you are dealt one card from a standard 52 card deck. Find the probability of being dealt a heart

Final answer:

The work needed to pump out half the water from the given aquarium is 17,150 Joules. This is determined by applying physical principles to calculate the mass of the water, the effect of gravity, and using an integral to find the total work done against gravity.

Explanation:

The student is asking about the work required to pump half of the water out of an aquarium with dimensions 7 m long, 1 m wide, and 1 m deep using physics concepts involving work, force, and Riemann sums. To approach this problem, we must consider the work done against gravity to move the water from its initial position to the top of the aquarium. The density of water (ρ) is 1000 kg/m³, and the acceleration due to gravity (g) is 9.8 m/s².

First, calculate the volume of water to be pumped out, which is half the aquarium volume: V = ½ × 7 m × 1 m × 1 m = 3.5 m³. Convert this volume to mass using the density of water, m = ρV = 1000 kg/m³ × 3.5 m³ = 3500 kg.

The work done to pump out half the water can be calculated using the concept of the center of mass of the water being lifted, which is at a height h/2 from the top of the water when the tank is half full, where h is the depth of the tank. Therefore, the work is W = mgh/2 = 3500 kg × 9.8 m/s² × 0.5 m = 17150 J.

To approximate the required work using a Riemann sum, consider the small amount of work to lift a thin layer of water δx from a depth x to the top of the tank, dW = ρgAdx(x), where A is the area of the tank's surface. We set up the integral ∫ W = ρgA ∫ xdx from 0 to h/2, and find the limit as the number of partitions goes to infinity. The integration gives us the same work W = 17150 J.

The total amount of candy left after adding the candy received from the grandmother and the sister, and subtracting the candy eaten by the friend, is approximately 0.63 pounds.

First, we add up the amounts of candy you received. You started with 1/3 pound from your grandmother and received an additional 1/2 pound from your sister, for a total of 5/6 pound of candy. However, because your friend ate some, we subtract 1/5 pound from this total. To do this, we need to convert all fractions to have a common denominator, which is 30 in this case. Therefore, 5/6 becomes 25/30, and 1/5 becomes 6/30. Subtraction gives us (25-6)/30 = 19/30 or approximately 0.63 pounds of candy left.

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Answer:

1.To add and subtract polynomials, the horizontal technique of deleting parenthesis, collecting like terms, and simplifying is the simplest. When there are negative terms, it gets more difficult since one must ensure that the term remains negative when gathering comparable terms. The vertical approach of building up a box and adding vertically takes longer to set up, but once completed, there is a clear depiction of where all of the similar terms are.

2.Because the product of a linear factor and a quadratic factor is a cubic product, algebra tiles cannot be used to simulate the multiplication of a linear polynomial by a quadratic polynomial.

3.Distribute the first polynomial's terms to the second polynomial's terms. When multiplying two terms together, remember to multiply the coefficients (numbers) and add the exponents. However, with Integers, you multiply two integers with opposite signs.

4.Before combining like terms, there will be four terms. When a monomial is multiplied by a trinomial, the result is six. There will be nine terms in two trinomials.

Step-by-step explanation:

Anne's savings will be worth $11,636.924 at the end of the five years.

We have,

PV= $40

Future Value = Present Value (1 + Interest Rate[tex])^{Time[/tex]

For the first two years:

Present Value = $40 per week * 52 weeks/year * 2 years = $4,160

Interest Rate = 3% = 0.03

Time = 2 years

Future Value (first two years) = $4,160 (1 + 0.03)²= $ 4,413.344

For the last three years:

Present Value = $40 per week * 52 weeks/year * 3 years = $6,240

Interest Rate = 5% = 0.05

Time = 3 years

Future Value (last three years) = $6,240 (1 + 0.05)³ = $ 7,223.58

Then, Total Future Value = Future Value (first two years) + Future Value (last three years)

= 4413.344 + 7223.58

= $ 11,636.924

Therefore, Anne's savings will be worth $11,636.924 at the end of the five years.

Learn more about Future Value here:

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Answer:  The required number of ways is 184756.

Step-by-step explanation:  Given that a test consists of 20 problems and students are told to answer any 10 of these questions.

We are to find the number of different ways in which the students choose 10 questions.

