given an exponential function for compounding interest, A(x)=P(1.02)^x, what is the rate of change?

1. The algebraic property of equality that is represented is the multiplication property. The answer is letter P.

2. The algebraic property of equality that is represented is the subtraction property. The answer is letter E.


3. The algebraic property of equality that is represented is the substitution property. The answer is letter M.


4. The algebraic property of equality that is represented is the division property. The answer is letter A.


5. The algebraic property of equality that is represented is the addition property. The answer is letter A.


6. The algebraic property of equality that is represented is the distributive property. The answer is Letter E.


7. The algebraic property of equality that are represented is the transitive property. The answer is letter is E.


8. The algebraic property of equality that are represented is the symmetric property. The answer is Letter A.


9. The algebraic property of equality that are represented is the reflexive property. The answer is letter C.


10. The algebraic property of equality that are represented by is the multiplication property. The answer is letter P.


11. The algebraic property of equality that are represented is the subtraction property. The answer is letter L.

12. The algebraic property of equality that are represented is the reflexive property. The answer is letter S.

13. The algebraic property of equality that are represented is the transitive property. The answer is letter U.


14. The algebraic property of equality that are represented is the symmetric property. The answer is letter K.

The given is 3 5 14 2  4 1 10 11 6 12 8 13 9 7

Plugging in the letters we got for each number, gives us MAKE APPLESAUCE.

As the number of apples is less than the total number of friends among which the apples need to be distributed so, no one will get a complete piece of apple so, 5 apples can be shared in between 7 friends as [tex]\frac{5}{7}[/tex].

According to the question, the total number of apples are 5 and the total number of friends are 7.

So, each friend will get (bu unitary method)-

[tex]\rm{Each\;friend\;will\;get}=\dfrac{5}{7}\;\rm apples[/tex]

Learn more about the unitary method here:

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Answer:

200 mi.

Step-by-step explanation:

200 mi. is more or greater than 200 km. Because, 200 mi. in km will be 321 km so that is greater than 200 km.  

I hope this helps you.

By calculating the volume of each ball bearing and considering the box's capacity, we determine that a box can theoretically hold 17,276 ball bearings. However, accounting for random packing efficiency, a box can actually hold approximately 11,059 ball bearings.

To determine how many ball bearings can fit in a box, we need to calculate the volume each ball bearing occupies and then see how many of these volumes can fit into the box's total capacity.

Additionally, taking into account the closest random packing efficiency of approximately 0.64 for loose ball bearings:

Answer:

[tex]y=25[/tex]

Step-by-step explanation:

We have been given that y varies inversely as x and [tex]y=50[/tex], when [tex]x=10[/tex].

We know that when y varies inversely with x, then the equation is: [tex]y=\frac{k}{x}[/tex], where, k represents constant of variation.

First of all, we will find constant of variation using point [tex]y=50[/tex], and [tex]x=10[/tex] as shown below:

[tex]50=\frac{k}{10}[/tex]

[tex]50*10=\frac{k}{10}*10[/tex]

[tex]500=k[/tex]

Upon substituting [tex]k=500[/tex] in inversely proportion we will get:

[tex]y=\frac{500}{x}[/tex]

To find the value of y, we will substitute [tex]x=20[/tex] in above equation as:

[tex]y=\frac{500}{20}[/tex]

[tex]y=\frac{50}{2}[/tex]

[tex]y=25[/tex]

Therefore, the value of y is 25.

the pool is 12 x 15

 add 6ft to each dimension ( 3 feet on each side)

12+6 = 18

15+6 = 21

21 *2 = 42

18*2 = 36

42 +36 = 78 feet perimeter

Answer:

quotient=2,731

Remainder=0

Step-by-step explanation:

893) 2,438,783 (2,731

      - 1 786

      ---------

          6527

        - 6251

        ---------

            2768

          - 2679

         -----------

                893

             -  893

             ---------

                 0

Answer and step-by-step explanation:

Yes, it does have practical significance.

If the method was ineffective, the chance of an increase this large was 3/10, or 30%.  This means with an ineffective teaching method, there was a 100-30 = 70% chance of not seeing a score increase this large.

Since we did see a result this large, and 30% is not a very large chance, we can say that the method was most likely effective.

The difference in SAT scores is statistically significant because there's only a 3/10 chance of observing such an increase by chance. However, it might lack practical significance because a 6-point increase, although positive, is relatively small and may not have a meaningful impact in real-world terms.

The question you're asking relates to the concept of statistical significance and practical significance in Mathematics. Statistical significance means that the difference in results (in this case, the increase in SAT scores from 475 to 481) isn't likely due to chance. You've mentioned that there's a 3 in 10 chance of observing such an increase, so statistically, the new teaching method seems to have a significant effect.

However, practical significance refers to whether the difference is meaningful in real-world terms. This depends on context and judgement. In this case, a 6-point increase in the SAT score is a positive change, but it's relatively small and might not make a substantial difference to university admissions or student's mathematical competence. Therefore, one could argue that while the result is statistically significant, it may lack practical significance.

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An equation with all real numbers as solutions is one that, regardless of what value a variable might have, both sides will always be equal, such as 2x+3 = 2x+3. An equation with no real solutions is a statement where the two sides cannot ever be equal, like 2x+5 = 2x+7.

An example of an equation with variables on both sides that has all real numbers as the solution is a statement where both sides are identical. For instance, 2x+3 = 2x+3. All real numbers can satisfy this equation for x because both sides are always equal, regardless of what your value of x is.

