How do I solve a-5/5=5

4.50 /2 = 2.25

11.25/5 = 2.25

22.50/10 = 2.25

 the constant is the price per pound which is 2.25

Answer:

c

Step-by-step explanation:

The set of all points p(x,y) in a coordinate plane that satisfy the given condition refers to the graph that represents the change of y as a function of x.

In a rectangular (Cartesian) xy-coordinate system, a point in a plane is described by a pair of coordinates (x, y). The set of all points p(x,y) that satisfy the given condition refers to the plotted or sketched graph that represents the change of y as a function of x. Using the given table of points (1,5), (2,10), (3,7), and (4,14), we can plot these points on a graph and connect them to see how y values change with different x values.

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sqrt(86) = 9.273

which is between 9 & 10

The correct symbol to complete the expression is,

⇒ 5 + 10 + 13 = 13 + 10 + 5

A relation by which we can compare two or more mathematical expression is called an inequality.

Given that;

The expression is,

⇒ 5 + 10 + 13 ? 13 + 10 + 5

Now,

Since, The expression is,

⇒ 5 + 10 + 13 ? 13 + 10 + 5

From LHS;

⇒ 5 + 10 + 13 = 28

From RHS;

⇒ 13 + 10 + 5 = 28

So, We get;

The correct symbol to use is,

⇒ 5 + 10 + 13 = 13 + 10 + 5

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Answer:

35% of the time

Step-by-step explanation:

A P E X

divide the two:

52.50 / 7 = 7.50 per hour

The area of the given rectangle is 1200 square units.

Given that, width = 30 units and length = 40 units.

The formula of the area of a rectangle is used to find the area occupied by the rectangle within its boundary. The area of a rectangle is obtained by multiplying its length and width. Thus, the formula for the area, 'A' of a rectangle whose length and width are 'l' and 'w' respectively is the product "l × w".

Now, the area of a rectangle = 40 × 30

= 1200 square units

Therefore, the area of the given rectangle is 1200 square units.

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To express the vector v = (-7,8) in the form (|v|cosθ, |v|sinθ), you first correctly found the magnitude |v| using the Pythagorean theorem. Then, to find the angle θ, use the arctangent function applied to the y/x components of your vector. Ensure the angle is in the correct quadrant by modifying it if necessary.

To find the magnitude of the vector v = (-7,8), you've correctly done the Pythagorean theorem and found sqrt(113), which is the absolute value of your vector, written as |v|. Now, to express the vector in the form (|v|cosθ, |v|sinθ), we need to find the angle θ. This is found using the arctangent function (also known as the inverse tangent or atan). θ = atan(y/x), where x and y are the x and y components of your vector respectively. Be careful with the sign as atan can only return values from -π/2 to π/2.

In your case θ = atan(8/-7), but since this gives a negative answer and the vector is in the second quadrant where angles are positive, you should add π to your result to get the correct angle. Then you can substitute |v| and θ into the expressions for the x and y components.

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The correct expression is equivalent to 18a+12b+9a+24b is,

⇒ 27a + 36b

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

The expression is,

⇒ 18a + 12b + 9a + 24b

Now, We can simplify as;

⇒ 18a + 12b + 9a + 24b

⇒ 18a + 9a + 12b + 24b

⇒ 27a + 36b

Thus, The correct expression is equivalent to 18a+12b+9a+24b is,

⇒ 27a + 36b

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Answer:

9.86 milligrams

Step-by-step explanation:

It is given that 1 liter f elixir contains 2 grams of phenobarbital elixir.

Since 202.884 teaspoons = 1 liter

so 1 teaspoon of elixir =  [tex]\frac{1}{202.884}[/tex] liter

∵ 1 liter of elixir contains phenobarbital = 2 gms.

∴  [tex]\frac{1}{202.884}[/tex] liter of elixir contains phenobarbital =  [tex]\frac{2*1}{202.884}[/tex] = 0.009858 grams

or 0.009858 × 10³ milligrams

 = 9.86 milligrams.

Therefore, 9.86 milligrams of phenobarbital will be required.

Final answer:

The milligrams of phenobarbital in one teaspoon of elixir, multiply the milligrams per milliliter (2 mg/mL) by the volume of one teaspoon (5 mL), resulting in 10 milligrams per teaspoon.

Explanation:

The milligrams of phenobarbital in one teaspoon of elixir by first understanding the concentration of the solution and then using the volume conversion. Since 1 liter of the phenobarbital elixir contains 2 grams (2000 mg), we need to find out how many milligrams are contained in a smaller volume, specifically in one teaspoon, which is 5 milliliters. milligrams, so 2 grams = 2000 milligrams.

Next, consider the volume conversion: There are 1000 milliliters in 1 liter. So if 2000 mg is in 1000 milliliters, there would be 2 mg in 1 milliliter (since 2000 ÷ 1000 = 2).

