What are some examples of distrbutive property, asosiative property and idenity property. i always get them mixed up Help!!

Final answer:

The ordered pair (-1, -4) satisfies the second equation (7x - y = -3) but doesn't satisfy the first equation (x - y = 3), so it is not a solution to the system of linear equations.

Explanation:

When evaluating whether the ordered pair (-1, -4) is a solution to the system of equations x - y = 3 and 7x - y = -3, we must substitute x and y into both equations:

Given these results, the ordered pair (-1, -4) satisfies the second equation but not the first.

There are 3 students in the class of 27 who neither like video games nor cartoons.

To find the number of students who do not like either video games or cartoons, we first need to understand that some students may like both. According to the data given, 16 students like video games, 20 like cartoons and 12 like both. So, the total liking either or both is 16 (video games) + 20 (cartoons) - 12 (both), which is 24 students. The number of students who do not like either would then be the total number of students in the class minus those who like either or both. Therefore, the number of students who do not like either video games or cartoons is 27 (total students) - 24 (liking either or both) = 3 students.

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Final answer:

To find out what percent 70.59 is of 19.5, divide 70.59 by 19.5 and multiply the result by 100. Doing this calculation, 70.59 is 362.05% of 19.5.

Explanation:

To determine what percent 70.59 is of 19.5, we use the formula:

Percentage = (Part / Whole)  imes 100

In this case, the 'part' is 70.59 and the 'whole' is 19.5. Plugging in the numbers:

Percentage = (70.59 / 19.5)  imes 100

Calculating this gives us:

Percentage = (3.6205)  imes 100

Percentage = 362.05%

So, 70.59 is 362.05% of 19.5.

Will can use his fund-raising thermometer-like graphics to compare which basketball team is better at free throws by seeing which thermometer is filled up more, representing a higher success rate. For perspective, a player like Helen who makes 75 percent of her free throws is considered highly proficient, so the team with a similar or higher percentage would be better. However, Will should also consider the number of attempts to ensure the comparison is fair.

To help Will determine which basketball team is better at free throws using the pictures he drew, we can look at the statistical method of comparing percentages. If Will has drawn two fund-raising thermometer-like graphics to represent free throw success rates for the two teams, these can be compared directly to see which team has a higher percentage of successful free throws. Assuming the thermometers are accurately scaled, the one that is filled up more represents the team with the higher success rate.

Let's use Helen as an example. Helen makes her free throws 75 percent of the time, which means that if she were to take 100 shots, she would make approximately 75 of them. If Will's thermometers for the two teams show different percentages, the team with the percentage closer to Helen's or higher would be considered better at free throws.

In addition to comparing percentages, to fully understand the effectiveness of the teams at free throws, Will might also want to consider the number of attempts each team has made, as a team with a high percentage but a very low number of attempts might not be as proficient as a team with a slightly lower percentage but a much higher number of attempts. This is because percentages can sometimes be misleading when the sample size is small.

Will can compare the free throw percentages of two basketball teams using his pictorial thermometer method to determine which team is better at free throws, similar to how we use Helen's 75% free throw success rate to assess her performance.

The subject of the question is Mathematics, specifically revolving around percentages and probability in the context of basketball free throws. To help Will decide which team is better at free throws using a pictorial method like the fund-raising thermometer, we must understand how to use percentages to compare performance. Helen, for example, makes her free throws 75 percent of the time, so if she were to attempt two shots, the probability that she makes the first shot, which we are calling event C, would be 0.75 or 75%. If Will has drawn similar thermometers for two teams that show their free throw percentages, he would simply need to compare the percentages to determine which team is more successful at free throws. The higher the percentage, the more proficient the team is at making free throws.

To find the reduction in Kerrin's weight as a percentage of his initial weight, we convert the weights to pounds, find the weight difference, then this difference is divided by the initial weight and multiplied by 100. Kerrin lost about 7.41% of his initial weight.

To calculate the reduction in Kerrin's weight as a percentage of his initial weight, we first need to convert the weights from stones and pounds into just pounds. At the start of the diet, Kerrin weighed 13 stone 9 pounds. As there are 14 pounds in a stone, this equates to (13 x 14) + 9 = 189 pounds. Three months later, he weighed 12 stone 7 pounds which equates to (12 x 14) + 7 = 175 pounds.

The total amount of weight lost is then calculated by subtracting his later weight from his starting weight which gives us 189 - 175 = 14 pounds. The change in weight as a percentage of the initial weight can then be calculated by dividing the weight difference by the initial weight, then multiply the result by 100 to give the percentage. The resulting calculation is (14 / 189) x 100 = 7.407407407407407 which is approximately 7.41%. Therefore, Kerrin lost about 7.41% of his initial weight after three months on the diet.

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To represent 5.5 divided by 5, you draw a line with 5 equally spaced hash marks, place 5.5 dots evenly across the marks (with an extra half dot marking the .5), showing each section contains 1.1, which is the result of the division.