On the other hand, an equation with no real solutions is one where it's impossible to find any real number that satisfies the equation. An example can be 2x+5 = 2x+7. No matter what value you put in place of x, the two sides of the equation will never be equal. Hence, this equation has no real solution.

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Answer:

The correct option is A.

Step-by-step explanation:

The given is a isosceles trapezium because the length of two non-parallel sides are equal.

An isosceles trapezium has a axis of symmetry which divides the trapezium in two equal and symmetric parts.

The axis of symmetry intersecting the parallel line at their midpoint.

From the below figure we can say that the sides AB and CD are parallel and y=-3 is the axis of symmetry.

So, when we respect the figure across y= -3, then we get the same figure.

Rotation 108 degree across any point will never maps the figure onto itself.

Therefore the reflection across the line y = -3 maps the figure onto itself and option A is correct.

Final answer:

To find the three consecutive odd integers whose sum is between 27 and 51, set up an inequality with n representing the smallest integer. Solve the inequality to find the possible values for n, which are then used to derive the consecutive integers.

Explanation:

To find three consecutive odd integers whose sum lies between 27 and 51, we need to establish variables for the integers and then solve for them. Let's call the smallest integer n, the next integer will then be n+2 (since odd numbers are two units apart), and the largest integer will be n+4. The equation will be n + (n+2) + (n+4), which simplifies to 3n + 6.

Since their sum must be between 27 and 51, our inequalities are:
27 < 3n + 6 < 51
Subtract 6 from all parts of the inequality:
21 < 3n < 45
Now, divide everything by 3:
7 < n < 15

Since n must be an odd integer, we have the options 9, 11, or 13 for n. Therefore, our sets of three consecutive odd integers could be:
(9, 11, 13), (11, 13, 15), or (13, 15, 17).

1. x= 11.7  μ = 7

2. z11.7= 1.3

3. Is 11.7 within a z-score of 3?

a. Yes because z11.7 < 3.

4. Which statement is true of z11.7?

b. z11.7 is between 1 and 2 standard deviations of the mean.

The value of the z-score is 1.33.

The z-score is defined as the mean (average) values, measured in terms of standard deviations from the mean.

The formula is used to calculate z-score is;

Z-score = X-μ / σ

= 11.7-7 / 3.52

= 4.7/3.52

= 1.33

Hence, the value of the z-score is 1.33.

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The y-intercept of the function g(x) = (1/4)^x is (0,1), and the horizontal asymptote is at y=0.

The correct option is (c).

Given the function g(x) = (1/4)^x, we need to find the y-intercept and the equation of the horizontal asymptote.

The y-intercept occurs where x = 0. Substituting 0 for x in g(x), we get:

g(0) = (1/4)^0 = 1

So the y-intercept is at the point (0,1).

For horizontal asymptotes, we look at the behavior of the function as x goes to positive or negative infinity. In this case, as x goes to infinity, (1/4)^x approaches 0. Therefore, the horizontal asymptote of g(x) is at y = 0.

The correct answer is c. The y-intercept is (0,1) and the asymptote is at y=0.

The constraints for this problem are represented by the inequalities x > 3y and 15x + 25y ≥ 700. The combinations that satisfy the constraints is (45,10) as it meets both conditions.

The constraints for the art store situation are represented by the inequalities x > 3y and 15x + 25y ≥ 700. The first inequality, x > 3y, represents the condition that the number of small prints made must be more than three times the number of large prints sold. The second inequality, 15x + 25y ≥ 700, characterizes the fact the store needs to make a daily profit of at least $700 in order to cover costs.

With these inequalities, we can determine which pairs of small print and large print quantities satisfy this system of constraints. Checking each pair, we find that the combination (45,10) satisfies both inequalities as 45 is more than three times 10, and 15*45 + 25*10 = $775, which is greater than $700. Thus, selling 45 small prints and 10 large prints will satisfy the constraints.

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Answer:

-1/4 should be the answer

Step-by-step explanation:

12 months per year

3200/12 = 266.66 per month

jan/feb = 2 months

266.666 x 2 = 533.33 round off to 533 cars sold

If 3,200 cars were sold all year, then approximately 533 cars were sold in the Jan/Feb period.

We need to calculate the proportion of the year that these two months represent and then apply that proportion to the total number of cars sold in the year.

Since there are 12 months in a year, the Jan/Feb period represents [tex]\(\frac{2}{12}\)[/tex] of the year. We can simplify this fraction to [tex]\(\frac{1}{6}\)[/tex] by dividing both the numerator and the denominator by 2.

To find the number of cars sold in the Jan/Feb period, we multiply the total number of cars sold in the year by the fraction representing Jan/Feb:

[tex]\[ \text{Number of cars sold in Jan/Feb} = \text{Total cars sold} \times \frac{2}{12} \][/tex]

Substituting the given total number of cars sold:

[tex]\[ \text{Number of cars sold in Jan/Feb} = 3,200 \times \frac{2}{12} \][/tex]

[tex]\[ \text{Number of cars sold in Jan/Feb} = 3,200 \times \frac{1}{6} \][/tex]

[tex]\[ \text{Number of cars sold in Jan/Feb} = \frac{3,200}{6} \][/tex]

Performing the division:

[tex]\[ \text{Number of cars sold in Jan/Feb} = 533.\overline{3} \][/tex]

[tex]\[ \text{Number of cars sold in Jan/Feb} \approx 533[/tex]