Use the following information to answer the exercise. A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour.What are the dependent and independent variables?

Answer:

F(x=11)= (-31)

Step-by-step explanation:

for the function

F(x) = x² - 13*x + 22/x-11 , for x ≠ 11

then in order to define F(x=11) so  F is continuous (see Note below) . By definition of continuity of a function:

F(x) is continuous in x=11 if lim F(x)=F(a) when x→a

then

when x→a , lim x² - 13x + 22/x-11 = lim 11² - 13*11 + 22/11 -11 = -3*11 + 2 = -31 = F(x=11)

then

F(x=11)= (-31)

Note:

F is not continuous in all x since

when x→0⁺ ,  lim (0⁺) ² - 13*0⁺  + 22/0⁺ -11 = (+∞)

when x→0⁻,  lim (0⁻) ² - 13*0⁻  + 22/0⁻ -11 = (-∞)

then

limit F(x) , when x→0 does not exist since the limit from the left and from the right do not converge → since the limit does not exist , the function is not continuous  in x=0

Answer:

8.99 days elapsed. Option (C) is correct

Step-by-step explanation:

The distribution  has density function k/t+1 for a constant k and t between 0 and 6.5 . Since the distribution is symmetrical in 6.5, the area it forms between 0 and 6.5 should be 1/2, thus

[tex]\frac{1}{2} = \int\limits_0^{6.5} \frac{k}{t+1} \, dt = k *(ln(t+1) \, |_0^{6.5}) = k * (ln(7.5)-ln(1)) = k*ln(7.5)[/tex]

Hence k = 1/(2ln(7.5)), approx 1/4.

We need to find the percentil 0.6, since the integral of the random variable is 1/2 over the first half, we need to find t such that the integral of the random variable between o and 6.5 + t is 0.6. This is equivalent to find t such that the integral between 6.5 and 6.5+t is 0.1. Due to the  over 6.5, this t should satisfy that the integral between 6.5-t and 6.5 is also 0.1. Lets compute the integral and find t

[tex]\int\limits^{6.5}_{6.5-t} {\frac{k}{t+1}} \, dx = \frac{1}{2ln(7.5)}*(ln(t+1) \, |_{6.5-t}^{6.5} \, ) = \frac{1}{2ln(7.5)} * (ln(7.5)-ln(7.5-t)) = \\\frac{1}{2} - \frac{ln(7.5-t)}{2ln(7.5)} = 0.1[/tex]

Therefore,

[tex]\frac{ln(7.5-t)}{2ln(7.5)} = 0.4\\\\ln(7.5-t) = 0.8*ln(7.5)\\\\7.5-t = e^{0.8*ln(7.5)}\\\\t = 7.5-e^{0.8*ln(7.5)} = 2.49[/tex]

As a result, the student sould have registered 2.49 days after the day 6.5, thus it should have registeredd at day 8.99. Option (C) is correct.

Answer:

51

Step-by-step explanation:

The possible components to sum up to 101 can only be divided into 2 groups, 1 is larger than 50 and the other is less than or equals to 50

For example

50 + 51 = 101

52 + 49 = 101

60 + 41 = 101

...

99 + 2 = 101

100 + 1 = 101

Therefore, the worst case scenario is to pick all numbers from only 1 group, either all number less than or equal to 50, which there are 50 of them from 1 to 50, or greater than 50, which there are 50 of them from 51 to 100.

So k has to be at least 51 to guarantee that at least one pair of the selected integers will sum to 101

The smallest value of k that guarantees at least one pair of the selected integers will sum to 101 is 100.

In order to guarantee that at least one pair of the selected integers will sum to 101, we need to find the largest value of k under which it is still possible for all pairs of integers to have a sum less than or equal to 100.

Let's consider the scenario where k is equal to 99. We can select 99 integers between 1 and 100, and it is still possible for none of the pairs to have a sum of 101. Every integer can be paired with one other integer to form a sum less than or equal to 100 (e.g., 1 and 100, 2 and 99, 3 and 98, and so on).