We know that

the number of ways in which r things can be chosen from n different things is given by

[tex]N=^nC_r.[/tex]

Therefore, the number of ways in which students chose 10 questions from 20 different questions is given by

[tex]N\\\\=^{20}C_r\\\\\\=\dfrac{20!}{10!(20-10)!}\\\\\\=\dfrac{20\times19\times18\times17\times16\times15\times14\times13\times12\times11\times10!}{10!\times 10\times9\times8\times7\times6\times5\times4\times3\times2\times1}\\\\\\=184756.[/tex]

Thus, the required number of ways is 184756.

Answer:

d

Step-by-step explanation:

Answer: 108 lbs.

Step-by-step explanation:

Given : A cube is packed with decorative pebbles. If the cube has a side length of 6 inches.

Volume of cube = [tex](side)^3[/tex]

i.e. Volume of cube = [tex](6)^3=216\text{ cubic inches}[/tex]

Since , each pebble weighs on average 0.5 lb per cubic inch.

Then, the total weight of the pebbles in the cube will be

= 0.5 x Volume of cube

= [tex]0.5\times216=108\text{ lb}[/tex]

Hence,  the total weight of the pebbles in the cube =108 lbs.

The numbers that fit the conditions of the question are x = 2, y = 2, and z = 2. This sums to 6 and minimizes the sum of their squares (12).

The subject of this problem involves real numbers and their sums and squares. The intention is to find three real numbers (x, y, and z) such that their sum equals 6, and the sum of their squares is minimized. By symmetry, it is preferable if these three numbers are equal. Therefore, x = y = z = 6/3 = 2 is the optimal solution.

So the three real numbers are x = 2, y = 2, and z = 2, which sum to 6 and the sum of their squares is as small as possible (12).

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In the process of calculating probabilities of events on a pair of dice, we found P(F), the probability of rolling a total of two, to be 1/36. P(E), the probability of rolling an even total, to be 1/2. However, when determining P(F|E), the probability of rolling a two given we've rolled an even total, the probability changes to 1/18 due to the reduced sample space.

The concepts involved in this question are related to probability, specifically the principles governing dice rolls. In this particular scenario, the events are rolling two dice and getting a total of two (Event F), and rolling an even total on the dice (Event E).

In this specific scenario, event F (the total is two) can only occur in one way - when both dice show 1. Since there are 36 potential outcomes when two dice are rolled (6 possibilities for the first die and 6 for the second), the probability of event F, P(F), is 1/36.

Event E (an even total) can occur in 18 ways (2,4,6 for the first die and 1,3,5 for the second or 1,3,5 for the first die and 2,4,6 for the second). So, P(E) = 18/36 = 1/2. However, when considering P(F|E) (the probability of event F given that event E has occurred), you need to adjust your consideration of 'total possibilities' based only on event E. Since P(E) = 1/2, your total possibilities now become 18. From these 18, only one will result in a total of two. Therefore, P(F|E) = 1/18.

Of course, there's different perspectives to consider how adding the condition that E had occurred would change the probability of event F. Essentially, by narrowing down the potential outcomes to only those that involve event E, you're working with a reduced sample space. This in turn affects the likelihood of event F occurring, hence the alteration in probability.

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Given:

24,000 books

Between 3 and 4 books will have a printing error in every 3000 printed books

Find: in the 24000 books, the total number of books that will have a printing error

Solution:

Based from the given, we need to know how many sets of 3000 we

have in the 24000 books so:

24000 / 3000 = 8

Now, in each set of 3000 we have between 3 and 4 errors and we have 8 sets of 3000 books:

minimum errors 3 * 8 = 24

maximum errors 4 * 8 = 32

Therefore, at this rate, the number of books that will have a printing error in the 24000

will be between 24 and 32 or 24 < E < 32.

In 24,000 books, the number of printing errors could be between 24 and 32. This is calculated based on an error rate of 3 to 4 errors per 3,000 books.

To determine the number of printing errors in 24,000 books, we need to understand the error rate. The problem states that there are between 3 and 4 books with errors per 3,000 printed books.

First, find the range of error rates per 3,000 books:

Next, scale this up to 24,000 books:

Therefore, the number of books with printing errors in 24,000 books will be between 24 and 32.