A dog trainer compared mean numbers of lessons for two

groups of dogs to learn a new trick.

Mean number of lessons Mean Absolute Deviation (MAD)

Group A 15

3

Group B 9

3

Which statement about the data is true?

A. The MAD for group A is more than the MAD for group B.

B. The mean number of lessons for group A is greater than the mean

for group B by 6 MADs.

C. The mean number of lessons for group A is greater than the mean

for group B by 2 MADs.

D. On average, the dogs in group B required more lessons than the

dogs in group A.

Answer:

110.92

Step-by-step explanation:

Assuming that the parameters (P), (h), and (B) all represent dimensions of the given prism, then based on the given information, the following can be concluded:

P = 6.6

h = 4.5

B = 2.2

The surface area is the two-dimensional area around a three-dimensional surface. In other words, if one was going to wrap the figure, the surface area is the amount of paper one would need. One can find the surface area by finding the area of each individual side and then adding all the results together. To find the area of a 2-dimensional figure by multiplying the length by the width.

(4.5) * (6.8) = 30.6

(4.5) * (6.8) = 30.6

(2.2) * (6.8) = 14.96

(2.2) * (6.8) = 14.96

(4.5) * (2.2) = 9.9

(4.5) * (2.2) = 9.9

Now add up all of the values,

30.6 + 30.6 + 14.96 + 14.96 + 9.9 + 9.9

= 110.92

Answer:

36 kilograms

Step-by-step explanation:

Since she made 8 kilograms of dough over the span of 2 hours, you divide 8 by 2 and get 4 then you have to multiply 9 hours by 4 kilograms of dough to get 36 kilograms of dough.

Yes, if ƒ is differentiable on (0, 1), then ƒ is uniformly continuous on (0, 1).

In mathematics, the concept of differentiability plays a crucial role in understanding the behavior of functions. If a function ƒ is differentiable on the interval (0, 1), it means that the derivative ƒ'(x) exists for every point x in that interval.

The answer states that if a function is differentiable on (0, 1), then it is uniformly continuous on the same interval.

To understand this result, we need to consider the properties of differentiability and uniform continuity.

Differentiability implies that the function has a well-defined tangent line at every point within the interval. This implies that the function cannot exhibit abrupt changes or discontinuities, as it must be smooth and continuous.

Uniform continuity, on the other hand, deals with the behavior of a function as the input values get arbitrarily close to each other. It ensures that the function does not exhibit extreme fluctuations or rapid oscillations.

If a function is differentiable on (0, 1), then it satisfies the conditions required for uniform continuity. This is because the derivative of the function acts as a measure of its rate of change.

If the derivative is bounded (i.e., it does not become infinitely large or small), then the function can be guaranteed to be uniformly continuous.

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Answer:

(6, 4 )

Step-by-step explanation:

Given the 2 equations

9x - 2y = 46 → (1)

x + 2y = 14 → (2)

Adding the 2 equations term by term will eliminate the y- term

10x + 0 = 60

10x = 60 ( divide both sides by 10 )

x = 6

Substitute x = 6 into either of the 2 equations and solve for y

Substituting into (2)

6 + 2y = 14 ( subtract 6 from both sides )

2y = 8 ( divide both sides by 2 )

y = 4

solution is (6, 4 )

Answer:

system is:

y = 1/2x

y = 1/2x - 2

No Solution to this system

Step-by-step explanation:

Value of x corresponding to P81 is 59.06.

A normal distribution has a mean u = 67.3 and a standard deviation of o=9.3.

The task is to find P81, which separates the bottom 81% from the top 19%.

For any normally distributed variable z with mean u and standard deviation o, the cumulative distribution function is defined as the probability of a standard normal variable being less than or equal to z.

A standard normal distribution has a mean of 0 and a standard deviation of 1.

That is, the variable z can be calculated as: z = (x - u) / o

The value P(z < z0) can be read off a standard normal table for any value z0.

As the normal distribution is symmetric, we can use the fact that P(z < -z0) = 1 - P(z < z0).

We now calculate z as follows: z0 = (P81 + 1) / 2 = 0.9051

From a standard normal table, we can see that P(z < 0.9051) = 0.8186.

Therefore, P(z < -0.9051) = 1 - P(z < 0.9051) = 0.1814.

