Is the sum of two monomials always a monomial? Is their product always a monomial?

Final answer:

There are 252 different starting lineups possible if positions are not assigned, and 30,240 starting lineups are possible if positions are assigned.

Explanation:

For part A of the question, where positions are not assigned, we can calculate the number of different starting lineup combinations by using the combination formula. Since we are choosing 5 players out of 10 without regard to order, we use:

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

Where n is the total number of players to choose from, and k is the number of players we want to choose. Therefore, we have:

C(10, 5) = 10! / (5! (10-5)!) = 252

There are 252 different starting lineups possible if positions are not assigned.

For part B, where positions are assigned, we use the permutation formula because the order in which we select the players matters. This formula is:

P(n, k) = n! / (n-k)!

Since the order in which the players are chosen matters, we have:

P(10, 5) = 10! / (10-5)! = 30,240

There are 30,240 different starting lineups possible with positions assigned.

There are 9000 possible 4-sequences that do not begin with the digit 0.

The question asks how many 4-sequences on the set {0..9} do not begin with 0. To solve this, we need to consider the number of possible options for each of the four positions in the sequence, knowing that the first digit cannot be 0.

For the first position, there are 9 options (1-9), since 0 is not allowed. For the next three positions, each can be any of the 10 digits (0-9), since there are no restrictions. Thus, the total number of such sequences is the product of the number of options for each position.

9 options (1st position) x 10 options (2nd position) x 10 options (3rd position) x 10 options (4th position) = 9 x 10 x 10 x 10 = 9000.

So, to find how many 4-sequences do not begin with 0, multiply the number of choices for each position: 9 (first position, cannot be 0) x 10 (second position) x 10 (third position) x 10 (fourth position), resulting in 9000 such sequences.

Final answer:

The experienced roofer and the beginning roofer together can complete the job in approximately 15.6 hours, which is found by adding their work rates and taking the reciprocal of the sum.

Explanation:

To find how long it takes for both an experienced and a beginning roofer to roof a house together, we need to sum their work rates and then find the reciprocal of that sum. We'll start by considering the work rates of both the experienced and the beginning roofer. The experienced roofer can roof a house in 26 hours, which means the roofer completes 1/26 of the job per hour. The beginning roofer completes the job in 39 hours, which is 1/39 of the job per hour.

Now, we can calculate their combined work rate:
1/26 + 1/39 = (39 + 26) / (26 * 39) = 65 / 1014 = 1/15.6

These two roofers together complete 1/15.6 of the job per hour. To find out how long it will take them to complete the job together, we take the reciprocal of this rate:
1 / (1/15.6) = 15.6 hours.

Therefore, it would take the experienced and beginning roofer combined around 15.6 hours to roof the house together.

The formula to find circumference of circle is given by

[tex] C =2\pi r [/tex]

Where r is radius of circle and pi =3.14

We are given Circumference here as 361π

Plugging the value in the formula :

[tex] 361\pi =2\pi r [/tex]

r=361/2 = 180.5

radius of circle is 180.5.

We know that Diameter is double of radius.

So diameter = 2*180.5 = 361

Answer: Diameter of circle is 361.

Answer:

The factor form of given expression is 3(x+8) so the required number is 3.

Step-by-step explanation:

The given expression is

[tex]3x+24[/tex]

Write the factors of each term.

[tex]3x+24=3\times x+3\times 8[/tex]

In both terms, 3 is a common factor.

Take out the common factor 3.

[tex]3x+24=3(x+8)[/tex]

The factor form of given expression is 3(x+8). The factor (x+8) is given, therefore the required number is 3.

To write the decimal number 0.8(repeating) as a fraction, use the concept of an infinite geometric series and convert it to an equation. Multiply both sides of the equation by 10 to eliminate the decimal point and subtract the original equation to eliminate the repeating part. Finally, divide by 9 to get the equivalent fraction form.

To write the decimal number 0.8(repeating) as a fraction, we can use the concept of an infinite geometric series. Let's assume the repeating decimal as x: x = 0.8(repeating). Now, multiply both sides of the equation by 10 to remove the decimal point: 10x = 8(repeating). Next, subtract the original equation from this new equation to eliminate the repeating part: 10x - x = 8(repeating) - 0.8(repeating). Simplifying these expressions gives us: 9x = 7.2. Finally, divide both sides of the equation by 9: x = 7.2/9. Therefore, the equivalent fraction form of 0.8(repeating) is 7.2/9.

Noel is more likely to draw two cards with a sum that is greater than 13.

To determine which outcome is more likely, let's count the number of combinations that meet each criterion.

1. **Sum that is a multiple of 3**:

  - Possible sums: 3, 6, 9, 12

  - Counting the combinations:

    - 3: 1 combination

    - 6: 2 combinations

    - 9: 3 combinations

    - 12: 4 combinations

  - Total combinations: 1 + 2 + 3 + 4 = 10 combinations.

