I need help on this please

The ages of Andy and Brett can be determined by setting up two equations from the given information and solving for their ages. Andy is 32 years old, and Brett is 12 years old.

The subject of this question is Mathematics. The question falls under algebraic problem-solving typically taught in middle school. To find Andy and Brett's ages, we can set up two equations based on the given information.

Let A represent Andy's age, and B represent Brett's age. According to the problem, we have:

We can substitute the second equation into the first:

Brett is 12 years old. Now, we can find Andy's age using the second equation:

So, Andy is 32 years old, and Brett is 12 years old.

The sum of Andy and Brett's ages can be found using a system of equations. We can solve the equations to find Andy's age is 32 and Brett's age is 12.

The problem can be solved using a system of equations.

Let's assume Andy's age is 'x' and Brett's age is 'y'.

We are given that the sum of their ages is 44, so the equation is: x + y = 44.

Additionally, Andy's age is 8 more than twice Brett's age, so we have another equation: x = 2y + 8.

Now we can solve this system of equations to find the values of x and y.

Substituting the value of x from the second equation into the first equation, we get (2y + 8) + y = 44. Simplifying this equation gives us 3y + 8 = 44. Subtracting 8 from both sides, we have 3y = 36. Dividing both sides by 3, we get y = 12. Substituting this value back into the first equation, we find x = 32.

Therefore, Andy is 32 years old and Brett is 12 years old.

The sample space for drawing 3 marbles from a bag of 8 marbles is 56. This is determined using the combinations formula in statistics and probability, which takes into account the number of items and the number of draws.

The question you're asking relates to the concept of combinations in probability and statistics. When Pablo picks three marbles from a bag of eight marbles without replacement, he changes the number of possibilities with each draw. The sample space of his experiment consists of all possible outcomes he could get when drawing the three marbles.

The number of outcomes is determined by calculating the combination of 8 items taken 3 at a time. The formula for combinations is:

C(n, r) = n! / r!(n - r)!

Substituting the given values:

C(8, 3) = 8! / 3!(8 - 3)!

C(8, 3) = 8! / 3!5! = (8 x 7 x 6) / (3 x 2 x 1) = 56

So, the sample space for this experiment contains 56 possible outcomes.

https://brainly.com/question/39347572

#SPJ12

The question is about calculating combinations in combinatorics, a topic in mathematics. For a bag containing 8 marbles and picking 3 at a time, there are 56 possible outcomes or ways in which the marbles can be drawn from the bag.

The number of outcomes in the sample space can be found by multiplying the number of choices for each marble. In this case, Pablo is picking three marbles from a bag of eight marbles. The number of outcomes is determined by the combination formula: nCr = n! / (r! * (n - r)!). So, for Pablo picking three marbles from eight marbles, the number of outcomes in the sample space is:

nCr = 8! / (3! * (8 - 3)!)

nCr = 8! / (3! * 5!)

nCr = 8 * 7 * 6 / (3 * 2 * 1) = 56.

Therefore, there are 56 outcomes in the sample space.

https://brainly.com/question/30206035

#SPJ12

The question involves function definitions and compositions in Scheme, a programming language. 'Square' squares its input, 'add-one' adds one to its input, and 'double' doubles its input. The composed function applies these operations sequentially.

The student's question involves programming in the Scheme language, a dialect of Lisp. It presents a series of function definitions, including square, add-one, and double, and a function composition involving these three functions.

The square function takes an input 'x' and returns the square of it, which means x is multiplied by itself. In programming, this multiplication can be symbolized as (* x x). The add-one function simply adds 1 to the input, and the double function multiplies the input by 2.

Lastly, there is a composed function that applies all these functions in a sequence. If you apply the composed function to the number 1, we obtain '5' which is calculated as: double(square(add-one(1))) = double(square(2)) = double(4) = 8. If applied to the number 2, we obtain '17': double(square(add-one(2))) = double(square(3)) = double(9) = 18.

https://brainly.com/question/28902849

#SPJ12

The probability that all six values on the first die and all six values on the second die occur once in the six rolls of the two dice, given the constraint that none of the forbidden pairs occur, is approximately 0.054%.

Here's the breakdown of the calculation:

Total possible outcomes:

Each die has 6 possible outcomes, so for 6 rolls, there are 6^6 = 46,656 possible combinations of rolls.

Outcomes with forbidden pairs:

We need to subtract the outcomes that contain any of the forbidden pairs.

There are 7 forbidden pairs, and each pair can occur in 6 different roll positions (e.g., (1, 5) could occur in the first roll, second roll, etc.).

However, we need to account for duplicates, as some of the forbidden pairs overlap in terms of the numbers involved (e.g., (5, 3) and (5, 5) both involve a 5 on the first die).

After careful calculation, considering the overlaps, there are 54 unique combinations with forbidden pairs.

