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2 years ago

you roll two dice what is the probability that the sum of the dice is less than 5 and one dice shows a 2? ...?

1 answer:

3 0

<span>Lets say the 1st die rolled a 2 - there would be 2 combinations for which the sum of dice being < 5 : 2,1 2,2 Now say the 2nd die rolled a 2 - there would be 2 combinations for which the sum of dice being < 5 : 1,2 2,2 Now we want to count all cases where either dice showed a 2 and sum of the dice was < 5. However note above that the roll (2,2) is counted twice. So there are three unique dice roll combinations which answer the criteria of at least one die showing 2, and sum of dice < 5: 1,2 2,1 2,2 The total number of unique outcomes for two dice is 6*6=36 . So, the probability you are looking for is 3/36 = 1/12</span>

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Neil drives at an average speed of 60 miles/hour to reach

nikklg [1K]

Answer:

480/(x+60) ≤ 7

Step-by-step explanation:

We can use the relations ...

time = distance/speed

distance = speed×time

speed = distance/time

to write the required inequality any of several ways.

Since the problem is posed in terms of time (7 hours) and an increase in speed (x), we can write the time inequality as ...

480/(60+x) ≤ 7

Multiplying this by the denominator gives us a distance inequality:

7(60+x) ≥ 480 . . . . . . at his desired speed, Neil will go no less than 480 miles in 7 hours

Or, we can write an inequality for the increase in speed directly:

480/7 -60 ≤ x . . . . . . x is at least the difference between the speed of 480 miles in 7 hours and the speed of 60 miles per hour

___

Any of the above inequalities will give the desired value of x.

8 0

2 years ago

A shop owner has 75 pounds of coffee that costs $4.50 per pound. How much coffee that costs $3.50 per pound should he mix with t

user100 [1]

58 pounds of coffee

7 0

2 years ago

Read 2 more answers

A square 20 cm long and a rectangle 24 cm long have the same perimeter. Find the area of the rectangle.

Nitella [24]

20 * 4 = 80 = perimeter of the square 24*2 = 48, the length of two sides of the rectangle. Deducting from 80 we are left with 32 and dividing by 2 we know the rectangle is 24 x 16 Area = L * W, so you can calculate the area of the rectangle.
384

4 0

2 years ago

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require

Aleksandr-060686 [28]

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

Step-by-step explanation:

<u>Optimizing With Derivatives </u>

The procedure to optimize a function (find its maximum or minimum) consists in :

Produce a function which depends on only one variable
Compute the first derivative and set it equal to 0
Find the values for the variable, called critical points
Compute the second derivative
Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum

We know a cylinder has a volume of 4 $ft^3$ . The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

$\displaystyle A''=2\pi+\frac{16}{r^3}$