Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirred solution flowing out at the same rate.
a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution:

Distance, Velocity - time functions are linked easily through derivation and integration:

Distance - time function → derivation → Velocity - time function

Velocity - time function → derivation → Acceleration - time function

(and vice versa)

Let's assume we have a distance - time function:

[tex]s(t) = 4t^{2} - 2t +7[/tex]

where s is measured in feet and t in seconds.

a) To find velocity at time t, we simply derivate the distance - time function:

[tex]\frac{ds}{dt} = v(t) = 8t - 2[/tex]

b) To find velocity at t-3, we simply substitute 3 in the velocity - time function:

[tex]v (t) = 8t -2\\v(3) = 8(3) -2\\v(3) = 22 \ ft/sec[/tex]  

c) A particle will be at rest when it's velocity is zero. Thus, we substitute v = 0 in the velocity - time function:

[tex]v (t ) = 8t -2\\8t -2 = 0\\8t = 2\\\\t = \frac{2}{8}\\\\t= \frac{1}{4} seconds[/tex]

Hence, at time t = 1/4 seconds, the object will be at rest.

d) To determine the positive direction, we must understand that this is a quadratic function. Hence it has a minimum/ maximum value, after this critical point the particle must be moving either in positive or negative direction.

Hence, we find this critical point. A critical point of any function is it's derivative equalled to zero.

The derivative of distance - time function is a velocity - time function. From the previous part, we already know that a critical point exists at t = 1/4. Now, we substitute, t = 1/4, in the distance - time function to find the other co-ordinate:

[tex]s (t) = 4t^{2} - 2t +7\\s(\frac{1}{4}) = 4(\frac{1}{4})^{2} - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = 4(\frac{1}{16}) - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = \frac{1}{4} - \frac{2}{4}+\frac{28}{4}\\\\s(\frac{1}{4}) = \frac{27}{4} \\\\[/tex]

The function will be positive after [tex](\frac{1}{4}, \frac{27}{4})[/tex]

e) The total distance travelled in first 8 seconds can be determined by substituting t = 8 in distance - time function:

[tex]s(t) = 4t^{2} - 2t+7\\\\s(8) = 4(8)^{2} - 2(8)+7\\\\s(8) = 4 (64) - 2 (8) +7\\\\s(8) = 247 feet[/tex]

The velocity of the particle any time t is v(t) = 4t^3 - 7 ft/sec. The velocity at t=3 seconds is 98 ft/sec. The particle is at rest at t=1.323. It moves in the positive direction when t < 1.323 or t > 1.323. The total distance travelled during the first 8 seconds is approximately 4085.6 feet.

The first step here is to find the velocity of the particle at any given time t. Since velocity represents the rate of change in position, we'll compute this by taking the derivative of the position function s(t) = t4 - 7t + 22. This gives us the velocity function v(t) = 4t3 - 7.

Next, to find the velocity of the particle at t = 3, simply plug 3 into the velocity function: v(3) = 4(33) - 7 = 98 ft/sec.

The particle is at rest when its velocity is zero, so we set v(t) = 0, or 4t3 - 7 = 0. Solving for t reveals that the particle is at rest when t = 1.323.

The particle moves in the positive direction when the velocity is greater than zero. Looking at v(t), we see that this is the case when t < 1.323 or t > 1.323. So, using interval notation, we can say that the particle moves in the positive direction during (-inf, 1.323) and (1.323, inf).

Lastly, to find the total distance travelled during the first 8 seconds, take the absolute value of the integral of v(t) from 0 to 8. Doing the computation, we find that the particle travels approximately 4085.6 feet during this time interval.

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Answer:

Step-by-step explanation:

The equation that can be used to approximate and to predict average

beginning salaries for candidates with bachelor's degrees is expressed as

s = 2806.6t + 32558

Where

s represents the starting salary and t is the number of years after 1995.

To determine when the beginning salaries for candidates will be greater than $60,000, the expression would be

2806.6t + 32558 > 60000

2806.6t > 60000 - 32558

2806.6t > 27442

t > 27442/2806.6

t > 9.78

Rounding to the nearest whole number, it becomes

t > 10

Therefore, beginning salaries for candidates will be greater than $60,000 after 2005

Beginning salaries for candidates will be greater than $60,000 in the year 2005.

To determine the year when beginning salaries will exceed $60,000, we need to solve for 't' when 's' is greater than $60,000 for the equation:

60,000 < 2806.6t + 32,558

Subtracting 32,558 from both sides gives us:

27,442 < 2806.6t

Dividing both sides by 2806.6 yields:

t > 9.78

Since 't' represents the number of years after 1995, we round up to the nearest whole year, resulting in t = 10. Therefore, beginning salaries for candidates are projected to exceed $60,000 approximately 10 years after 1995.