Finally, since 1 teaspoon equals 5 milliliters, we multiply the amount of milligrams in 1 milliliter by 5.

Mathematically: 2 mg/mL * 5 mL = 10 mg

Therefore, there are 10 milligrams of phenobarbital in 1 teaspoon of elixir.

The store sold a total of 948 balls on Saturday and Friday.

Given that,
A sporting good store sells golf balls and boxes each box has 4 compartments of golf balls each compartment has three golf balls the store sold 37 boxes on Friday and 42 boxes on Saturday how many golf balls did the store sell is to be detrmined.

In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,

Here,
A sporting goods store sells golf balls and boxes each box has 4 compartments of golf balls each compartment has three golf balls
Implies
each box has = 3 x 4 = 12 balls

Total box sold = 37 + 42 = 79
Total balls = 79 * 12 = 948

Thus, the store sold a total of 948 balls on Saturday and Friday.

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"The correct counterexample to disprove the inequality [tex]\( n^3 \leq 3n^2 \)[/tex]for real numbers [tex]\( n \)[/tex]is any real number[tex]\( n \)[/tex]greater than 3.

To find a counterexample, we need to show that there exists at least one real number[tex]\( n \)[/tex] for which the inequality [tex]\( n^3 \leq 3n^2 \)[/tex] does not hold. Let's solve the inequality to find the values of[tex]\( n \)[/tex] that satisfy it:

 [tex]\[ n^3 \leq 3n^2 \][/tex]

[tex]\[ n^3 - 3n^2 \leq 0 \][/tex]

[tex]\[ n^2(n - 3) \leq 0 \][/tex]

The critical points of this inequality are [tex]\( n = 0 \) and \( n = 3 \)[/tex]. We can divide the real number line into three intervals:[tex]\( n < 0 \), \( 0 \leq n \leq 3 \)[/tex], and[tex]\( n > 3 \).[/tex]Let's test each interval:

 1. For[tex]\( n < 0 \), \( n^2 \)[/tex] is positive, but [tex]\( n - 3 \)[/tex] is negative, so the product [tex]\( n^2(n - 3) \)[/tex]is negative, and the inequality holds.

 2. For [tex]\( 0 \leq n \leq 3 \)[/tex], both [tex]\( n^2 \) and \( n - 3 \)[/tex] are non-negative, so the product [tex]\( n^2(n - 3) \)[/tex]is non-positive, and the inequality holds.

3. For [tex]\( n > 3 \)[/tex], both [tex]\( n^2 \)[/tex] and [tex]\( n - 3 \)[/tex] are positive, so the product[tex]\( n^2(n - 3) \)[/tex] is positive, and the inequality does not hold.

Therefore, any real number [tex]\( n \)[/tex] greater than 3 will serve as a counterexample to the inequality[tex]\( n^3 \leq 3n^2 \)[/tex]. For instance, if we take [tex]\( n = 4 \),[/tex] we get:

[tex]\[ 4^3 \leq 3 \cdot 4^2 \][/tex]

[tex]\[ 64 \leq 48 \][/tex]

 This statement is false, as 64 is not less than or equal to 48. Thus, \( n = 4 \)[tex]\( n = 4 \)[/tex] is a valid counterexample."

Final answer:

The t-distribution is employed when testing the difference between two population means if the data come from a simple random sample and the population is approximately normally distributed or the sample size is large. It is also used when the population standard deviation is unknown, and a one-tailed or two-tailed test is conducted based on expectations about the direction of the mean difference.

Explanation:

Conditions for Using the t-Distribution to Test Differences Between Two Means

The t-distribution is appropriately used to test the difference between two population means under specific conditions. Firstly, your data must be obtained from a simple random sample. Secondly, the sample should come from a population that is approximately normally distributed, or the sample size should be large enough to compensate for non-normality. This is because the t-test relies on the assumption of normality, especially with smaller sample sizes (<30). If the sample size is small and the population distribution is unknown or not normal, the use of a t-test may not be appropriate without transformation or other techniques to meet the assumptions.

It is also essential to use the t-distribution when the population standard deviation is unknown, and as a result, the sample standard deviation is used as an estimate. Moreover, the t-test can adapt to various research scenarios, considering that it is designed to handle different levels of variability and data types. This flexibility makes the t-distribution a key tool for many scientific studies where the variance is not known ahead of time.

In hypothesis testing, the direction of the mean difference is considered, such as when conducting a one-tailed test or a two-tailed test. A one-tailed test is conducted when an expectation about the direction of the difference exists, while a two-tailed test is used to determine if there is any statistically significant difference between group means without a predefined direction. Lastly, when employing the t-test for paired data, the assumption is that the population from which the differences are drawn is normally distributed.

2.5 hours at a rate of 2.5 miles per hour = 2.5 * 2.5 = 6.25 miles

the trail was 6.25 miles long

Lucy rode at 6 1/5 mile per hour

 so 6.25 / 6 1/5 = 1.008 = 1.01 hours