Now we calculate the corresponding value of x:

z = (x - u) / o-0.9051 = (x - 67.3) / 9.3x = 59.06

Therefore, P81 corresponds to the value x = 59.06.

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Answer:

if x is 16 then 3/4(16) - 12

when you simplify 4 by 16

3(4)-12

12-12=0

A wardrobe with 3 pants, 5 shirts, and 7 ties, has a possible outcome of 105 outfits and not 15. So the answer is False

False. The number of total possible outfits is not 15. To calculate the number of possible outfits, we need to multiply the number of choices for each item together. In this case, we have 3 choices for pants, 5 choices for shirts, and 7 choices for ties. Therefore, the total number of possible outfits would be 3 x 5 x 7 = 105.

The statement incorrectly states that there are only 15 possible outfits. It's important to consider that when selecting multiple items, the total number of combinations is found by multiplying the number of choices for each item together. In this scenario, with 3 pants, 5 shirts, and 7 ties, there are 105 possible outfits, not 15.

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It is not possible for a matrix to have the vector (3, 1, 2) in its row space and (2, 1, 1)T in its null space. Let's explain why.

Let A be an m × n matrix, and let x be a nonzero vector in the null space of A, so Ax = 0. We can also say that x is in the null space of A transpose. So x is an element of N(AT).Let’s prove the contradiction that arises from the initial claim by assuming that 3,1,2 is a row vector in the row space of A and 2,1,1 is a column vector in N(AT).We have that A[3 1 2]T = 0 and 2,1,1 is in the null space of A transpose. We also know that if a vector v is in the row space of A, then there exists a vector y such that v = A*y, where y is a column vector. So in this case, we can say that 3,1,2 is in the row space of A if there is a column vector y such that A * y = [3 1 2]T. But if that's the case, then we have the following equation: A* y = [3 1 2]. This can be written as: TA* = [3 1 2]If we then take the transpose of both sides, we have: A* y = [3 1 2]T and TA = [3 1 2]. However, this implies that TA* = TA, which can only be true if A is a symmetric matrix. But A is an m × n matrix, where m and n are not equal, so A cannot be a symmetric matrix. Therefore, it is not possible for a matrix to have the vector (3, 1, 2) in its row space and (2, 1, 1)T in its null space.

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mode is 4 and the median is also 4

mode is the number that appears the most which is 4 and median is the number that is in the middle when you line up all the numbers in order  (0,2,2,3,3,4,4,4,4,5,5,5,6,6,10)

Based on this, a 90% confidence interval for the proportion, p. of adult residents who are parents in this county is 0.787 < p < 0.873.

The point estimate for the proportion is calculated by dividing the number of adults with kids by the total sample size. In this case, the point estimate is 166/200 = 0.83.

To construct the confidence interval, we can use the formula:

[tex]p \pm z \times \sqrt{\frac{p \times (1 - p )}{n}}[/tex]

Where:

p is the point estimate

z is the z-score corresponding to the desired confidence level (90% confidence corresponds to a z-score of approximately 1.645)

n is the sample size

Substituting the values into the formula, we get:

[tex]= 0.83 \pm 1.645 \times \sqrt{\frac{0.83 \times (1-0.83)}{200}}[/tex]

Calculating the values, we can obtain the 90% confidence interval for the proportion of adult residents who are parents.

To construct a 90% confidence interval for the proportion of adult residents who are parents in the county, we can use the sample proportion and the standard error formula. Out of the 200 adult residents sampled, 166 had kids.

we calculate the sample proportion, p-hat:

p-hat = 166 / 200

         = 0.83

Next, we calculate the standard error using the formula:

SE = √((p-hat × (1 - p-hat)) / n)

SE = √((0.83 × (1 - 0.83)) / 200) ≈ 0.025

To construct the confidence interval, we use the formula:

p-hat ± (z × SE)

where, z is the z-score corresponding to the desired confidence level.

For a 90% confidence interval, the z-score is approximately 1.645.

Substituting the values into the formula, we get:

= 0.83 ± (1.645 × 0.025)

Calculating the upper and lower bounds:

Lower bound = 0.83 - (1.645 × 0.025) ≈ 0.787

Upper bound = 0.83 + (1.645 × 0.025) ≈ 0.873

Therefore, the 90% confidence interval for the proportion of adult residents who are parents in the county is approximately 0.787 < p < 0.873.