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Answer:

the probability that a real emergency situation exists is 0.16 (16%)

Step-by-step explanation:

defining the event A= the alarm sounds ,we have

P(A)= probability that an emergency situation exists * probability that the alarm sounds given that an emergency situation exists + probability that a emergency situation does not exist * probability that the alarm sounds given that an emergency situation does not exist = 0.004* 0.95+ 0.996 * 0.02 = 0.02372

then if we use the theorem of Bayes for conditional probability and define the event E= a emergency situation exists , then

P(E/A)= P(E∩A)/P(A)= 0.004* 0.95/0.02372 =0.16 (16%)

where

P(E∩A)= probability that an emergency situation exists and the alarm sounds

P(E/A) =  probability that an emergency situation exists given that the alarm has sounded

Answer:

1) is not possible

2) P(A∪B) = 0.7

3) 1- P(A∪B) =0.3

4) a) C=A∩B' and P(C)= 0.3

b)  P(D)= 0.4

Step-by-step explanation:

1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4  . Thus the maximum possible value of P(A∩B) is 0.4

2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by

P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7

P(A∪B) = 0.7

3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3

4) the event C that the selected student has a visa card but not a MasterCard is given by  C=A∩B'  , where B' is the complement of B. Then

P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3

the probability for the event D=a student has exactly one of the cards is

P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4

Answer:

(a) Ratio of A to B = 17566 : 5827

(b) Ratio of B to A = 5827 : 17566

(c) A is 301.46% percent of B.

Step-by-step explanation:

We are given A = 52,698 number of deaths due to a deadly disease in the United States in 2005   and   B = 17,481 number of deaths due to the same disease in the United States in 2009.

(a) Ratio of A to B = [tex]\frac{A}{B}[/tex] = [tex]\frac{52,698}{17,481}[/tex] = 17566 : 5827

(b) Ratio of B to A = [tex]\frac{B}{A}[/tex] = [tex]\frac{17,481}{52,698}[/tex] = 5827 : 17566

(c) Let A is x% of B so the equation formed will be;

    A = x% of B

     52,698 = x% of 17481

  Therefore x = [tex]\frac{52,698}{17,481}*100[/tex] = 301.46%

Hence, A is 301.46% of B.

The ratio of A to B is 3:1, the ratio of B to A is 1:3, and A is approximately 301.45 percent of B.

a. Find the ratio of A to B:

We can find the ratio of A to B by dividing A by B, which gives us 52,698/17,481. Evaluating this division gives a ratio of approximately 3:1, meaning that for every 3 deaths in 2005, there was 1 death in 2009.

b. Find the ratio of B to A:

To find the ratio of B to A, we divide B by A, which gives us 17,481/52,698. Simplifying this division gives a ratio of approximately 1:3, which is the inverse of the previous ratio.

c. Complete the sentence: A is ____ percent of B:

To find the percentage of A relative to B, we divide A by B, then multiply by 100. Evaluating this division gives us (52,698/17,481) * 100, which is approximately 301.45%. Therefore, we can complete the sentence by saying 'A is approximately 301.45 percent of B'.

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Answer:

The mean of function provided is 1.186.

The variance of the provided f(x) is 0.198

Step-by-step explanation:

It is provided that the probability mass function is,

f(x)= (214/43)×(1/6)ˣ; x=1,2,3

The mean is calculated as,

E(X)=∑  x × f(x)

        x

=1×(216/43)×(1/6)¹ + 2 × (216/43)×(1/6)² × 3 × (216/43)×(1/6)³

=36/43 + 12/43  +3/43

​  =1.186

​  

The mean of function provided is 1.186

Explanation | Common mistakes | Hint for next step

The expected value of the probability mass function,f(x)= (216/43×(1/6)ˣ

 is 1.1861.186 .

Step 2 of 2

To calculate the variance, first calculate  E(X²)=∑ x² × f(x)

= 1² ×(216/43) × (1/6)¹ + 2² × (216/43) × (1/6)² × 3² × (216/43) ×(1/6)³

=36/43 +24/43 +9/43

=1.605

​  

The variance is calculated as,

V(X) =E(X²) - [E(X)]²

=1.605 -(1.186)²

= 0.198

The variance of the provided f(x) is 0.198

Explanation | Common mistakes

The variance of function f(x)=(216/43) × (1/6)ˣ ; x =1,2,3 is 0.198

The mean and variance of the random variable with the given probability mass function is 1.186 and 0.198 respectively and this can be determined by using the formula of mean and variance.