However, if we select one more integer to make k equal to 100, there will be at least one pair that adds up to 101, since we have one extra integer that cannot form a sum less than 101 with any of the other integers.

Therefore, the smallest value of k that guarantees at least one pair of the selected integers will sum to 101 is 100.

Answer:

1) The solution of the system is

[tex]\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right[/tex]

2) The solution of the system is

[tex]\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right[/tex]

Step-by-step explanation:

1) To solve the system of equations

[tex]\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right[/tex]

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

[tex]\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]

Step 2: Swap rows 1 and 2

[tex]\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]

Step 3:  [tex]\left(R_1=\frac{R_1}{6}\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]

Step 4: [tex]\left(R_3=R_3-R_1\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right][/tex]

Step 5: [tex]\left(R_2=\frac{R_2}{3}\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right][/tex]

Step 6: [tex]\left(R_3=R_3+R_2\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right][/tex]

Step 7: [tex]\left(R_3=\left(\frac{6}{37}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]

Step 8: [tex]\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]

Step 9: [tex]\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]

Step 10: Rewrite the system using the row reduced matrix:

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right[/tex]

2) To solve the system of equations

[tex]\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right[/tex]

using the row reduction method you must:

Step 1:

[tex]\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]

Step 2: [tex]\left(R_1=\frac{R_1}{4}\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]

Step 3: [tex]\left(R_2=R_2-\left(2\right)R_1\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]

Step 4: [tex]\left(R_3=R_3-R_1\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]

Step 5: [tex]\left(R_2=\left(2\right)R_2\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]

Step 6: [tex]\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]

Step 7: [tex]\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right][/tex]

Step 8: [tex]\left(R_3=\left(- \frac{2}{117}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]

Step 9: [tex]\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]

Step 10: [tex]\left(R_2=R_2-\left(15\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]

Step 11:

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right[/tex]

Answer:

a) 117.4

b) 117.9

c) Option A)  When there is rounding or grouping, the median can be highly sensitive to small change

Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]

n = 9

a) Median of the reported blood pressure values

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4

Median =

[tex]\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4[/tex]

b) New median of the reported values

Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

[tex]\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9[/tex]

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A)  When there is rounding or grouping, the median can be highly sensitive to small change

The median of the reported blood pressure values is 117.4. If the blood pressure of the second individual is changed to 117.9, the new median would be 117.9. This shows that the median is not sensitive to small changes in rounding or grouping.

(a) What is the median of the reported blood pressure values?

To find the median, we need to arrange the blood pressure values in numerical order: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4. Since there are 9 values, the middle value is the 5th value, which is 117.4.

(b) Suppose the blood pressure of the second individual is 117.9 rather than 117.4. What is the new median of the reported values?

If we change the second individual's blood pressure to 117.9, the new order is: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4. Now, there are 9 values and the middle value is the 5th value, which is 117.9. So the new median would be 117.9.

(c) What does this say about the sensitivity of the median to rounding or grouping in the data?

C. When there is rounding or grouping, the median is not sensitive to small changes.

https://brainly.com/question/33918236

#SPJ3

Answer:

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

Step-by-step explanation:

For each of the entries which are letters, there are 26 possible outcomes.

For each of the entries which are digits, there are 10 possible outcomes.

These outcomes can be permutated.

For example, ABC1234 is a different outcome than A1B2C34. This means that we need to use the permutations formula.

The number of permutations of n, divided into two groups of size a and b, is:

[tex]P_{a,b}^{n} = \frac{n!}{a!b!}[/tex].

In this problem, we have a permutation of 7, divided into a group of 4(digits) and 3(letters).

How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits?

This is [tex]P_{3,4}^{7}*(26)^{3}*10^{4} = 35*(26)^{3}*10^{4} = 6,151,600,000[/tex]

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

Final answer:

The number of different 7-place license plates possible with 3 letters and 4 digits, where repetition is allowed, is 175,760,000.