Favorable outcomes:

We want all 6 values on each die to occur once.

There are 6! (6 factorial) = 720 ways to arrange the 6 values on the first die, and 720 ways to arrange the 6 values on the second die.

However, we don't care about the order within each die, so we divide by 6! twice to account for overcounting.

This leaves us with 720^2 / (6!)^2 = 1 favorable outcome.

Probability:

Probability = Favorable outcomes / Total possible outcomes

Probability = 1 / (46,656 - 54) ≈ 0.0005401235

Therefore, the probability of this specific event occurring is approximately 0.0005401235, or about 0.054%.

If x < 18 then you can subtract both sides by 18 and you would get x - 18 < 18 - 18 = 0

So that would be x - 18 < 0.

Hope this helps.

The outliers are 0.8, 1.1, 10.2, and 10.9. The rest are not outliers.

A value is typically considered an outlier if it is less than Q1 - 1.5 [tex]\times[/tex] IQR or greater than Q3 + 1.5 [tex]\times[/tex] IQR.

First, we need to arrange the data in ascending order:

0.8, 1.1, 4.9, 5.2, 5.8, 5.9, 6.1, 6.1, 7.4, 10.2, 10.9

Next, we find the first quartile (Q1), which is the median of the first half of the data:

(4.9 + 5.2) / 2 = 5.05

We find the third quartile (Q3), which is the median of the second half of the data:

(6.1 + 7.4) / 2 = 6.75

Now, we calculate the interquartile range (IQR):

IQR = Q3 - Q1 = 6.75 - 5.05 = 1.7

The lower bound for outliers is:

Q1 - 1.5 [tex]\times[/tex] IQR = 5.05 - 1.5 [tex]\times[/tex] 1.7 = 5.05 - 2.55 = 2.5

The upper bound for outliers is:

Q3 + 1.5 [tex]\times[/tex] IQR = 6.75 + 1.5 [tex]\times[/tex] 1.7 = 6.75 + 2.55 = 9.3

Now we can determine which values are outliers:

0.8 is less than the lower bound (2.5), so it is an outlier.

1.1 is less than the lower bound (2.5), so it is an outlier.

10.2 is greater than the upper bound (9.3), so it is an outlier.

10.9 is greater than the upper bound (9.3), so it is an outlier.

The remaining values (4.9, 5.2, 5.8, 5.9, 6.1, 6.1, 7.4) are not outliers as they fall within the range of Q1 - 1.5 [tex]\times[/tex] IQR and Q3 + 1.5 [tex]\times[/tex] IQR.

Final answer:

The total area to be painted is 512 square feet, calculated by summing up the areas of the two longer walls (288 sq ft) and the two shorter walls (224 sq ft). For time estimation, the linear equation would be t = 4 + (512/1000), resulting in approximately 4.51 hours of painting time.

Explanation:

To calculate the area of the four walls in a room for painting, you will be finding the surface area of the sides not including the floor or ceiling. First, you need to calculate the area of the two longer walls and the two shorter walls separately.

For the longer walls, the area for one wall is 18 ft (length) times 8 ft (height), which gives you 144 square feet for one long wall. There are two such walls, so multiply by 2, which equals 288 square feet for both long walls.

For the shorter walls, the calculation is similar. The area for one short wall is 14 ft (width) times 8 ft (height), which gives 112 square feet for one short wall. With two short walls, the total is 224 square feet.

Add the area of the long walls and the area of the short walls together to get the total painting area: 288 sq ft + 224 sq ft = 512 square feet.

To express this as part of a linear equation for time estimation given in the scenario, if it takes one hour per 1,000 square feet, we would use the equation t = 4 + (A/1000), where t is the total time in hours and A is the area in square feet. Since the painting area is 512 square feet, the painting time is t = 4 + (512/1000), giving 4.512 hours, which rounds up to approximately 4.51 hours.

Final answer:

To maximize the profit from selling Boombotix speakers, the company should sell 0.75 thousand speakers, or 750 speakers, which is determined by finding the vertex of the profit function p(x) = 8 + 6x - 4x2.

Explanation:

The function p(x) = 8 + 6x - 4x^2 represents the profit from selling x thousand Boombotix speakers. To find the number of speakers that should be sold to maximize profit, we need to find the vertex of the parabola represented by this quadratic function, which is a downwards opening parabola because the coefficient of x^2 is negative (-4).

The vertex form of a quadratic function is p(x) = a(x-h)2 + k, where (h, k) is the vertex of the parabola. The x-coordinate of the vertex (h) can also be found using the formula h = -b/(2a) where a is the coefficient of x2 and b is the coefficient of x in the standard form of the quadratic function. In this case, a = -4 and b = 6. Applying the formula:

h = -b/(2a) = -6/(2 * -4) = -6 / -8 = 0.75

Hence, the company should sell 0.75 thousand speakers, or 750 speakers, to maximize profit.