Given :

[tex]f(x) = \left(\dfrac{216}{43}\right)\times \left(\dfrac{1}{6}\right)^x[/tex]

The mean can be evaluated by using the following calculation:

[tex]\rm E(x) = \sum x\times f(x)[/tex]

[tex]\rm E(x) = 1\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^1+ 2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^2+ 3\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^3[/tex]

[tex]\rm E(x) = \dfrac{36}{43}+\dfrac{12}{43}+\dfrac{3}{43}[/tex]

E(x) = 1.186

The variance can be evaluated by using the following calculation.

[tex]\rm E(x^2)=\sum x^2 f(x)[/tex]

[tex]\rm E(x^2) = 1^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^1+ 2^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^2+ 3^2\times \left(\dfrac{216}{43}\right) \times \left(\dfrac{1}{6}\right)^3[/tex]

[tex]\rm E(x^2) = \dfrac{36}{43}+\dfrac{24}{43}+\dfrac{9}{43}[/tex]

[tex]\rm E(x^2) = 1.605[/tex]

Now, the variance is given by:

[tex]\rm V(x) = E(x^2)-[E(x)]^2[/tex]

[tex]\rm V(x) = 1.605-(1.186)^2[/tex]

V(x) = 0.198

The variance is 0.198 and the mean is 1.186.

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Answer:

1 / √17

Step-by-step explanation:

to solve cos(tan^-1(4))

we break it into simpler terms

tan^-1(4) ------ these will be taken as an angle when dealing with cos

tan Ф = opposite / adjacent = 4 / 1 = 4

Using Pythagoras Theorem

Hypothenus ² = opposite² + adjacent ²

h² = 4² + 1²

h²  = 16 + 1

h² = 17  

a = √17

cos Ф = adjacent / hypothenus = 1 / √17

cos(tan^-1(4)) =  1 / √17

Answer:

-2; Inferior good

Step-by-step explanation:

Given that,

Initial Quantity = 10 boxes

New Quantity = 8 boxes

Percentage increase in Sally's income = 10%

Change in consumption:

= 8 boxes - 10 boxes

= - 2 boxes

Percentage change in quantity demanded:

= (Change in quantity demanded ÷ Initial quantity) × 100

= (-2 ÷ 10) × 100

= - 20%

Therefore,

Income elasticity of demand:

= percentage change in quantity demanded ÷ Percentage change in income

= - 20% ÷ 10

= -2

Inferior goods are generally have a negative income elasticity of demand which means that an increase in the income of the consumer will lead to reduce the quantity demanded for inferior good and vice versa.

Hence, the good is a inferior type of good.

Answer:

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Answer:

[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the diastolic blood pressure readings of adult males

From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=84, Sd(X)=9[/tex]

So we can assume [tex]\mu=84 , \sigma=9[/tex]

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

So we need values such that

[tex]P(X<\mu -\sigma)=P(X <75)=0.16[/tex]    

[tex]P(X>\mu +\sigma)=P(X >93)=0.16[/tex]  

[tex]P(X<\mu -2*\sigma)P(X<66)=0.025[/tex]    

[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]

[tex]P(X<\mu -3*\sigma)=P(X<57)=0.0015[/tex]

[tex]P(X>\mu +3*\sigma)=P(X>211)=0.0015[/tex]

So for this case the answer would be:

[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]

The empirical rule indicates that about 2.5% of adult males have diastolic blood pressure readings greater than 102 mmHg, as 102 mmHg is two standard deviations above the mean diastolic blood pressure of 84 mmHg.

The empirical rule states that for a bell-shaped distribution:

Approximately 68% of the data falls within one standard deviation of the mean.

Approximately 95% of the data falls within two standard deviations of the mean.

Approximately 99.7% of the data falls within three standard deviations of the mean.

Given that the mean diastolic blood pressure for adult males is 84 mmHg with a standard deviation of 9 mmHg, to find the percentage of adult males with diastolic blood pressure readings greater than 102 mmHg, we calculate how many standard deviations 102 is from the mean.

To calculate this, use the following formula for the z-score: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

For 102 mmHg:

z = (102 mmHg - 84 mmHg) / 9 mmHg = 2

This means that 102 mmHg is two standard deviations above the mean. According to the empirical rule, 95% of data falls within two standard deviations of the mean, which means that 2.5% falls above this range as the data is symmetric about the mean.

Thereby, approximately 2.5% of adult males have diastolic blood pressure readings greater than 102 mmHg.