Explanation:

To calculate the number of different 7-place license plates possible with 3 letters and 4 digits where repetition is allowed, we need to use the fundamental counting principle. For the three letters, assuming an English alphabet of 26 letters, each position can be filled in 26 different ways. Since there are 3 positions for letters, the number of combinations for the letter part is 26 × 26 × 26.

For the four digits, each position can be filled in 10 different ways (0 to 9). Therefore, the number of combinations for the digit part is 10 × 10 × 10 × 10.

To find the total number of combinations for the license plates, we multiply the combinations of letters by the combinations of digits: (26 × 26 × 26) × (10 × 10 × 10 × 10) = 175,760,000 different 7-place license plates possible.

[tex]x''+4x'+53x=15[/tex]

has characteristic equation

[tex]r^2+4r+53=0[/tex]

with roots at [tex]r=-2\pm7i[/tex]. Then the characteristic solution is

[tex]x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)[/tex]

For the particular solution, consider the ansatz [tex]x_p=a_0[/tex], whose first and second derivatives vanish. Substitute [tex]x_p[/tex] and its derivatives into the equation:

[tex]53a_0=15\implies a_0=\dfrac{15}{53}[/tex]

Then the general solution to the equation is

[tex]x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}[/tex]

With [tex]x(0)=8[/tex], we have

[tex]8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}[/tex]

and with [tex]x'(0)=-19[/tex],

[tex]-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}[/tex]

Then the particular solution to the equation is

[tex]\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}[/tex]

The given equation is a second order linear differential equation. However, there seems to be an inconsistency with the constant right-hand side. In a well-formulated equation, one would solve this by using characteristic roots and trivial solutions, finding the general solution, then applying initial conditions.

The given equation is a second order linear differential equation. Given the initial conditions, including the fact that x(0) = 8 and x˙ = −19, one must adjust for these when solving the differential equation. By utilizing the characteristic equation for determining the roots, the solution and consequent constants are then determined.

However, please note that without some form of driving force (right-hand side function), this is a simple harmonic oscillator equation. Since the right hand side function (15 in this case) is constant, there's an inconsistency in the problem. In this context, for a correct form, it should have a time-dependent function on the right side. If we assume that an inconsistency has occurred and the right side is zero, a full solution could be given.

In a correct equation scenario, one would be able to solve the initial value problem with characteristic roots and trivial solutions, find the particular solution, then a general solution and apply initial value conditions to find specific constants.

Learn more about Differential Equation here:

https://brainly.com/question/33433874

#SPJ2

Answer:

11)y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]

12)y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]

13)y = [tex]5e^{-(x+1)}[/tex]

14)y = 0

Step-by-step explanation:

Given data:

[tex]y=c_{1} e^{x} +c_{2} e^{-x}[/tex]

y''-y=0

The equation is

[tex]m^{r}[/tex]-1 = 0

(m-1)(m+1) = 0

if  above equation is zero then either

m - 1 = 0 or  m + 1 = 0

m = 1        ,    m  = - 1

11)

y(0) = 1 , y'(0) = 2

[tex]y'=c_{1} e^{x} -c_{2} e^{-x}[/tex]

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] = 1   (y(0) = 1) (1)

[tex]c_{1}[/tex] -  [tex]c_{2}[/tex] = 2   (y'(0) = 2)  (2)

adding 1 & 2

2[tex]c_{1}[/tex] = 3

[tex]c_{1}[/tex] = 3/2

3/2 +  [tex]c_{2}[/tex] = 1

[tex]c_{2}[/tex]  = 1 -  3/2

[tex]c_{2}[/tex] = - 1/2

y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]

12)

y(0) = 1 , y'(0) = e

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] = 0 (y(0) = 1) (3)

[tex]c_{1}[/tex] = - [tex]c_{2}[/tex]

[tex]e=c_{1} e -c_{2} e^{-1}[/tex]   (y'(0) = 2)  (4)

[tex]e=c_{1} e -\frac{c_{2} }{e} }[/tex]

