A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate.
(a) How much sugar is in the tank at the begining?
(b) Find the amount of sugar after t minutes.
(c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit y(t) as t approcahes infinity.

Answer:

With a skewed distribution and data with outliers.

Step-by-step explanation:

There are three measures of central tendency.

The median is not affected by outliers and skewed data as compared to the mean.

Thus, median likely to produce a better measure of central tendency than the mean with a skewed distribution and data with outliers.

Answer:

a) D.)H0: pF = pM versus Ha: pF ≠ pM

b) [[tex]z=\frac{0.537-0.384}{\sqrt{0.449(1-0.449)(\frac{1}{73}+\frac{1}{54})}}=1.7138[/tex]  

c) [tex]p_v =2*P(Z>1.7138)=0.0866[/tex]  

d)   B.)The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the P-value is greater than 0.01.

e) [tex](0.537-0.384) - 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=-0.0755[/tex]  

f) [tex](0.537-0.384) + 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=0.3815[/tex]  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{M}=28[/tex] represent the number of men that favored war with Iraq

[tex]X_{W}=29[/tex] represent the number of women that favored war with Iraq

[tex]n_{M}=73[/tex] sample of male selected  

[tex]n_{W}=54[/tex] sample of female selected  

[tex]p_{M}=\frac{28}{73}=0.384[/tex] represent the proportion of men that favored war with Iraq

[tex]p_{W}=\frac{29}{54}=0.537[/tex] represent the proportion of women that favored war with Iraq

[tex]\alpha=0.01[/tex] represent the significance level

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the value for the test (variable of interest)  

Part a

We need to conduct a hypothesis in order to checkif the proportion of females that favored war with Iraq was significantly different from the proportion of males that favored war with Iraq , the system of hypothesis would be:  

Null hypothesis:[tex]p_{M} = p_{W}[/tex]  

Alternative hypothesis:[tex]p_{M} \new p_{W}[/tex]  

The best option is:

D.)H0: pF = pM versus Ha: pF ≠ pM

Part b

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{W}-p_{M}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex] (1)  

Where [tex]\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{28+29}{73+54}=0.449[/tex]  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.537-0.384}{\sqrt{0.449(1-0.449)(\frac{1}{73}+\frac{1}{54})}}=1.7138[/tex]  

Part c

We have a significance level provided [tex]\alpha=0.01[/tex], and now we can calculate the p value for this test.  

Since is a one two sided test the p value would be:  

[tex]p_v =2*P(Z>1.7138)=0.0866[/tex]  

Part d

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the best conclusion would be:

 B.)The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the P-value is greater than 0.01.

Part e

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_W -\hat p_M) \pm z_{\alpha/2} \sqrt{\frac{\hat W_A(1-\hat p_W)}{n_W} +\frac{\hat p_M (1-\hat p_M)}{n_M}}[/tex]  

For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=2.58[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.537-0.384) - 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=-0.0755[/tex]  

Part f

[tex](0.537-0.384) + 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=0.3815[/tex]  

The proper hypotheses are H0: pF = pM and Ha: pF ≠ pM. The test statistic is -0.087. The p-value is 0.1651.

a) The proper hypotheses for this situation are:

H0: pF = pM

Ha: pF ≠ pM

Hence, the correct option is D.

b) The test statistic is computed by subtracting the male statistic from the female statistic:

Test statistic = pF - pM = 28/73 - 29/54 = -0.087

c) The p-value for the test is computed using the test statistic and the appropriate test statistic distribution. Based on the given values, the p-value is found to be 0.1651.

d) Using a 0.01 level of significance, the conclusion should be: The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the p-value is greater than 0.01. Hence, the correct option is D.

e) The lower endpoint of a 99% confidence interval for the difference between the proportion of females that favor the war and the proportion of males that favor the war is -0.2302.

f) The upper endpoint of a 99% confidence interval for the difference between the proportion of females that favor the war and the proportion of males that favor the war is 0.0566.

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Answer:

Step-by-step explanation:

The simplified formula for payback period (assuming even cashflow) is as follows:

Payback Period (in years) = Cost / cashflow per year

Generally, the choice which gives the shorter payback period is desired

For project A,

Cost = $250,000 and  cashflow per year = $75,000

Payback Period for project A

= 250,000 ÷ 75,000 = 3.333 years

For Project B

Cost = $150,000 and  cashflow per year = $52,000

Payback Period for project B

= 150,000 ÷ 52,000 = 2.88 years

comparing the PP for A & B, it is clear that B has the shorter payback period, and hence choice B is more desirable

The simplified formula for the payback period (assuming even cashflow) is as follows:

Payback Period (in years) = Cost / cashflow per year

Generally, the choice which gives the shorter payback period is desired

For project A,

Cost = $250,000 and  cashflow per year = $75,000

Payback Period for project A

= 250,000 ÷ 75,000 = 3.333 years

For Project B

Cost = $150,000 and  cashflow per year = $52,000

Payback Period for project B

= 150,000 ÷ 52,000 = 2.88 years

comparing the PP for A & B, it is clear that B has the shorter payback period, and hence choice B is more desirable

Word problem usually consists of mathematical modeling questions, where statistics and information about a sure machine are given and a student is required to expand a version. for example, Jane had $five.00, then spent $2.00. How lots does she have now?