Answer:

Θ =tan⁻¹ (4KQ²/mgr²), Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]

Step-by-step explanation:

initially the angle Θ=0° ,the vertical forces were equal to product of mass and gravity(m*g) and there was no horizontal or lateral force in action. But after the displacement of balls new forces are induced.

X-Axis:

Fe = TsinΘ

[KQ²/(r/2)²] = TsinΘ         where r₁=r/2, r₁ = new distance

(4KQ²/r²) = TsinΘ

Y-Axis

TcosΘ = mg

As we know that tanΘ=sinΘ/cosΘ

We have, tanΘ = 4KQ²/mgr²

By adjusting this equation and putting K=1/4π∈₀ we get,

Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]

Answer:a) observational study

b) Randomization

c)Observational study

Step-by-step explanation:

a) The participants in (a) are monitored closely and data collected are reviewed.

b) Participants are randomly assigned to work and study.

a) The research method described is an observational study. b) The research method described is a randomized experiment. c) The research method described is an observational study.

a) The research method described is an observational study. The students are simply keeping a log of the hours they spend studying in a group and reporting it after the course is completed. The researcher is not manipulating any variables or assigning students to different groups.

b) The research method described is a randomized experiment. The students are randomly assigned to study groups and the teacher tells each group how often to meet. This allows for the manipulation of the independent variable (number of hours studying in a group) and the measurement of its effect on the dependent variable (final course grade).

c) The research method described is an observational study. The students voluntarily join groups based on how often the groups will meet. The researcher is not manipulating any variables or assigning students to different groups. The students' choice of groups is a naturally occurring phenomenon that is being observed.

Answer:

y = 2x

Step-by-step explanation:

Claire is hungry. She buys 2 donuts each costing x $. How much should she pay?

Since one donut costs x $ 2 donuts cost 2x $.

Therefore, total amount Claire should pay, call it y = 2x

Hence, we have y = 2x.

Answer:

[tex]$ \textbf{a}_{\textbf{17}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{55} $[/tex]

Step-by-step explanation:

The [tex]$ n^{th} $[/tex] term of an arithmetic sequence is given by:

                       [tex]$ \textbf{a}_{\textbf{n}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{a} \hspace{1mm} \textbf{+} \hspace{1mm} \textbf{(n - 1)d} $[/tex]

where a is the first term of the sequence

and     d is the common difference.

We are given the [tex]$ 3^{rd} $[/tex] and the [tex]$ 13^{th} $[/tex] term of the sequence.

We are asked to find the [tex]$ 17^{th} $[/tex] term.

From the formula, we can write

[tex]$ a_3 = a + (3 - 1)d $[/tex]

[tex]$ \implies 13 = a + 2d \hspace{6mm} \hdots (1) $[/tex]  

Also, [tex]$ a_{13} = a + (13 - 1)d $[/tex]

[tex]$ \implies 43 = a + 12d \hspace{6mm} \hdots (2) $[/tex]

Now, we solve Equation (1) and (2) for a and d.

Solving we get:

a = 7; d = 3

Therefore, [tex]$ 17^{th} $[/tex] term, [tex]$ a_{17} $[/tex] can now be calculated.

[tex]$ a_{17} = a + (17 - 1)d $[/tex]

[tex]$ \implies a_{17} = 7 + 16(3) $[/tex]

[tex]$ \implies \textbf{a}_{\textbf{17}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{55} $[/tex]

Therefore, the [tex]$ 17^{th} $[/tex] term of the sequence is 55.

Hence, the answer.

Answer:

(a) Yes, f(x) is a valid probability mass function.

(b) The probability that you will walk at least two dogs this week = 0.74.

(c)  The expected number of dogs you will walk this week = 3 dogs.

(d) The expected value of X2 = 11.29

(e) Var[x] = E(X2) – (E[X])2 = 3.2811

Step-by-step explanation:

We are given with the probability mass function (pmf),f(x), o'is defined as follows:

Firstly let X = Number of dogs you walk this week

    X                 P(X = x)

    0                      0.14

    1                       0.12

    2                      0.15

    3                      0.23

    4                      0.18

    5                      0.09

    6                      0.08

    7                       0.01

(a) Now f(x) to be a valid probability mass function, two conditions should be    met :

So, First condition is already met as all values are positive and for second condition = 0.14 + 0.12+ 0.15+ 0.23+ 0.18+ 0.09+ 0.08+ 0.01 = 1

Hence both the conditions are satisfied so f(x) is a valid probability mass function.

