Algebraically determine if the relation y=10x is symmetric with respect to the x-axis, y-axis, the origin or has no symmetry
A.) y-axis
B.)no symmetry
C.)the origin
D.)x-axis

Answer:

Standard deviation of the distribution of sample means = 7.0855

Step-by-step explanation:

We are given that a sample of size 44 is drawn from a population with mean 36 and standard deviation 47.

Using Central Limit Theorem, it is stated that;

Standard deviation of the distribution of sample means = [tex]\frac{Population S.D.}{\sqrt{n} }[/tex]

         =  [tex]\frac{\sigma}{\sqrt{n} }[/tex]  = [tex]\frac{47}{\sqrt{44} }[/tex] = 7.0855

the answer is (6,-3)

When the square sheet of cardboard is cut and folded to create a rectangular box with a lid, the maximum possible volume is achieved when x = 20 inches. All other values will result in a smaller volume for the box.

Here, we have,

1. Calculate the side lengths of the box and lid:

Box: 40 inches - 2x = 40 - 2x

Lid: 2x

2. Calculate the volume of the box with a lid when x = 20 inches:

Box Volume:

(40 - 2x) x (40 - 2x) x (40 - 2x) = (40 - 2*20) x (40 - 2*20) x (40 - 2*20) = (20) x (20) x (20)

= 8000 in^3

Lid Volume: 2x x 2x x 2x = 2*20 x 2*20 x 2*20

= 8000 in^3

Total Volume: 8000 in^3 + 8000 in^3 = 16000 in^3

3. Calculate the volume of the box with a lid when x = 20/3 inches:

Box Volume: (40 - 2x) x (40 - 2x) x (40 - 2x) = (40 - 2*20/3) x (40 - 2*20/3) x (40 - 2*20/3) = (33 1/3) x (33 1/3) x (33 1/3)

= 7157.7 in^3

Lid Volume: 2x x 2x x 2x = 2*20/3 x 2*20/3 x 2*20/3

= 1365.3 in^3

Total Volume: 7157.7 in^3 + 1365.3 in^3

= 8523 in^3

4. Comparison:

When x = 20 inches, the box has a minimum possible volume of 16000 in^3, while when x = 20/3 inches, the box has a volume of 8523 in^3. Therefore, the statement is true.

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Complete question:

The figure above represents a square sheet of cardboard with side length 40 inches. The sheet is cut and pieces are discarded. When the cardboard is folded, it becomes a rectangular box with a lid. The pattern for the rectangular box with a lid is shaded in the figure. Four squares with side length xx and two rectangular regions are discarded from the cardboard. Which of the following statements is true? (The volume V of a rectangular box is given by V=lwh).

A. When x = 20 inches, the box has a minimum possible volume.

B. When x = 20 inches, the box has a maximum possible volume.

C. When x = 20/3 inches, the box has a minimum possible volume.

D. When x = 20/3 inches, the box has a maximum possible volume.

Answer: 49

Step-by-step explanation:

We need to remember that the median of a set of data is the middle value in the set.

To find the median of a set of data the first step is to arrange the data in order from least to greatest, but, in this case, the set of data given is already arranged from least to greatest:

26,34,38,49,65,75,81

Therefore, you can oberve that the middle value in the set is the following:

26,34,38,49,65,75,81

Then, the median for this set of data is: 49

The answer is provided in the image attached.

To solve for x in 19-24, x = -5. For the angle measurement in question for 25-28, more information is needed.

To solve for x in 19-24, we subtract 24 from 19, which gives us x = -5.

Regarding the measurement of the angle indicated in bold for 25-28, the provided options (a, b, c) seem to be unrelated to this question. Could you please provide more information or clarify your query?

The answer to this question is A.

In a box plot, the values outside box are less likely to occur than the values inside the box. The store a person MOST LIKELY pay $18.75 for a pair of shoes from is given by: Option A: Store A

A box plot has 5 data description.

For all the box-plots plotted in the image attached, we can see that only the first box plot is such that the amount $18.75 is falling inside the box's boundaries.

The box is contained with more likely elements than those which are outside of it.  It is because, the box contains those elements which are between first and third quartile. Since the selection was done randomly, so mean, median and mode all lie approximately on the same point due to the distribution tending to normal distribution(sample size is large enough (50 > 30, a needed sample size for consideration of that sample belonging from normal distribution)). Now, for the first graph, the median is closest to $18.75, and the more close a value is to center of a normal distribution, the more probable it is.

That's why, the store a person MOST LIKELY pay $18.75 for a pair of shoes from is given by: Option A: Store A

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Answer:163.46

Step-by-step explanation:

Group terms that contain the same variable

(x²-8x)=39

Complete the square. Remember to balance the equation by adding the same constants to each side

(x²-8x+16)=39+16

Rewrite as perfect squares

(+/-)]x-4]=√55

(+)]x-4]=√55--------> x=4+√55

(-)]x-4]=√55-------> x=4-√55

the answer is

 (x-4)²-55=0

Answer: (x-4)²=55

Step-by-step explanation:

This is apex answer

Answer:

There can be only one solution

Step-by-step explanation:

The only soution is zero, because anything else is unequal.

i feel like its A but i am dumb

General Idea:

Vertical line test is a test used to determine if a relation is a function. As per vertical line test, a relation is a function if there are no vertical lines that intersect the graph at more than one point.

Applying the concept:

Kelly ask Tyrese to help her use the vertical line test to determine whether or not the curve that she was given it is a function Tyrese informed Kelly that an order for the given curve to be designated as a function of a vertical line drawn through any x value must intersect the Curve [tex] ONE [/tex] time.

Final answer:

The height of the smallest doll in a sequence is 4.116 inches found using the formula based on geometric progression, with each smaller doll having a height of 7/10 the height of the next larger one, starting from the largest at 12 inches.