To solve 5.5 divided by 5 using a model, one approach is to draw a line and label it with five equally spaced hash marks to represent the number 5. Then, place 5.5 dots on the line, disseminated evenly across the five hash marks. The model shows that each of the 5 sections contains 1 dot, and the final 0.5 dot is placed halfway between the fifth hash mark and the sixth (unmarked) section. We do this because 5.5/5 equals 1.1, and this model visually represents the fact that 5.5 divided by 5 equals approximately 1.1.

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Answer:

15

Step-by-step explanation:

The number 48 is 30 percent of the number 160.

We have to find 30% of what number is 48.

let the unknown number be x.

so, we can set up the equation as

30% of x= 48

30/100 x = 48

30x = 4800

Divide both side by 30

30x/30= 4800/ 30

Now, simplify

x= 120

Thus, 48 is 30% of 160.

Learn more about Percentage here:

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Final answer:

To determine the length of any side of the square bedroom with an area of 169 square feet, the square root of the area is calculated, revealing that each side is 13 feet long.

Explanation:

To find the length of any side of the square bedroom, since the area of Davis' square bedroom is 169 square feet, we need to take the square root of the area. A square room has all four sides of equal length, meaning the length can be found by the equation:

Area = length × length or length2

Taking the square root of both sides, we get:

length = √Area = √169

The square root of 169 is 13, so the length of any side wall in Davis' bedroom is 13 feet.

We can use ratio

If the person works for 12 hours and earns $510, for one hour:

12*1=510*x

x=510/12

x= $42.5

Thus the person earns 42.5 dollars per hour.

Your answer is A) $42.50/1 hour.

⁷/₁₀ kg

Given:

Question:

How many kilograms of cherries did they buy altogether?

The Process:

Since Linda bought [tex]\frac{1}{10} \ kg[/tex] of cherries less than Tiffany, we subtract [tex]\frac{1}{10} \ kg[/tex] from the [tex]\frac{2}{5} \ kg[/tex] of cherries Tiffany bought.

The least common multiple (LCM) of 5 and 10 is 10.

[tex]\boxed{ \ \frac{2}{5} - \frac{1}{10} = ? \ }[/tex]

[tex]\boxed{ \ = \bigg( \frac{2 \times 2}{5 \times 2} \bigg) - \frac{1}{10} \ }[/tex]

[tex]\boxed{ \ = \bigg( \frac{4}{10} \bigg) - \frac{1}{10} \ }[/tex]

[tex]\boxed{ \ = \bigg( \frac{3}{10} \ }[/tex]

Therefore, Linda bought [tex]\frac{3}{10} \ kg[/tex] of cherries.

Let us find out how many kilograms of cherries did they buy altogether.

[tex]\boxed{ \ Tiffany's cherries + Linda's cherries =? \ }[/tex]

[tex]\boxed{ \ \frac{2}{5} + \frac{3}{10} = ? \ }[/tex]

[tex]\boxed{ \ = \bigg( \frac{2 \times 2}{5 \times 2} \bigg) + \frac{3}{10} \ }[/tex]

[tex]\boxed{ \ = \bigg( \frac{4}{10} \bigg) + \frac{3}{10} \ }[/tex]

[tex]\boxed{\boxed{ \ = \bigg( \frac{7}{10} \ }}[/tex]

Thus, they bought [tex]\frac{7}{10} \ kg[/tex] of cherries altogether.

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Using Diagram

[tex]\boxed{Tiffany's cherries: \frac{2}{5} = \frac{4}{10}}[/tex]

[tex]Tiffany's cherries: \boxed{\circ}\boxed{\circ}\boxed{\circ}\boxed{\circ}[/tex] or 4 of 10 units.

[tex]\boxed{Linda's cherries = \frac{1}{10}}[/tex] less than Tiffany's, so we subtract 1 from 4 units of Tiffany's, that is [tex]\boxed{\circ}\boxed{\circ}\boxed{\circ}[/tex]

[tex]\boxed{ \ Tiffany's cherries + Linda's cherries =? \ }[/tex]

[tex]\boxed{\circ}\boxed{\circ}\boxed{\circ}\boxed{\circ} + \boxed{\circ}\boxed{\circ}\boxed{\circ} = \boxed{\circ}\boxed{\circ}\boxed{\circ}\boxed{\circ}\boxed{\circ}\boxed{\circ}\boxed{\circ}[/tex]

Thus, the number of the two diagrams is 7 out of 10 units or [tex]\frac{7}{10} \ kg[/tex] of cherries.

Keywords: Tiffany, bought, 2/5, kg of cherries, Linda, 1/10, less than, how many, kilograms, they buy altogether, 3/10, 7/10, LCM

A new exhibit at a local zoo contains 12 ostriches. This is four more than twice the number of elephants. How many elephants are in the exhibit?

Answer: To make $2,670,474 you would need the following number of coins:

• 10,681,896 quarters

• 26,704,740 dimes

• 53,409,480 nickels and

• 267,704,740 penny's.

Step-by-step explanation:

Answer:

Nonlinear

Step-by-step explanation:

I took the k12 test and I have proof.