Suppose a fair die is rolled eight times. find numerical values for the expectation

The dimensions of the box that minimizes material usage, enclosing 100 m³ with no top, are approximately 5.848 meters for the square base side length and 2.682 meters for the height.

Let's denote the side length of the square base as x meters and the height of the box as h meters. Since the box has no top, the volume (V) of the box is given by the product of the area of the square base and the height:

[tex]\[ V = x^2 \cdot h \][/tex]

Given that [tex]\(V = 100 \, \text{m}^3\)[/tex], we have the equation:

[tex]\[ 100 = x^2 \cdot h \][/tex]

Now, we want to minimize the amount of material needed to construct the box, which is the surface area (A) of the box. The surface area is the sum of the area of the square base and the areas of the four sides:

[tex]\[ A = x^2 + 4xh \][/tex]

To minimize A, we can express h in terms of x from the volume equation and substitute it into the surface area equation:

[tex]\[ h = \frac{100}{x^2} \]\[ A(x) = x^2 + 4x\left(\frac{100}{x^2}\right) \]\[ A(x) = x^2 + \frac{400}{x} \][/tex]

Now, to find the minimum amount of material, we take the derivative of A with respect to x and set it equal to zero:

[tex]\[ \frac{dA}{dx} = 2x - \frac{400}{x^2} = 0 \][/tex]

Multiply through by x^2 to get rid of the fraction:

[tex]\[ 2x^3 - 400 = 0 \][/tex]

Solve for x:

[tex]\[ x^3 = 200 \]\[ x = \sqrt[3]{200} \][/tex]

Now that we have x, we can find h using the volume equation:

[tex]\[ h = \frac{100}{x^2} \]\[ h = \frac{100}{(\sqrt[3]{200})^2} \]\[ h = \frac{100}{\sqrt[3]{400}} \][/tex]

The dimensions of the box that will minimize the amount of material needed are approximately [tex]\(x \approx 5.848\)[/tex] meters and [tex]\(h \approx 2.682\)[/tex] meters.

Answer:

Customers older than 45 years are 11 in number.

Step-by-step explanation:The age of store customers is represented by the stem ad leaf plot.The stem represents the tens digit while leaf denotes the unit digit.The question is asking us to find the number of customers who are older than 45 so 45 is not considered.

The age of customers more than 45 are:

48,50,50,51,55,56,62,64,65,65,73.There are 11 customers in all .

Answer:k=2

Step-by-step explanation:

The four families consisting of 12 distinct individuals can be seated in 12! (479,001,600) ways. There are no seating restrictions, so each person can occupy any of the twelve seats.

The subject of this question is combinatorics, a branch of mathematics. In this problem, you have four families, each consisting of 1 male and 2 females, and you want to know in how many ways they can be seated in a row of twelve seats.

Given there is no restriction about the seating pattern, each member can occupy any seat. So, consider each family member as a distinct person; you then have 12 people to be seated in 12 different ways. This can be done in 12 factorial (12!) ways. Factorial implies the product of all positive integers up to that number. A simple way to calculate 12 factorial is: 12*11*10*9*8*7*6*5*4*3*2*1, which equals 479,001,600.

So, the four families can be seated in a row of twelve seats in 479,001,600 ways. This principle of arrangements is a key part of combinatorics and discrete mathematics.

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Considering each family as a single unit, the total number of ways they can be seated in a row of twelve seats is given by the formula: 12! divided by (2!) to the power of 4.

This is actually a problem that is solved using the principles of permutations and combinations in mathematics. Given we have 4 families each with 1 male and 2 females, we have a total of 12 people. Now, if we are to arrange these 12 people in a row of twelve seats, we have 12! (factorial) ways to do it. However, each family group is to be considered as a single unit and within each unit, arrangements don't count, so we must divide by the number of ways to arrange the 2 females within each family of 4, which is 2!. Hence, the total number of ways to seat the group is given by (12! / (2!)^4).

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Final answer:

By setting up and solving a system of equations, we find that there are 13 birds and 19 lions in the zoo.