[tex]e =\frac{c_{1} e^{2} -c_{2} }{e} }[/tex]

[tex]e^{2} ={c_{1} e^{2} -c_{2} }[/tex]

replace [tex]c_{2}[/tex] = [tex]c_{1}[/tex] by equation 3

[tex]e^{2} ={c_{1} e^{2} -c_{1} }[/tex]

taking common [tex]c_{1}[/tex]

[tex]e^{2} =c_{1} ({e^{2} -1 })[/tex]

[tex]\frac{e^{2} }{({e^{2} -1 })} =c_{1}[/tex]

[tex]-\frac{e^{2} }{({e^{2} -1 })} =c_{2}[/tex]

∴ y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]

13)

y(-1) = 5 , y'(-1) = -5

[tex]c_{1}[/tex][tex]e^{-1}[/tex] +  [tex]c_{2}[/tex][tex]e^{1}[/tex] = 5   (y(-1) = 5 ) (5)

[tex]c_{1}[/tex][tex]e^{-1}[/tex] -  [tex]c_{2}[/tex][tex]e^{1}[/tex] = -5    (y'(-1) = -5)  (6)

Adding 5&6

2[tex]c_{1}[/tex] [tex]e^{-1}[/tex] = 0

[tex]c_{1}[/tex] = 0

[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - [tex]c_{1}[/tex][tex]e^{-1}[/tex]

[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - 0

[tex]c_{2}[/tex]= 5/e

y = [tex]5e^{-1} e^{-x}[/tex]

y = [tex]5e^{-(x+1)}[/tex]

14)

y(0) = 0 , y'(0) = 0

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] =  0 (y(0) = 0) (7)

[tex]c_{1}[/tex] -  [tex]c_{2}[/tex] = 0   (y'(0) = 0)  (8)

Adding 7 & 8

2[tex]c_{1}[/tex] = 0

[tex]c_{2}[/tex] =

y = 0

Answer:

-$80

Step-by-step explanation:

Assuming that the variation in the number of shirts per day is -1 hundred shirs, the variation in profit with respect to time is given by the derivate of the profit equation:

[tex]p=-40x^2+581x-520\\\frac{dp}{dx}=-80x+581[/tex]

Let x be the number of shirts sold in a day, then x-1 is the number of shirts sold in the following day, the change in profit is:

[tex]p'(x) - p'(x-1)=-80x+581 - (-80(x-1)+581)\\p'(x) - p'(x-1)=-80(x-x+1) = -80[/tex]

The profit is changing by -$80 per day.

Final answer:

Using the pigeonhole principle, we can prove that among 502 positive integers, at least two will have the same remainder when divided by 1000, implying their difference is divisible by 1000.

Explanation:

The question asks to prove that among 502 positive integers, there are always two integers so that either their sum or their difference is divisible by 1000. This statement can be understood through the pigeonhole principle, which in basic terms means if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon.

In this case, consider the remainders when these integers are divided by 1000. Since there are only 1000 possible remainders (from 0 to 999), and we have 502 numbers, at least two of them must have the same remainder when divided by 1000, according to the pigeonhole principle.

Let these two numbers be a and b, where without loss of generality, a ≥ b. If a and b have the same remainder when divided by 1000, then a - b is divisible by 1000. Alternatively, if we had a case where the sum is considered, assuming complementary pairs mod 1000, a similar argument involving the pigeonhole principle can conclude that there must be at least one pair whose sum or difference gives a number divisible by 1000, satisfying the initial claim.

Answer:

yes

Step-by-step explanation:

the rate of change is constant

Answer:

The margin of error is 3.97 percentage points.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 600, p = 0.56[/tex]

95% confidence interval

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 - 1.96\sqrt{\frac{0.56*0.44}{600}} = 0.5203[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 + 1.96\sqrt{\frac{0.56*0.44}{600}} = 0.5997[/tex]

The margin of error is the upper limit subtracted by the proportion, or the proportion subtracted by the lower limit. They are the same values.

