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2 years ago

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A group of 4 students went to the movie theater. They each bought a ticket for $x and spent $10.50 each on snacks. A group of 3

adults went to the movie theater and paid twice as much for each of their tickets as each student. The group of adults also spent $10.00 each on snacks. The group of students spent the same amount as the group of adults. What is the cost of each adult ticket?

1 answer:

Anit [1.1K]2 years ago

8 0

i don't know how you can answer it, they do not say at all how much the tickets cost they just say how much the snacks cost

Step-by-step explanation:

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If a sphere is 120 feet in diameter what is the volume? <br><br> Help please

natulia [17]

Answer:

120

Step-by-step explanation:

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3 years ago

Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

s344n2d4d5 [400]

The various answers to the question are:

To answer 90% of calls instantly, the organization needs four extension lines.
The average number of extension lines that will be busy is Four
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$$

Generally, the equation for is mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

$P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$$

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

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3 0

2 years ago

Please help I have 9 min

Harrizon [31]

I believe the answer is D because the pathagreom theorum proves it true (excuse my terrible grammar)

5 0

3 years ago

A basketball-player scores a free-throw with probability 0.9. What is the probability that her first miss occurs on the 6th shot

-BARSIC- [3]

Answer:

$Pr = 0.059049$

Step-by-step explanation:

Given

$p = 0.9$ --- probability of scoring

Required

Probability that his first miss is his 6th shot

Let q represent the event that he did not score.

Using complement rule:

$q = 1 - p = 1 - 0.9 = 0.1$

The event that his first miss is his 6th is represented as:

p p p p p q ---- That he scoress the first 5 attempts

So, the probability is:

$Pr = p^5 * q$

$Pr = 0.9^5 * 0.1$

$Pr = 0.059049$

7 0

3 years ago

For what value of c does x^2−2x−c=4 have exactly one real solution?<br> PLEASE HELP!!!!!!!

Paul [167]

Answer:

-5

Step-by-step explanation:

Moving all terms of the quadratic to one side, we have

$x^2-2x-(c+4)=0$ .

A quadratic has one real solution when the discriminant is equal to 0. In a quadratic $ax^2+bx+d$ , the discriminant is $\sqrt{b^2-4ad}$ .

(The discriminant is more commonly known as $\sqrt{b^2-4ac}$ , but I changed the variable since we already have a $c$ in the quadratic given.)

In the quadratic above, we have $a=1$ , $b=-2$ , and $d=-(c+4)$ . Plugging this into the formula for the discriminant, we have

$\sqrt{(-2)^2-4(1)(-(c+4))$ .

Using the distributive property to expand and simplifying, the expression becomes

$\sqrt{4-4(-c-4)}=\sqrt{4+4c+16}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{20+4c}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{4}\cdot\sqrt{5+c}\\~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{c+5}.$

Setting the discriminant equal to 0 gives

$2\sqrt{c+5}=0$ .

We can then solve the equation as usual: first, divide by 2 on both sides:

$\sqrt{c+5}=0$ .

Squaring both sides gives

$c+5=0$ ,

and subtracting 5 from both sides, we have

$\boxed{c=-5}.$

3 0

2 years ago