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2 years ago

Martin's car had 86,456 miles on it. Of that distance, Martin's wife drove 24,901 miles,

Mathematics

1 answer:

navik [9.2K]2 years ago

3 0

Answer:

Martin drove 53,558 miles.

Step-by-step explanation:

If you add up the amount of miles his son and wife drove, you get 32,898. To get the answer, you would subtract that from the total number of miles to get 53,558.

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Josiah kept track of how many songs of each genre were played in an hour from his MP3 player. The counts are displayed in the ta

Evgen [1.6K]

Answer:

1) He should have found the average of the number of rock songs by averaging 4 and 6 to get 5.

2) He did not multiply the numerator and denominator by the correct number to equal 1,500.

5) He should have multiplied the numerator and denominator by 75, not 30, because 20 x 75 = 1,500.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A university is trying to determine what price to charge for tickets to football games. At a price of $ per ticket, attendance

Anika [276]

This question is incomplete, the complete question is;

A university is trying to determine what price to charge for tickets to football games. At a price of $24 per ticket, attendance averages 40,000 people per game. Every decrease of $4 adds 10,000 people to the average number. Every person at the game spends an average of $4 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price?

Answer:

a) price per ticket = $18

b) 55,000 will attend at that price

Step-by-step explanation:

Given that;

price of ticket = $24

so cost of ticket = 24 - x

every decrease of $4 adds 10,000 people

for $1 adds 2,500 people

Total people = 40000 + 2500x

Revenue function = people × cost + people × 4

R(x) = (40000 + 2500x)(24 - x) + (40000 + 2500x)4

R(x) = 960,000 - 40000x + 60000x - 2500x² + 160,000 + 10000x

R(x) = 1120000 + 30000x - 2500x²

R(x) = -2500x² + 30000x + 1120000

now for maximum revenue; R'(x) is to zero

R'(x) = d/dx( -2500x² + 30000x + 1120000 ) = 0

⇒ -5000x + 30000 + 0 = 0

⇒ -5000x + 30000 = 0

⇒ 5000x = 30000

x = 30000 / 5000

x = 6

R'(X) = -5000 < 0 ⇒ MAXIMUM

∴ Revenue is maximum at x = 6

so cost of ticket ;

price per ticket = (24 - x) = 24 - 6 = $18

so price per ticket = $18

Number of people = 40000 + 2500x

= 40000 + (2500 × 6)

= 40000 + 15,000

= 55,000

Therefore, 55,000 will attend at that price

8 0

2 years ago

Each day, X arrives at point A between 8:00 and 9:00 a.m., his times of arrival being uniformly distributed. Y arrives independe

astraxan [27]

Answer:

Y will arrive earlier than X one fourth of times.

Step-by-step explanation:

To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:

$f_X(x) = \frac{1}{L}$

Where L stands for the length of the interval over which the variable is distributed.

Now, as X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:

$f_X(x) = 1\\\\f_Y(y)=2$ .

Now, the probability of an event is equal to the integral of the density probability function:

$\iint_A f_{X,Y} (x,y) dx\, dy$

Where A is the in which the event happens, in this case, the region in which Y<X (Y arrives before X)

It's useful to draw a diagram here, I have attached one in which you can see the integration region.

You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.

Let's set up the integration:

$\iint_A f_{X,Y} (x,y) dx\, dy\\\\\iint_A f_{X} (x) \, f_{Y} (y) dx\, dy\\\\2 \iint_A dx\, dy$

We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:

$2 \iint_A dx\, dy= 2 \times \frac{0.5 \times 0.5 }{2} \\\\2 \iint_A dx\, dy= \frac{1}{4}$

That means that one in four times Y will arrive earlier than X. This result can also be seen clearly on the diagram, where we can see that the triangle is a fourth of the rectangle.