So the margin of error is 0.5997 - 0.56 = 0.56 - 0.5203 = 0.0397 = 3.97 percentage points.

Answer:

6 is a factor of n

Step-by-step explanation:

2 is a factor of n and 3 is a factor of n means

n = 2×3×k

  = 6×k

then  n = 6×k

then 6 is a factor of n

Final answer:

If both 2 and 3 are factors of a number n, then 6 must also be a factor of n, because the product of unique prime factors is always a factor of that number.

Explanation:

The product of unique prime factors of a number will be a factor of that number. Since 2 and 3 are prime factors and both are factors of n, their product (which is 6) must also be a factor of n.

For example, consider the number 12. 12 is divisible by 2 and 12 is divisible by 3, and indeed, 12 is divisible by 6 as well. This holds true for any number n that has 2 and 3 as factors. Thus, we can conclude that 6 is a factor of n if both 2 and 3 are factors of n.

Answer:

No, because it is not the tens digit that goes up by 1 in these numbers, it is the unit digit.

Step-by-step explanation:

It is important to know the concepts of units, tenths and cents.

For example

1 = 1 unit

10 = 1*10 + 0 = The tens digit is one the unit digit is 0

21 = 2*10 + 1 = The tens digit is two and the unit digit is 1.

120 = 1*100 + 2*10 + 0 = The cents digit is 1, the tens digit is two and the unit digit is 0.

So

Adding 1 is the same as the unit digit going up by 1.

Adding 10 is the same as the tens digit going up by 1.

Adding 100 is the same as the cents digit going up by 1.

In this problem, we have that:

864,865,866,867,868,869

Each value is the 1 added to the previous value, that is, the unit digit goes up by 1.

Mariel thinks the tens digit goes up by 1 in these numbers. Do you agree?

No, because it is not the tens digit that goes up by 1 in these numbers, it is the unit digit.

Answer:

disagree, its the unit number that goes up by 1

Step-by-step explanation:

Answer:

A.

sigma > 22500

B.

Null hypothesis:sigma = 22500

Alternative hypothesis:sigma > 22500

Step-by-step explanation:

A.

The claim states that the standard deviation of high school teachers income  is more than 22,500. This can be represented in the symbolic form as sigma > 22500.

B.

The null hypothesis and alternative hypothesis for the given scenario can be written as

Null hypothesis: Standard deviation of income of high school teachers is 22,500.

The standard deviation is represented as sigma.Symbolically it can be written as

Null hypothesis: sigma = 22500

Alternative hypothesis: Standard deviation of income of high school teachers is more than 22,500.

Symbolically it can be written as

Alternative hypothesis: sigma > 22500

Answer:

the explanation is given below.

Step-by-step explanation:

from this, the lowest number is 1 and the highest number is 100

x² + y² + z² - 10x - 4y - 14z + 74 = 0

Step-by-step explanation:

The general equation of a sphere is (x-a)² + (y-b)² + (z-c)² = r²

Where x, y, and z are the coordinates of points on the surface of the sphere.

a, b, and c represents the center of the sphere

r is the radius of the sphere. Note that the radius is always the same for all points on the sphere,

In this equation, the radius r is the largest radius that stays in the octant.

In the given question, (5,2,7) is the center of the sphere.

Therefore, substitute this into the general equation to get:

(x-5)²+(y-2)²+(z-7)² = r² ---------------------------------------------(i)

To find the radius r, we have to look at the distance from the center coordinate to each bounding planes xy-plane, xz-plane, and yz-plane.

The distance from the center to the xy-plane is the center of the z coordinate which is 7. The distance from the center to the xz-plane is the center of the y coordinate which is 2. The distance from the center to the yz-plane is the center of the x coordinate which is 5.

Therefore, to determine the radius contained in the first octant, we need to choose the smallest distance so as not to cross into a second octant. That will also be the largest possible radius for it not to cross into a different octant.