2. **Sum greater than 13**:

  - Possible sums: 14, 15, 16, 17, 18, 19, 20, 21, 22, 23

  - Counting the combinations:

    - 14: 1 combination

    - 15: 2 combinations

    - 16: 2 combinations

    - 17: 3 combinations

    - 18: 2 combinations

    - 19: 3 combinations

    - 20: 2 combinations

    - 21: 3 combinations

    - 22: 1 combination

    - 23: 1 combination

  - Total combinations: 1 + 2 + 2 + 3 + 2 + 3 + 2 + 3 + 1 + 1 = 20 combinations.

Comparing the total combinations, we see that there are more combinations with a sum greater than 13 (20 combinations) than there are combinations with a sum that is a multiple of 3 (10 combinations). Therefore, Noel is more likely to draw two cards with a sum that is greater than 13.

The probable question may be:

Noel is playing a game where he draws one playing card each out of two stacks of 5 cards. The table below shows all possible sums for the two numbers on the cards.

3, 6, 7, 9, 1

1 = 4, 7, 8, 10, 12

4 = 7, 10, 11, 13, 15

6 = 9, 12, 13, 15, 17

8 = 11, 14, 15, 17, 19

12 = 15, 18, 19, 21, 23

Is Noel more likely to draw two cards with a sum that is a multiple of 3 or two cards with a sum that is greater than 13?

Noel is more likely to draw two cards with a sum greater than 13, as there are 37 combinations compared to 11 combinations for multiples of 3.

To determine whether Noel is more likely to draw two cards with a sum that is a multiple of 3 or two cards with a sum that is greater than 13, we can analyze the table provided.

1. Sum that is a multiple of 3:

  - There are several combinations of card values that result in a sum that is a multiple of 3. We can count these combinations from the table:

    - (3, 4), (3, 7), (3, 9), (3, 12)

    - (6, 7), (6, 10), (6, 12)

    - (7, 8), (7, 11)

    - (9, 10), (9, 13)

    - (11, 12)

2. Sum greater than 13:

  - We need to count the combinations of card values that result in a sum greater than 13:

    - (6, 8), (6, 10), (6, 11), (6, 12), (6, 13), (6, 15), (6, 17)

    - (7, 8), (7, 9), (7, 10), (7, 11), (7, 12), (7, 13), (7, 15), (7, 17), (7, 19)

    - (9, 10), (9, 11), (9, 12), (9, 13), (9, 15), (9, 17), (9, 19)

    - (11, 12), (11, 13), (11, 15), (11, 17), (11, 19)

    - (12, 13), (12, 15), (12, 17), (12, 19)

    - (13, 15), (13, 17), (13, 19)

    - (15, 17), (15, 19)

    - (17, 19)

After counting the combinations, we can compare the number of combinations in each category:

1. Sum that is a multiple of 3:

  - Total combinations: 11

2. Sum greater than 13:

  - Total combinations: 37

Comparing the totals, we can see that Noel is more likely to draw two cards with a sum that is greater than 13, as there are more combinations resulting in this outcome.

Complete Question:

Answer: C. 4, 1, 2, 3.

I hope this helps :)

Answer: Third option is correct.

Step-by-step explanation:

Since there are four steps in solving one's personal financial challenges:

1) First he needs to assessing the needs and wants .

2) Then, he will consider opportunity costs, i.e. next best alternative.

3) Then, he will assessing the risks and returns associated with the opportunity costs.

4) Last, but not the least, he will be setting short and long term goals.

So, The correct order of these steps is 4,1,2,3.

Hence, Third option is correct.

Answer:

w + 2w = 60

Step-by-step explanation:

w + 2w = 60

This equation will not give you the correct value for W because W is the siez or represents the size of the smaller side, since you have 4 sides, 2 small and 2 big, the sum of this 4 sides will be the perimeter, or the total meters of fence needed, since the total meter of fence need is 60 and in the equation you just have 2 sides, the value for W will be double than the real.

Answer:

74,088 cm³

Step-by-step explanation:

[tex]whl = V \\ \\ [4,2]^{3} = 74,088[/tex]

I am joyous to assist you anytime.

Answer:

36 cm^2

Step-by-step explanation:

Answer:

Option D - $5370.30

Step-by-step explanation:

Given : If the APY of a savings account is 5.3%, and if the principal in the savings account is $5100 for an entire year.

To find : What will be the balance of the savings account after all the interest is paid for the year?          

Solution :

First we find the interest he paid.

The interest formula is [tex]I=P\times R\times T[/tex]

Where, P is the principal P=$5100

R is the rate of interest R=5.3%=0.053

T is the time T=1 year

Substituting the values,

[tex]I=5100\times 0.053\times 1[/tex]              

[tex]I=\$270.3[/tex]                  

The balance of the saving account is the sum of interest paid and principal value.

Amount = Interest +Principal

Amount = $270.3 +$5100          

Amount = $5370.3

Therefore, Option D is correct.