find the limit?

coldgirl [10]

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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3 years ago