(b) Probability that we will walk at least two dogs this week = P(X>=2)

      = 1 - P(X = 0) - P(X = 1) = 1 -  0.14 - 0.12 = 0.74

(c) To Compute the expected number of dogs you will walk this week we will use expectation formula which says:

           E(X) = [tex]\frac{\sum X\times P(X=x)}{\sum P(X=x)}[/tex] = [tex]\frac{0*0.14 + 1*0.12 + 2*0.15 + 3*0.23 + 4*0.18 + 5*0.09 + 6*0.08 + 7*0.01}{1}[/tex]

                                           = 2.83 or 3 after rounding off.

Therefore the expected number of dogs you will walk this week are 3 dogs.

(d) The expected value of X2 [E(X2)] =   [tex]\frac{\sum X^{2} \times P(X=x)}{\sum P(X=x)}[/tex]

      = [tex]\frac{0^{2} *0.14 + 1^{2} *0.12 + 2^{2} *0.15 + 3^{2} *0.23 + 4^{2} *0.18 + 5^{2} *0.09 + 6^{2} *0.08 + 7^{2} *0.01}{1}[/tex] = 11.29

(e) Var[x] = E(X2) – (E[X])2

     We have E(X2) = 11.29 and E(X) = 2.83

      Var[x] = [tex]11.29 - (2.83)^{2}[/tex] = 3.2811.

The random variable X represents the number of dogs walked in a week for a dog walking business. By verifying that the sum of the probabilities equals 1, we've confirmed the pmf is valid. The probability of walking at least two dogs is 0.74, and calculations for the expected number and variance involve using the given probabilities and applying the respective formulas.

Understanding Probability Mass Functions

In the context of a dog walking business, let us define the random variable X as the number of dogs walked in a week. The values that X can take on are 0, 1, 2, 3, 4, 5, 6, and 7. The probabilities of these events are given by P(X = x), where x are the respective values X can take.

To verify if the given probability mass function (pmf) is valid, the sum of all probabilities P(X = x) should be equal to 1. We calculate the sum as follows:

Adding them gives 1 which confirms it is a valid pmf.

To find the probability of walking at least two dogs this week, which is P(X ≥ 2), we need to add up the probabilities for all events where X is 2 or more:

P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)
= 0.15 + 0.23 + 0.18 + 0.09 + 0.08 + 0.01
= 0.74

To compute the expected number of dogs walked in a week, or E[X], we use the formula:

E[X] = Σ x * P(X = x)

For E[X2], we calculate the expectation of X-squared by multiplying each value of x by itself and then by its probability, before summing these products.

Finally, the variance Var[X] of the random variable X is given by E(X2) - (E[X])2. This involves first calculating E[X2] as explained, then squaring the previously found E[X] value, and subtracting that square from E[X2].

Answer:

B) Method 2 will increase the standard deviation of the students' scores.

Step-by-step explanation:

Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.

The average score for his students was 43 and the standard deviation of the scores was 5 points.

And now he is considering two different strategies for rescaling the exam results of which:

Method 1 = Add 17 points to everyone's score.

Method 2 = Multiply everyone's score by 1.7 .

And we have to check what will be the impact of these methods on the standard deviation of the students' scores.

For this let us consider a simple example to understand this:

Firstly, Formula for calculating Standard Deviation =  [tex]\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}[/tex]

Suppose,

     X             X - Xbar       [tex](X - Xbar)^{2}[/tex]

     3              3 - 6 = -3         -3 * -3 = 9

     5              5 - 6 = -1          -1 * -1 = 1

     10            10 - 6 = 4          4 * 4 = 16

Mean of above data, Xbar = [tex]\frac{3+ 5+10}{3}[/tex] = 6

Standard Deviation of data = [tex]\sqrt{\frac{26}{3-1} }[/tex] = 3.6055

Now let us suppose that we multiply each value of above data with 2 so the new data will be:

     X                X - Xbar           [tex](X - Xbar)^{2}[/tex]

 3*2 = 6        6 - 12 = -6         -6 * -6 = 36

 5*2 = 10       10 - 12 = -2       -2 * -2 = 4

10*2 =20      20 - 12 = 8         8 * 8 = 64

Mean of new data, Xbar = [tex]\frac{6+ 10+20}{3}[/tex] = 12

Standard Deviation of new data = [tex]\sqrt{\frac{104}{3-1} }[/tex] = 7.2111

Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as

  3.6055 * 2 = 7.2111

Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.

And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.