The smallest distance therefore is 2. So we substitute r = 2 into equation (i) above to get:

(x-5)²+(y-2)²+(z-7)² = 2²

(x-5)²+(y-2)²+(z-7)² = 4

Therefore (x-5)²+(y-2)²+(z-7)²-4 = 0  ----------------------------------------(ii)

A monic formula is a formula where the highest power of its single variable has a coefficient of 1.

Therefore, we expand equation (ii) in form of a monic formula to get

x² + y² + z² - 10x - 4y - 14z + 74 = 0

The highest power of x², y², and z² is 1

The probability that a single randomly selected salary is at least $134,000 is approximately 0.5120 or 51.20%.

To find the probability that a single randomly selected salary is at least $134,000, we need to calculate the z-score and use the standard normal distribution.

1. Calculate the z-score:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case, x = $134,000, μ = $133,000, and σ = $31,000.

z = (134000 - 133000) / 31000

z = 1 / 31000

z ≈ 0.0323

2. Find the probability associated with the z-score:

We can use a z-table or a calculator to find the probability.

From the z-table, we find that the probability corresponding to a z-score of 0.0323 is approximately 0.5120.

Therefore, the probability that a single randomly selected salary is at least $134,000 is 0.5120 or 51.20%.

Answer:

a. 24ways

b.256ways

Step-by-step explanation:

the letters IOWA contains for letters, since we are to arrange without repeating any letter, we permutate the letters.

For permutation of n object in r ways is expressed as

P(n,r)=n!/(n-r)!

hence for n=4 and r=4, we have P(4,4)=4!/(4-4)!

P(4,4)=4!/(0)!

P(4,4)=4*3*2*1=24ways

b. To arrange the letters such that each letter can be repeated, we can arrange the letter I in four ways, letter O can be arrange in four ways, letter W can be arranged in four ways and letter A can be arranged in four ways ..

Hence we arrive at

4*4*4*4=256ways

Answer:

The answer to your question is

a) 400

b) The function has two real solutions (-2 and 8)

Step-by-step explanation:

Process

1.- Discriminant = b² - 4ac

                         = 12² -4(-2)(32)

                         = 144 + 256

                        = 400

2.- Solutions (using the general formula)

                   x = [tex]\frac{- b +/- \sqrt{b^{2}- 4ac}}{2a}[/tex]

                   x = [tex]\frac{- 12 +/- \sqrt{400}}{2(-2)}[/tex]

                   x = [tex]\frac{-12 +/- 20}{2(-2)}[/tex]

   x₁ = [tex]\frac{-12 + 20}{2(-2)} = \frac{-12 + 20}{-4} = \frac{-8}{4} = - 2[/tex]

   x₂ = [tex]\frac{- 12 - 20}{- 4} = \frac{-32}{- 4} = 8[/tex]

This function has two real solutions (-2, 8)

Answer: C- The value of the discrimination is 400.

E- The function has 2 real solutions.

Step-by-step explanation: Hope that helped!

Answer:

a) [tex]P(X=12)=(12C12)(0.2)^{12} (1-0.2)^{12-12}=4.096x10^{-9}[/tex]

b) [tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313 [/tex]

c) [tex] E(X) = np = 12*0.2= 2.4[/tex]

d) [tex] Sd(X) = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386[/tex]

e) [tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558[/tex]

f)  [tex] P(X\leq 10) =1.1x10^{-9}[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=12, p=0.2)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Part a

For this case we want to find this probability:

[tex]P(X=12)=(12C12)(0.2)^{12} (1-0.2)^{12-12}=4.096x10^{-9}[/tex]

Part b

[tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0) [/tex]

[tex]P(X=0)=(12C0)(0.2)^{0} (1-0.2)^{12-0}=0.0687[/tex]

[tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313 [/tex]

Part c

The expected value is given by:

[tex] E(X) = np = 12*0.2= 2.4[/tex]

Part d

The standard deviation is given by:

[tex] Sd(X) = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386[/tex]

Part e

If we want the probability that the number classified as deceptive would be lower than the mean we want:

[tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)[/tex]

