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GarryVolchara [31]

3 years ago

9

Ticket Taker Tony collects 90 tickets in 5 minutes. Ticket Taker Tina collects 128 tickets in 8 minutes. Which ticket taker coll

ects more tickets per minute?

1 answer:

SpyIntel [72]3 years ago

6 0

Answer:

Ticket Taker Tony collects more tickets per minute.

Step-by-step explanation:

In order to find out which ticket taker collects more tickets per minute, you need to solve for how many tickets each Ticket Taker collects per minute to compare. Since 90:5 and 128:8 are different ratios, you need to find the unit rate, or how many tickets each collects per minute. To find unit rate, you divide the numerator by the denominator:

Ticket Taker Tony: $\frac{90 tickets}{5 minutes}$ = $\frac{18 tickets}{1 minute}$

Ticket Taker Tina: $\frac{128 tickets}{8 minutes}$ = $\frac{16 tickets}{1 minute}$

Every minute, Tony collects 18 tickets, while Tina only collects 16, so Ticket Taker Tony collects more.

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Emma uses a 250 meters roll of crepe paper to make steamers how many dekameters of crepe paper does Emma use

Ksivusya [100]

The dekameter (or decameter) first needs to be a translated into how many the meters. deka means 10, so a dekameter it must be 10 meters. So we have to divide 250 by 10, and then we will come up with 25 dekameters.
hope that helps!!

6 0

3 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of

steposvetlana [31]

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way

Step-by-step explanation:

From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26

A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace.

P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13

WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces?

P(AA) = (4/52)(3/51) = 1/221.

WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king?

P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed.

WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the

probability of drawing the first queen which is 4/52.

The probability of drawing the second queen is also 4/52 and the third is 4/52.

We multiply these three individual probabilities together to get P(QQQ) =

P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.

Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)

5 0

3 years ago

Which of the following rational functions is graphed below? A. F(x) = B. F(x) = C. F(x) = D. F(x) =

EastWind [94]

Answer:

The answer is D or A

Step-by-step explanation:

7 0

3 years ago

What is the domain of the function v = root(3, x)

vredina [299]

Answer:

The first answer. Negative infinity to positive infinity.

Step-by-step explanation:

As long as x is a real number, there will be a cube root

3 0

3 years ago

Read 2 more answers