Answer:

[tex]T(p) = \frac{1}{4}p[/tex]

The new price is $6.75

Step-by-step explanation:

The new price of the toy after the discount (T) is given by the original price (p) subtracted by the discounted amount (3/4 of p):

[tex]T(p) = p-\frac{3}{4}p \\T(p) = (1-\frac{3}{4})p\\T(p) = \frac{1}{4}p[/tex]

If the original price was $26.99, the new price is:

[tex]T= \frac{1}{4}*\$26.99\\T=\$6.75[/tex]

Answer:

a) [tex] W=0,2[/tex]

b) [tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]

Step-by-step explanation:

Part a

For this case we have the following differential equation:

[tex] W \sqrt{4-2W}=0[/tex]

If we square both sides we got:

[tex] W^2 (4-2W) =0[/tex]

And we have two possible solutions for this system [tex] W=0, W=2[/tex]

So then that represent the constant solutions for the differential equation.

So then the solution for this case is :

[tex] W=0,2[/tex]

Part b: Solve the differential equation in part (a)

For this case we can rewrite the differential equation like this:

[tex] \frac{dW}{dx} =W \sqrt{4-2W}[/tex]

And reordering we have this:

[tex] \frac{dW}{W \sqrt{4-2W}} = dx[/tex]

Integrating both sides we got:

[tex] \int \frac{dW}{W \sqrt{4-2W}} = \int dx[/tex]

Using CAS for the left part we got:

[tex] -tanh^{-1} (\frac{1}{2} \sqrt{4-2W})= x+c[/tex]

We can multiply both sides by -1 we got:

[tex] tanh^{-1} (\frac{1}{2} \sqrt{4-2W})=-x-c[/tex]

And we can apply tanh in both sides and we got:

[tex] \frac{1}{2} \sqrt{4-2W} = tanh(-x-c)[/tex]

By properties of tanh we can rewrite the last expression like this:

[tex]\frac{1}{2} \sqrt{4-2W} = -tanh(x+c)[/tex]

We can square both sides and we got:

[tex] \frac{1}{4} (4-2W) = tanh^2 (x+c) [/tex]

[tex] 1-\frac{1}{2}W = tanh^2 (x+c)[/tex]

And solving for W we got:

[tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]

And that would be our solution for the differential equation

Answer:

a. 4)

b. 4)

Step-by-step explanation:

Hello!

An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information.

An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).

(a) The U.S. Census Bureau tracks population age. In 1900, the percentage of the population that was 19 years old or younger was 44.4%. In 1930, the percentage was 38.8%; in 1970, the percentage was 37.9%; and in 2000, the percentage in the age group was down to 28.5% (The First Measured Century, T. Caplow, L. Hicks, B. J. Wattenberg).

In this item, the researcher merely obtained the records of the population that were 19 or younger over several years and compared the obtained percentages. This is a clear example of an observational study, the researcher did not manipulate any factor or variable, he just looked up the information and documented it.

(b) After receiving the same lessons, a class of 100 students was randomly divided into two groups of 50 each. One group was given a multiple-choice exam covering the material in the lessons. The other group was given an essay exam. The average test scores for the two groups were then compared.

In this item, the students were randomly assigned to one of two groups and each group was given a test, group one: multiple choice and group two: essay. This is an example of an experimental study, after the class, the researcher controlled the variable "type of test" giving one type to each group and the subjects were randomly selected fro the groups assuring that they will not receive biased results. And at the end of the experiment, the response variables "test scores" where compared.

I hope it helps!

Final answer:

The studies presented about U.S. Census Bureau population tracking and exam type comparison in a classroom represent an observational study and an experiment, respectively. The former is so because no treatment was applied, while the latter involved purposefully assigning different tests to observe outcomes.

Explanation:

When classifying the given studies as either an observational study or an experiment, it's important to understand the key characteristics of each. An observational study involves monitoring subjects without any intervention, while an experiment involves the deliberate imposition of a treatment to observe potential changes.

(a) The U.S. Census Bureau tracking population age is an example of an observational study because the data on the percentage of the population that was 19 years old or younger in various years was collected without any manipulation or treatment being applied to the subjects.

(b) Dividing a class into two groups and administering different types of exams constitutes an experiment. This is because a treatment (the type of exam) was deliberately imposed on the students to observe the difference in test scores, which we interpret as the outcome of interest.

Answer:

We conclude,  Mohammed's opportunity cost knitting one sweater is 48 cookies.

Step-by-step explanation:

We have that  Mohammed can knit 5 sweaters or bake 240 cookies.

In one week Aisha can knit 15 sweaters or bake 480 cookies.  

We calculate how much is Mohammed's opportunity cost knitting one sweater. We get  

\frac{240}{5}= 48.

We conclude,  Mohammed's opportunity cost knitting one sweater is 48 cookies.