[tex]P(X=0)=(12C0)(0.2)^{0} (1-0.2)^{12-0}=0.0687[/tex]

[tex]P(X=1)=(12C1)(0.2)^{1} (1-0.2)^{12-1}=0.2062[/tex]

[tex]P(X=2)=(12C2)(0.2)^{2} (1-0.2)^{12-2}=0.2835[/tex]

[tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558[/tex]

Part f

For this case our random variable would be:

[tex]X \sim Binom(n=200, p=0.2)[/tex]

And we want this probability:

[tex] P(X\leq 10) = P(X=0)+P(X=1)+ .......+P(X=10)[/tex]

And we can use the following excel code to find the answer:

"=BINOM.DIST(10;200;0.2;TRUE)"

And we got: [tex] P(X\leq 10) =1.1x10^{-9}[/tex]

Answer:

Mean = 68.375

Step-by-step explanation

The mean will be:

Mean= sum of scores / number of students

Mean=1641/24

Mean = 68.375

Answer:

standard deviation =3.11

Step-by-step explanation:

data given is ungrouped data

standard deviation =√ [∑ (x-μ)² / n]

mean (μ)=∑x/n

           = [tex]\frac{10+15+12+18+19+16+12}{7}[/tex]

          =14.57

x-μ   for data 10, 15, 12, 18, 19, 16, 12 will be

                     -4.57, 0.43, -2.57, 3.43, 4.43,1.43, -2.57

(x-μ)² will be 20.88, 0.1849, 6.6049, 11.7649,19.6249, 2.0449, 6.6049

∑ (x-μ)² will be = 67.7049

standard deviation = √(67.7049 / 7)

                      =3.11

Final answer:

The standard deviation for the given data is approximately 3.11.

Explanation:

To calculate the standard deviation (s) of the set of scores 10, 15, 12, 18, 19, 16, 12, you would follow these steps:

Here's how to carry out each step with the provided scores:

Therefore, the standard deviation of the scores is about 3.11.

Answer: individual

Step-by-step explanation:

An individual is a person or object that is a member of the population being studied. A population is defined as a group of individuals with a common characteristic living and interbreeding within a given area, in statistics, population is a collection of individuals to be studied. Individuals can also be referred to as the objects/person described by a set of data. For example: when studying the height of students in a school, the students attending that school are individuals.

Answer:

a) The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.

b) 16% of women over 20 have upper arm lengths less than 33.3 cm.

Step-by-step explanation:

The Empirical Rule(68-95-99.7 Rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 35.8 cm

Standard deviation = 2.5 cm

(a) What range of lengths covers almost all (99.7%) of this distribution?

This range is from 3 standard deviations below the mean to three standard deviations above the mean.

So from 35.8 - 3*2.5 = 28.3 cm to 35.8 + 3*2.5 = 43.3 cm

The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.

(b) What percent of women over 20 have upper arm lengths less than 33.3 cm?

68% of the women over 20 have upper arm length between 33.3 cm and 38.3 cm. The other 32% have upper arm length lower than 33.3 cm or higher than 38.3. The distribution is symmetric, so 16% of the have upper arm length lower than 33.3 cm and 16% have upper arm length higher than 38.3 cm

So 16% of women over 20 have upper arm lengths less than 33.3 cm.

Answer:  The required laplace transform of g(t) is [tex]\dfrac{5}{(s+5)^2}.[/tex]

Step-by-step explanation:  We are given to find the laplace transform of the following function :

[tex]g(t)=5te^{-5t}.[/tex]

We know the following formulas for laplace transform :

[tex](i)~L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}},\\\\(ii)~L\{cf(t)\}=cL\{f(t)\}.[/tex]

In the given function function, we have

c = 5,  n = 1  and  a = -5.