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Answer:

mean = 5+9+9+6+6+11+8+4/7 = 8.29

median = 6

mode = 6

range = 11 - 4 = 7

Answer:

Step-by-step explanation:

5 , 9 , 6 , 6 , 11 , 8 , 4

Mean = sum of all data ÷ number of data

         [tex]= \frac{5+9+6+6+11+8+4}{7}\\\\= \frac{49}{7}\\\\= 7[/tex]

Median: To find median, arrange in ascending order and medianis the middle term

4 , 5 , 6 , 6 , 8 , 9 , 11

Middle term = 4th term

Median = 6

Mode: a number that appears most often is mode

6 appears 2 times

Mode = 6

Range:

Range = maximum value - minimum value

          = 11 - 4

          = 7

The average fixed cost of each case will be $25

Average fixed cost = fixed cost / number of cases.

$4500/ 180= $25

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Answer:

[tex] \sin(m < q) = \frac{7}{9} \\ \sin(m < q) =(0.77777777778) \\ m < q = { \sin 0.77777777778}^{ - 1} \\ m < q = (51.05755873102)[/tex]

Answer:

27 1/6

Step-by-step explanation:

14 + 12=26

4+3=7

26 7/6= 27 1/6

9514 1404 393

Answer:

  (d)  512√3

Step-by-step explanation:

The area (A) of a regular hexagon can be computed from its side length (s) using the formula ...

  A = (3√3)/2·s²

Here, the side length is given as ...

  s = 32√3/3 = 32/√3

Then the area is ...

  [tex]A=\dfrac{3\sqrt{3}}{2}\cdot\left(\dfrac{32}{\sqrt{3}}\right)^2=\dfrac{3\sqrt{3}\cdot1024}{2\cdot3}=\boxed{512\sqrt{3}}[/tex]

The area of the hexagon is 512√3 square units.

Answer:  h < 6

Step-by-step explanation:

3h - 5 < 13

3h < 18

h < 6

Answer:

h<6

Step-by-step explanation:

Answer:

Upper right corner.

Step-by-step explanation:

I took the test and that one was right. Hope this helps!

In the given situation, "angle 4 is the angle of elevation from the radar tower to the person" is the description of angle 4.In trigonometry, an angle of elevation or inclination is the angle between the horizontal and the line of sight of an observer looking upwards. An angle of depression is the angle between the horizontal and the line of sight of an observer looking downwards.

In the given situation, angle 4 refers to the angle formed between the horizontal and the line of sight from the radar tower to the person. As the angle is formed while looking upwards from the radar tower to the person, it is called the angle of elevation. Hence, the correct description of angle 4 in this situation is "angle 4 is the angle of elevation from the radar tower to the person."

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Answer:

80 minutes.

Step-by-step explanation:

do I need to explain? I hate explaining :(

The principal should reject her original hypothesis, as the lower bound of the interval is less than 110.

The t-distribution is used when the standard deviation for the population is not known, and the bounds of the confidence interval are given according to the equation presented as follows:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

The variables of the equation are listed as follows:

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 20 - 1 = 19 df, is t = 2.8609.

The parameters for this problem are given as follows:

[tex]\overline{x} = 108, s = 10, n = 20[/tex]

Then the lower bound of the interval is given as follows:

[tex]108 - 2.8609 \times \frac{10}{\sqrt{20}} = 101.6[/tex]

The upper bound of the interval is given as follows:

[tex]108 + 2.8609 \times \frac{10}{\sqrt{20}} = 114.4[/tex]

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Test the claim that the proportion of people who own cats is significantly different than 70% at the 0.02 significance level. The p-value is 0.024.

The null and alternative hypotheses for the claim that the proportion of people who own cats is significantly different than 70% at the 0.02 significance level are:

H0: p = 0.7 (null hypothesis

)H1: p ≠ 0.7 (alternative hypothesis)

The test is a two-tailed test because the alternative hypothesis includes not equal to (<>) which means either p is less than 0.7 or greater than 0.7

Based on a sample of 500 people, 62% owned cats.

This means that the sample proportion, p = 0.62.

To calculate the p-value, we will use the z-test statistic.