Explanation:

To find the height of the smallest doll, we use a geometric sequence formula height of smallest doll = height of largest doll × (7/10)^n, with 'n' being the number of dolls smaller than the largest one.

To solve for a specific number of dolls, we'd need the value of 'n'. However, without knowing the total number of dolls in the sequence, we use the given formula to understand the pattern of the dolls' heights.

For illustrative purposes, if there were 3 dolls smaller than the largest, the height of the smallest doll would be 12 × (7/10)^3 = 12 × 0.343 = 4.116 inches. The exact height for the smallest doll would vary based on the total number of dolls in the set.

Answer:

y = 2 (x - 8 )2 + (-72)) .

The x-coordinate of the minimum is 8.

Step-by-step explanation:

I just took this test on plato and I got it correct.

The result of 3 times as much as the sum of 3/4 and 2/6 is; 13/4

First, we must evaluate the sum of 3/4 and 2/6; we have;

Using the lowest common multiple; 12

We have; (9 +4)/12 = 13/12.

Therefore, 3 times 13/12 = 39/12 = 13/4

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Approximately 0.2266, or 22.66%, is the probability that rolling the die more than 140 times is needed to exceed a total of 400.

To approximate the probability that rolling the die more than 140 times is needed to exceed a total of 400, we can use a normal approximation to the binomial distribution since the number of rolls is large.

First, let's calculate the mean (μ) and standard deviation (σ) of the number of rolls needed to exceed 400:

[tex]\[ \text{Mean (μ)} = \frac{\text{Total target}}{\text{Expected value per roll}} = \frac{400}{\frac{7}{2}} \][/tex]

[tex]\[ \text{Standard deviation (σ)} = \sqrt{\frac{\text{Total target} \times (\text{Sides}^2 - 1)}{12}} = \sqrt{\frac{400 \times (6^2 - 1)}{12}} \][/tex]

Now, we'll use the normal approximation and the z-score formula to find the probability:

[tex]\[ z = \frac{\text{X} - \text{μ}}{\text{σ}} \][/tex]

[tex]\[ z = \frac{140 - \text{μ}}{\text{σ}} \][/tex]

Then, we look up the z-score in a standard normal distribution table or use a calculator to find the probability associated with that z-score.

Let's calculate these values.

First, let's calculate the mean (μ) and standard deviation (σ):

[tex]\[ \text{Mean (μ)} = \frac{400}{\frac{7}{2}} = \frac{800}{7} \approx 114.29 \][/tex]

[tex]\[ \text{Standard deviation (σ)} = \sqrt{\frac{400 \times (6^2 - 1)}{12}} = \sqrt{\frac{400 \times 35}{12}} \approx \sqrt{\frac{14000}{12}} \approx \sqrt{1166.67} \approx 34.16 \][/tex]

Now, let's find the z-score for rolling the die more than 140 times:

[tex]\[ z = \frac{140 - \text{μ}}{\text{σ}} = \frac{140 - 114.29}{34.16} \approx \frac{25.71}{34.16} \approx 0.75 \][/tex]

Using a standard normal distribution table or calculator, we find the probability associated with a z-score of 0.75, which represents the probability that rolling the die more than 140 times is needed to exceed a total of 400.

The Correct Question is :

A six-sided die, in which each side is equally likely to appear, is repeatedly rolled until the total of all rolls exceeds 400. Approximate the probability that this will require more than 140 rolls.

The approximate probability that it will require more than 140 rolls for the total to exceed 400 is 0.005.

To find the approximate probability that it will require more than 140 rolls for the total to exceed 400, we can relate it to the probability that the sum of the first 140 rolls is less than 400.

Let X be the random variable representing the sum of the rolls. We want to find P(X > 400), which is the probability that it will require more than 140 rolls.

We can calculate this by finding the complement of the event that the sum of the first 140 rolls is less than 400.

Let A be the event that the sum of the first 140 rolls is less than 400. Then, P(A) is the probability that we're interested in.

Now, we calculate P(A):

Since each side of the die is equally likely, the expected value of the roll is [tex]\( \frac{1+6}{2} = 3.5 \)[/tex].

The expected value of the sum of the first 140 rolls is [tex]\( 140 \times 3.5 = 490 \)[/tex].

Therefore, P(A) can be approximated using the normal distribution, since the sum of the rolls follows approximately a normal distribution due to the Central Limit Theorem.

Using the properties of the normal distribution, we can standardize the value:

[tex]\[ Z = \frac{400 - 490}{\sqrt{140 \times \left(\frac{1}{12}\right)}} \][/tex]

Here, [tex]\( \frac{1}{12} \)[/tex] is the variance of a single roll of the die.

Now, we find P(A) using the standardized value of Z:

P(A) = P(X < 400) = P(Z > z)

We can then find the probability from a standard normal distribution table or calculator.

[tex]\[ P(A) \approx P(Z > -2.589) \][/tex]

From a standard normal distribution table, we find that [tex]\( P(Z > -2.589) \approx 0.995 \)[/tex].

So, the approximate probability that it will require more than 140 rolls for the total to exceed 400 is 1 - 0.995 = 0.005.

The probable question may be:

A six-sided die, in which each side is equally likely to appear, is repeatedly rolled until the total of all rolls exceeds 400. What is the approximate probability that this will require more than 140 rolls? (Hint: Relate this to the probability that the sum of the first 140 rolls is less than 400.)

Answer:

  h(x) = 25x^2 -4

Step-by-step explanation:

Fill in the given arguments and evaluate:

  h(x) = f(5x) -f(-2)

  = ((5x)^2 +3) - ((-2)^2 +3)

  h(x) = 25x^2 -4