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Marrrta [24]
2 years ago
8

Lionel is planning a one-day outing.

Mathematics
2 answers:
pychu [463]2 years ago
6 0

Answer:

Option B - y=5x+40 and  y=3x+60

Step-by-step explanation:

Given : The Thrill amusement park charges an entry fee of $40 and an additional $5 per ride, x. The Splash water park charges an entry fee of $60 and an additional $3 per ride, x.

To find : Which system of equations could be used to determine the solution where the cost per ride of the two amusement parks, y, is the same?

Solution :

Let x be the number of rides and

y be the cost per ride.

According to question,

The Thrill amusement park charges an entry fee of $40 and an additional $5 per ride.

The equation form is y=40+5x

The Splash water park charges an entry fee of $60 and an additional $3 per ride.

The equation form is y=60+3x

Therefore, The required system of equations form are

y=5x+40 and  y=3x+60

So,Option B is correct.

Lelechka [254]2 years ago
4 0

Answer:

y= 40+5x

y= 60+3x

Step-by-step explanation:

Thrill amusement park

Entry fee = $40

Cost of 1 ride = $5

Let x be teh no. of rides

Cost of x rides = 5x

So, Total cost = 40+5x

Let the total cost be y

So,Total cost =y= 40+5x

Splash water park

Entry fee = $60

Cost of 1 ride = $3

Let x be the no. of rides

Cost of x rides = 3x

So, Total cost = 60+3x

Let the total cost be y

So,Total cost =y= 60+3x

So, system of equations could be used to determine the solution where the cost per ride of the two amusement parks, y, is the same is :

y= 40+5x

y= 60+3x

Hence Option B is true .

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A jar contains sugar. The jar and the sugar have a total weight of 850 g. Anna uses 2/3 of the sugar. The jar and the sugar now
maria [59]

Answer:

J = 370g

Step-by-step explanation:

Given

Represent the weight of the Jar with J and the sugar with S

Initially, we have:

J + S = 850g

After 2/3 of sugar is removed we have:

J + S -\frac{2}{3}S= 530g

Required

Determine the weight of the jar

J + S = 850g --- (1)

J + S -\frac{2}{3}S= 530g --- (2)

Simplify (2)

J + S -\frac{2S}{3}= 530g

Take LCM

J + \frac{3S - 2S}{3}= 530g

J + \frac{S}{3}= 530g

Make S the subject in (1)

J + S = 850g

S = 850g - J

Substitute 850g - J for S in J + \frac{S}{3}= 530g

J + \frac{850g - J}{3}= 530g

Multiply through by 3

3 * J + 3*\frac{850g - J}{3}= 530g * 3

3J + 850g - J= 1590g

Collect Like Terms

3J  - J= 1590g-850g

2J= 740g

Make J the subject

J = \frac{1}{2} * 740g

J = 370g

<em>The jar weighs 370g</em>

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A simple random sample (SRS) of 100 of a certain popular car in 1997 found that 20 had a certain minor defect in the brakes. An
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Answer: The required interval is (0.679, 0.821).

Step-by-step explanation:

Since we have given that

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\hat{p_1}=\dfrac{p_1}{n_1}=\dfrac{20}{100}=0.2\\\\\hat{p_2}=\dfrac{p_2}{n_2}=\dfrac{50}{400}=0.125

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QUICK! 75 POINTS !!Select all that are part of the solution set of csc(x) &gt; 1 and over 0 ≤ x ≤ 2π.
Vladimir79 [104]

Answer:

\frac{\pi}{4}

\frac{5\pi}{6}

Step-by-step explanation:

The answer uses the unit circle and that sine and cosecant are reciprocals.

The first choice doesn't even fit the criteria that x is between 0 and 2\pi (inclusive of both endpoints) because of the x=\frac{-7\pi}{6}.

Let's check the second choice.

\csc(\frac{\pi}{4})=\frac{2}{\sqrt{2}} \text{ since } \sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}.

\csc(\frac{\pi}{4})>1 \text{ since } \frac{2}{\sqrt{2}}>1

\csc(\frac{\pi}{2})=1 \text{ since } \sin(\frac{\pi}{2})=1 which means \csc(\frac{\pi}{2})=1 which is not greater than 1.

So we can eliminate second choice.

Let's look at the third.

\csc(\frac{5\pi}{6})=2 \text{ since } \sin(\frac{5\pi}{6})=\frac{1}{2} which means \csc(\frac{5\pi}{6})>1.

\csc(\pi)  isn't defined because \sin(\pi)=0.

So we are eliminating 3rd choice now.

Let's look at the fourth choice.

\csc(\frac{7\pi}{6})=-2 \text{ since } \sin(\frac{7\pi}{6})=\frac{-1}{2} which means \csc(\frac{7\pi}{6}) and not greater than 1.

I was looking at the rows as if they were choices.

Let me break up my choices.

So we said x=-\frac{7\pi}{6} doesn't work because it is not included in the inequality 0\le x \le 2\pi.

How about x=0?  This leads to \csc(0) which doesn't exist because \sin(0)=0.

So neither of the first two choices on the first row.

Let's look at the second row again.

We said \frac{\pi}{4} worked but not \frac{\pi}{2}

Let's look at the choices on the third row.

We said \frac{5\pi}{6} worked but not x=\pi

Let's look at at the last choice.

We said it gave something less than 1 so this choice doesn't work.

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