The formula for calculating the z-test statistic is given as:

z = (p - P) / √(PQ/n) where P is the hypothesized proportion (P = 0.7), Q is the complement of P (Q = 1 - P), and n is the sample size.

Using the given values in the formula, we have; z = (0.62 - 0.7) / √(0.7 × 0.3 / 500) = -2.52

The p-value for a two-tailed test at 0.02 level of significance is obtained from the standard normal table.

The area in both tails beyond the z-score of 2.52 is 0.012.

Therefore, the p-value is:

p-value = 2 × 0.012 = 0.024

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Answer:

67-32=a

if his mom is 67 and he is 32 years younger 67 minus 32 would equal a which is his age

Answer:

The ratio of perimeter of ABCD to perimeter of WXYZ = [tex]\frac{2}{3}[/tex]

Step-by-step explanation:

First, we have to determine the multiplicative factor of the dimensions for both figures.

Considering sides AB and WX,

multiplicative factor = [tex]\frac{12}{8}[/tex]

                                = 1.5

So that:

XY = 6 x 1.5 = 9

YZ = 7 x 1.5 = 10.5

ZW = 7 x 1.5 = 10.5

Perimeter of ABCD = 6 + 7 + 7 + 8

                                = 28

Perimeter of WXYZ = 9 + 10.5 + 10.5 + 12

                                 = 42

The ratio of the perimeters of the two quadrilaterals can be determined as;

ratio = [tex]\frac{perimeter of ABCD}{Perirmeter of WXYZ}[/tex]

       = [tex]\frac{28}{42}[/tex]

       = [tex]\frac{2}{3}[/tex]

The ratio of the perimeter of ABCD to perimeter of WXYZ is [tex]\frac{2}{3}[/tex].

Answer:

I belive the answer is A

Step-by-step explanation:

So any answer with 22t would make sense, so you have A and C. In C though, it is subtracting 22, but since 6195 is the total it would have to include the 22 so it is A.

Our assumption that the average of four real numbers is less than all of the numbers is false. By contradiction, we conclude that the average of four real numbers is greater than or equal to at least one of the numbers.

To prove the statement using a proof by contradiction, we assume the opposite, namely, that the average of four real numbers is less than all of the numbers. Let's denote the four numbers as a, b, c, and d. We assume that the average of these numbers, which we'll denote as avg, is less than a, b, c, and d.

Now, let's consider the sum of these four numbers: a + b + c + d. The average of these numbers, avg, is calculated by dividing the sum by 4. Therefore, we have avg = (a + b + c + d)/4.

If avg is less than a, b, c, and d, then (a + b + c + d)/4 < a, (a + b + c + d)/4 < b, (a + b + c + d)/4 < c, and (a + b + c + d)/4 < d.

Now, let's consider the sum of these inequalities: (a + b + c + d)/4 + (a + b + c + d)/4 + (a + b + c + d)/4 + (a + b + c + d)/4 < a + b + c + d.

Simplifying the left-hand side, we have (a + b + c + d) + (a + b + c + d) + (a + b + c + d) + (a + b + c + d) < 4(a + b + c + d).

This simplifies to 4(a + b + c + d) < 4(a + b + c + d), which is a contradiction. The left-hand side is greater than the right-hand side, which contradicts our initial assumption.

Therefore, our assumption that the average of four real numbers is less than all of the numbers is false. By contradiction, we conclude that the average of four real numbers is greater than or equal to at least one of the numbers.

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If We consider the string w = [tex]CA^pG^pT^pC[/tex], then L ∉ RLs by pumping lemma.

To prove that L ∉ RLs using the RL pumping theorem, we assume L is a regular language and apply the pumping lemma for RLs. Let p be the pumping length of L.

We consider the string w = [tex]CA^pG^pT^pC[/tex], where |w| ≥ p. According to the pumping lemma, we can decompose w into uvxyz such that |vxy| ≤ p, |vy| > 0, and for all k ≥ 0, the string [tex]u(v^k)x(y^k)z[/tex] is also in L.

However, by examining the structure of L, we see that the number of A's and G's is dependent on each other and must match the number of T's.

Since pumping up or down would alter this balance, there is no way to satisfy the condition for all k, leading to a contradiction. Therefore, L cannot be a regular language, and we conclude that L ∉ RLs.

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