Explanation:

To solve the problem of determining the number of birds and lions in the zoo given 32 heads and 102 feet, we can set up a system of equations. Birds have 2 feet, and lions have 4 feet. Let B represent the number of birds and L represent the number of lions.

Equations:

B + L = 32 (total heads)

2B + 4L = 102 (total feet)

Solving this system, we'll first multiply the first equation by 2 to facilitate elimination:

2B + 2L = 64

2B + 4L = 102

Subtracting the first modified equation from the second equation gives us 2L = 38, so L=19. Substituting L=19 into the first equation, we find B = 32 - 19 = 13.

Therefore, there are 13 birds and 19 lions in the zoo.

i got the same question :D

In mathematics, two variables are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related by use of a constant multiplier. The constant is called the coefficient of proportionality or proportionality constant.

The height of the cylinder is 8mm.

To find the height of the cylinder, we need to first find the surface area of the sphere. The surface area of a sphere is given by the formula: 4πr^2. Since the diameter of the sphere is 16mm, the radius is half of that, which is 8mm. Plugging the value of the radius into the formula, we get: 4π(8^2) = 256π mm^2.

Next, we need to find the surface area of the cylinder. The base diameter of the cylinder is equal to the diameter of the sphere, which is 16mm. Therefore, the radius of the cylinder is also 8mm. The surface area of the cylinder is given by the formula: 2πrh + 2πr^2. Since the height is unknown, we'll use a variable 'h'.

Since the surface area of the sphere is equal to the surface area of the cylinder, we can set up an equation: 256π = 2πrh + 2πr^2. Plugging in the values, we get: 256π = 2π(8)(h) + 2π(8^2).

Cancelling out the common factor of 2π, we have the equation: 128 = 8h + 64. Subtracting 64 from both sides of the equation, we get: 64 = 8h. Dividing both sides by 8, we find that the height of the cylinder is: h = 8mm.

As the level of significance here is 0.05 because we are comparing the p-value with 0.05. Therefore the critical region boundaries here would be given as:

For 5 degrees of freedom, we get: ( from the t-distribution tables )

P(t_5 < 2.571) = 0.975

Therefore due to symmetry, we get:

P(-2.571 < t_5 < 2.571) = 0.95

Therefore the critical region here would be given as: +2.571 and -2.571

The critical values for the t-distribution are used to define the boundaries for the critical region in a hypothesis test. In this case, the boundaries are t < -2.073 or t > 2.073.

The critical values for the t-distribution are used to define the boundaries for the critical region in a hypothesis test. In this case, the results of the study are reported as t(22) = 2.12, p < .05, two tails. To find the boundaries for the critical region, we need to look up the critical value for a two-tailed test with 22 degrees of freedom and a significance level of 0.05.

Using a t-distribution table or a calculator, we find that the critical value is approximately 2.073. Therefore, the boundaries for the critical region in this study are t < -2.073 or t > 2.073.

Any calculated t-value that falls outside of these boundaries would lead to rejecting the null hypothesis and concluding that the variables are significantly correlated.

Answer : The fraction of pounds of fruits over the total pounds of fruit and vegetables is, [tex]\frac{1005}{2669}[/tex]

Step-by-step explanation :

As we are given that:

Total amount of vegetables = 208 pounds

Total amount of fruits = [tex]125\frac{5}{8}\text{ pounds}=\frac{1005}{8}\text{ pounds}[/tex]

Thus, the total amount of fruits and vegetables will be:

[tex]208+125\frac{5}{8}\\\\=208+\frac{1005}{8}\\\\=\frac{1664+1005}{8}\\\\=\frac{2669}{8}[/tex]

Now we have to calculate the fraction of pounds of fruits over the total pounds of fruit and vegetables :

[tex]\frac{\text{Pounds of fruits }}{\text{ total pounds of fruits and vegetables}}\\\\=\frac{(\frac{1005}{8})}{(\frac{2669}{8})}\\\\=\frac{1005}{8}\times \frac{8}{2669}\\\\=\frac{1005}{2669}[/tex]

Thus, the fraction of pounds of fruits over the total pounds of fruit and vegetables is, [tex]\frac{1005}{2669}[/tex]