Answer:

Step-by-step explanation:

let A represent the amount of the 10% fungicide solution

let B represent the amount of the 50% fungicide solution

B milliliters = 50% of B = (50/100)A = 0.5B

Total milliliters = 26% of 200 milliliters = 0.26 * 200 = 52

A milliliters + B milliliters = 200 milliliters

0.1A + 0.5B = 52

using substitution method to solve A and B

0.1(200-B) + 0.5B = 52

20 - 0.1B + 0.5B = 52

20 + 0.4B = 52

0.4B = 52 -20 = 32

0.4B = 32

B = 32/0.4 = 80

since B = 80

A = 200 -B = 200 - 80 = 120

Answer: he must use 120 milliliters of the 10% solution and 80 milliliters of the 50% solution.

Step-by-step explanation:

Let x represent the amount of 10% fungicide solution that Ngoc must use.

Let y represent the amount of 50% fungicide solution that Ngoc must use.

The total volume of the fungicide solution that he wants to create is 200 milliliters. It means that

x + y = 200

Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. This means that

(10/100 × x) + (50/100 × y) = (26/100 × 200)

0.1x + 0.5y = 52 - - - - - - - - - -1

Substituting x = 200 - y into equation 1, it becomes

0.1(200 - y) + 0.5y = 52

20 - 0.1y + 0.5y = 52

- 0.1y + 0.5y = 52 - 20

0.4y = 32

y = 32/0.4 = 80 milliliters

x = 200 - y = 200 - 80

x = 120 milliliters

Answer:

a.

There are multiple factors contained in the unobserved error  ε. These factors are of two types i.e, economic and social.

Social Factors:

1. Husbad's level of education.

2. Current marital status

3. Religion, norms and culture

4. Residence location

Economic Factor:

1. Total family networth

2. Husband's income

3. Wife's income

4. Future income and sources

Most of these factors are directly or indirectly related to the level of education of woman. As an educated man would most probably tend to marry an educated woman. Therefore, these factors are correlated in a sense with level of education.

b.

A simple regression analysis can uncover the ceteris paribus effect of education on fertility as a term for unobserved errors is already inducted in formula.

The variable ε in the model encapsulates unobserved factors influencing fertility other than education. These factors might be related to education level. A simple regression might not show the exclusive ceteris paribus effect of education on fertility due to the inability to control for these unobserved factors.

In the given model, ε represents unobserved factors that affect a woman's fertility, aside from her years of education. Examples of these factors could be health conditions, lifestyle, access to health care, or cultural beliefs, among others. It is indeed possible that these factors, summarized by ε, could be correlated with levels of education. For instance, a higher level of education may lead to better awareness and access to health resources, thus influencing fertility indirectly.

As for the second part of your question, a simple regression analysis of this model would not necessarily uncover the true ceteris paribus effect of education on fertility. The ceteris paribus assumption means that 'all other things held constant.' However, in a simple regression, other crucial factors varying with education (captured in ε) are not controlled for and could bias the estimated education effect on fertility. Therefore, a more thorough multiple regression analysis might be needed to adequately control for these other factors.

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Answer:

c. neither significantly low nor significantly high.

Step-by-step explanation:

We have been given that 1700 births are randomly selected and 857 of the births are girls. We are asked to choose the correct option about number of girls using subjective judgement.

We could expect that there would be 850 boys and 850 girls on average among 1700 births.

In our given scenario there are 857 girls, so there would be 843 boys. 857 girls compared to 850 girls are close.

Now probability of choosing 857 girls or more from 1700 births is approximately 0.5041 or 50%, therefore, the outcome is neither significantly low nor significantly high.

The number of girls in the study, which is about 50.4% of the total births, does not significantly deviate from the expected natural ratio of 100 girls to 105 boys, and therefore is described as neither significantly low nor significantly high.

When we talk about the natural ratio of births, we refer to the expected proportion of girls to boys in a population, under natural circumstances without human intervention. According to Newsweek, the natural ratio is 100 girls to 105 boys. If we look at the provided study data where 1700 births were randomly selected and 857 of the births are girls, we need to determine if this proportion of girls is unusually high or low compared to the natural ratio.

The expected number of girls, according to the natural ratio, would be about 50% of the population, given there is no gender preference or other factors affecting the gender birth rate. In this case, the proportion of girls is 857/1700, or approximately 50.4% which is very close to the expected 50%. Therefore, using subjective judgment, we would describe the number of girls in this study as c. neither significantly low nor significantly high.

It is important to remember that small deviations from the expected ratio can occur due to natural variability, and the number observed in this study does not seem to deviate significantly from the natural ratio.

Answer:

Infinitely many solutions

Step-by-step explanation:

The given system is

2y = 14 - 2x

y = -x + 7

Let us substitute the second equation into the first one to get:

2(-x+7)=14-2x

Expand to get:

-2x+14=14-2x

This means

x=x

This tells us that the system has infinitely many solutions.