Therefore, we get

[tex]L\{g(t)\}\\\\=L\{5te^{-5t}\}\\\\=5L\{te^{-5t}\}\\\\\\=5\times\dfrac{1!}{(s-(-5))^{1+1}}\\\\\\=\dfrac{5}{(s+5)^2}.[/tex]

Thus, the required laplace transform of g is [tex]\dfrac{5}{(s+5)^2}.[/tex]

The Laplace transform of the given function is [tex]\frac{5} { (s + 5)^2}[/tex].

The Laplace transform of a function g(t) is defined as:

[tex]L{(g(t))} = \int\limits^{\infty}_0 e^-^s^tg(t) dt[/tex]

We need to find the Laplace transform of [tex]g(t) = 5te^-^5^t[/tex]. To do this, we use the shifting theorem and known transforms.

First, recall the Laplace transform of [tex]nte^-^a^t[/tex] is:

[tex]L(nte^-^a^t)} = \frac{n! } {(s + a)^n^+^1}[/tex]

For our function [tex]g(t)[/tex] :

a = 5

n = 1

Applying the formula:

[tex]L{(5te^-^5^t)} = \frac{(5 * 1! )}{(s + 5)^2}[/tex]

After simplifying, we get:

[tex]L\leftparanthesis(\ 5te^-^5^t)\rightparanthesis\ = \frac{5 }{ (s + 5)^2}[/tex]

Answer:

P = 4s

Step-by-step explanation:

The perimeter of a geometric shape is simply the sum of all its sides length. Since the shape in question is a square, which means that all of the four sides have the same length 's', the perimeter can be expressed by:

[tex]P = s+s+s+s\\P=4s[/tex]

For any value of 's', the formula above expresses the perimeter 'P' as a function of 's'

Answer:

Step-by-step explanation:

The probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5 is 6/19,

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

It is given that:

The randomly chosen number between 1 and 100 is divisible by 3

Applying conditional probability:

Let A is the event: the numbers divisible by 3

Let B is the event: At least one digit equal to 5

P(A|B) = n(A∩B)/n(B)

P(A|B) = 6/19

Thus, the probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5 is 6/19.

Learn more about the probability here:

brainly.com/question/11234923

#SPJ2

Answer:

The probability that at most 3 attempts are required to produce a craft item with satisfactory quality is 0.9375

Step-by-step explanation:

Let E be a random variable denoting the event that an artist makes a craft item with satisfactory level.

Then the random variable E follows a Geometric distribution.

A Geometric distribution is defined as the number of failures (k) before the first success.

The probability function of Geometric distribution is:

[tex]P(X=k)=(1-p)^{k}p[/tex], p = Probability of success and k = 0, 1, 2, 3...

The probability of success is, p = 0.5 and the number of failures is, k = 3.

Compute the probability of at most 3 attempts before the first success is:

[tex]P(X\leq 3) =P(X=3)+P(X = 2)+P(X=1) +P(X = 0)\\=[(1-0.5)^{0}*0.5]+[(1-0.5)^{1}*0.5]+[(1-0.5)^{2}*0.5]+[(1-0.5)^{3}*0.5]\\=0.9375[/tex]

Therefore, the probability that at most 3 attempts are required to produce a craft item with satisfactory quality is 0.9375.

The tangent line approximation near x=0 for the function f(x) = \\sqrt{10 + x} is found by first calculating its derivative, then using that derivative to construct the equation of the tangent line at x=0, resulting in the linear approximation y = (1/2)(10)^{-1/2}x + \\sqrt{10}.

Finding the tangent line approximation for a function near a point involves using the function's derivative at that point. For the function f(x) = \\sqrt{10 + x}, the derivative at x = 0, denoted as f'(0), will provide the slope of the tangent. To find this, let's differentiate f(x) using the chain rule. The derivative of f(x) with respect to x is (1/2)(10 + x)^{-1/2}. At x = 0, this simplifies to 1/2(\\sqrt{10}), which is the slope of the tangent line at that point. Hence, the tangent line equation is y - f(0) = f'(0)(x - 0), which simplifies to y = (1/2)(10)^{-1/2}x + \\sqrt{10}. This form equation is the linear approximation of f(x) near x = 0.