The two lines coincide

Answer:

infinatly many

Answer:

there are 36 possible bag lunches

Step-by-step explanation:

Assuming that the possible sandwiches do not depend on the selection of the chips and fruits ( and the same for chips or fruits respect to the other food in the bag)

then

number of possible bag lunches= possible sandwiches * possible chips* possible fruits = 3 * 4 *3 = 36

then there are 36 possible bag lunches

To solve this problem, multiply the number of options for each category together. This method results in a total of 36 different possible lunches that can be made.

This problem is an example of a situation where the number of possibilities is determined by multiplying the number of options in each category. This is because in each lunch bag there is a sandwich, chips, and a fruit. The choice of sandwich, chips, and fruit are independent of each other.

There are 3 types of sandwiches: Peanut Butter, Cheese, and Roast Beef. There are 4 types of chips: Corn, Potato, Pita, and Pretzel. Finally, there are 3 types of fruit: Apple, Orange, and Pear.

So, to find the total number of possible lunches, we simply multiply the options together:

3 sandwiches * 4 chips * 3 fruits = 36 possible lunches

Therefore, there are 36 different combinations of lunches that can be made.

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yes, because 6 miles equals 1 hour, and they are asking the hours for 2 and 18 miles. So all you need to do is  6 x2+ 18 miles

The sample space consists of all sets of 5 numbers that can be picked from 40, and each outcome has an equal chance of occurring. The probability that exactly three out of five numbers are even is calculated as the combination of choosing three even numbers and two odd numbers, divided by the total number of outcomes.

(a) In this case, the sample space consists of all possible outcomes of the experiment. There are 40 distinct numbers, and we're picking 5. The total number of distinct sets of 5 that can be chosen from 40 is given by the combination formula C(n, k) = n! / [k!(n - k)!]. So, in this case, C(40, 5), which would give the total number of outcomes. The probability measure P assigns each outcome a probability. Since we're choosing the numbers randomly and uniformly, each outcome has an equal probability of occurring, which is 1 / C(40, 5).

(b) To find the probability that exactly three out of the five numbers are even, we first note that there are 20 even numbers and 20 odd numbers in the set 1, 2, ..., 40. The number of ways to choose three even numbers is C(20, 3) and the number of ways to choose two odd numbers is C(20, 2). The probability of this happening is thus [C(20, 3) * C(20, 2)] / C(40, 5).

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Answer:

Type I error: The correct option is (C).

Type II error: The correct option is (D).

Step-by-step explanation:

The type-I-error is the probability of rejecting the null hypothesis when the null hypothesis is true.

The type-II-error is the probability of filing to reject the null hypothesis when in fact it is false.

The hypothesis in this problem can be defined as follows:

Null hypothesis (H₀): The percentage of adults who have a job is equal to 88%.

Alternate Hypothesis (Hₐ): The percentage of adults who have a job is different from 88%.

The type-I-error in this case will be committed when we conclude that the percentage of adults who have a job is different from 88% when in fact it is equal to 88%.

The type-II-error in this case will be committed when we conclude that the percentage of adults who have a job is equal to 88% when in fact it is different than 88%.

Answer:

The probability that the good exam belongs to student X is 0.8571.

Step-by-step explanation:

It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,

Compute the probability that exactly one student does well in the exam as follows:

[tex]P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56[/tex]

Then the probability that X is the one who did well in the exam is:

[tex]P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571[/tex]

Thus, the probability that the good exam belongs to student X is 0.8571.

What's the opposite of division? Multiplication

So we multiply 2 on both sides to get A=-10

Hope this helped

[tex]\text{Hey there!}[/tex]

[tex]\mathsf{We\ can\ treat\ the\ value\ of\ A\ as\ invisible\ 1\ until\ we\ find\ the\ actual\ value\ of\ it}[/tex]

[tex]\mathsf{\dfrac{1}{2}a=-5}[/tex]

[tex]\mathsf{Multiply\ by\ 2\ on\ both\ of\ your\ sides}[/tex]

[tex]\mathsf{\dfrac{1}{2}a\times2=-5\times2}[/tex]

[tex]\text{Cancel out }\mathsf{\dfrac{1}{2}a\times2}\text{ because it gives you the value of 1}[/tex]

[tex]\text{Keep: }\mathsf{-5\times2}\text{ because it gives the value of a}[/tex]

[tex]\mathsf{-5\times2=-10}[/tex]

[tex]\boxed{\boxed{\mathsf{Answer: a=-10}}}[/tex]

[tex]\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{LoveYourselfFirst:)}[/tex]