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The 'Duration (amount of time)' in the study is an example of quantitative data. Quantitative data is numerical data that can be measured or counted.

The 'Duration (amount of time)' in the study is an example of quantitative data.

Quantitative data is numerical data that can be measured or counted. In this case, the duration of resident use of the local park is being measured in terms of time (amount of time).

Examples of other quantitative data include age and number of times per week.

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After converting the provided integral to polar coordinates, the value of integral is evaluated π/2.

When the Cartesian coordinates (x,y) are expressed in the polar coordinates (r, θ), then this form is called the polar form.

The given integral function in the problem is,

[tex]\int_0^{\sqrt{4}} \int\limits^x_{-x} dydx[/tex]

Let suppose, [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex]. Thus,

[tex]\sin\theta=\dfrac{y}{r}\\\cos\theta=\dfrac{x}{r}[/tex]

Limits are  y=x. From the trigonometry, the value of theta in the given triangle can be given as,

[tex]\dfrac{\sin\theta}{\cos\theta}=1\\\tan\theta=1\\\theta=\tan^{-1}1\\\theta=45^o\\\theta=\dfrac{\pi}{4}[/tex]

Similarly, for y=-x the value of angle,

[tex]\theta=-\dfrac{\pi}{4}[/tex]

Thus, the limits of theta are from -π/4 to π/4. From the Pythagoras theorem,

[tex]r^2=x^2+y^2\\r^2=(r\cos\theta)^2+(r\sin\theta)^2\\r^2=r^2(1)[/tex]

Thus, the limits of r is from 0 to 1. Convert the given integral in polar form as,

[tex]\int\limits^{\pi/4}_{-\pi/4} \int_0^{1} dt ds\\\int\limits^{\pi/4}_{-\pi/4} [1-0] dt \\\int\limits^{\pi/4}_{-\pi/4} dt \\\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4} \right) \\\dfrac{\pi}{2} \\[/tex]

Hence, after converting the provided integral to polar coordinates, the value of integral is evaluated π/2.

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To convert the given Cartesian integral to polar coordinates, identify the bounds in polar terms, then rewrite the integral accordingly. After setting up the new limits for r and θ, use the relationship between Cartesian and polar coordinates to express the area element, and integrate step-by-step.

To convert the integral ∫4√0∫x-xdydx to polar coordinates and evaluate it, we first need to describe the limits of integration and the region of integration in terms of polar coordinates (r, θ). The given integral ranges over a region bounded by the parabola y = √x and the x-axis from x=0 to x=4. Converted to polar coordinates, this region is bounded by the rays θ = 0 and θ = π/2 and the circles r = 0 and r = 4cos(θ).

So the double integral can be rewritten as ∫π/20∫4cos(θ)0 rdrdθ. To evaluate this integral, we integrate r from 0 to 4cos(θ), then integrate θ from 0 to π/2:

∫π/20 (∫4cos(θ)0 r dr) dθ = ∫π/20 [1/2 r^2]|^{4cos(θ)}_0 dθ = ∫π/20 8cos^2(θ) dθ
Using the double angle formula, cos^2(θ) = (1+cos(2θ))/2, the integral becomes:

8 ∫π/20 (1+cos(2θ))/2 dθ = 4 ∫π/20 (1+cos(2θ)) dθ

This can now be integrated directly to get the final result.

The total number of different license plates possible with 2 letters and 5 digits is 67,600,000. If no letter or digit can be repeated, the number reduces to 32,760,000.

This problem involves combinations and permutations, which are fundamental concepts in discrete mathematics. If a license plate has 2 letters followed by 5 digits, each chosen independently, we consider the product of the number of possibilities for each space. There are 26 possible letters (A-Z) for each of the 2 letter spaces and 10 possible digits (0-9) for each of the 5 digit spaces. Therefore, the total number of different license plates is 26 * 26 * 10 * 10 * 10 * 10 * 10 = 67,600,000.

However, if no letter or digit can be repeated, the number of possibilities for the second letter space becomes 25 (since one letter has already been used), and similarly for the digits, it gets reduced by one with each additional digit. Hence, the total number of different license plates without repeated letters and digits would be 26 * 25 * 10 * 9 * 8 * 7 * 6 = 32,760,000.

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There are 67,600,000 total possible license plates with two letters and five digits. When restricting to plates with no repeated letters or digits, there are 165,765,000 possible plates.

The possible license plates can be calculated using the principle of multiplication. This principle states that if there are m ways to do one thing, and n ways to do another, then there are m*n ways to do both. In this case, there are 26 options for the first letter, another 26 for the second letter, and 10 options for each of the 5 digits (0-9). So, we can multiply these options to get our answer.

To calculate the number of possible combinations, we use the formula 26*26*10*10*10*10*10 = 67,600,000. Therefore, there are 67,600,000 possible license plates.

For the plates with no repeated letters or numbers, we still use the principle of multiplication but reduce each subsequent option by one. It will be 26*25 (because one letter has been used already and it can't be reused) *10*9*8*7*6 (same as for digits). Our formula is 26*25*10*9*8*7*6 = 165,765,000 plates.

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The sampling technique described is called 'sampling without replacement'. In this case, names are drawn from the bag and not returned, decreasing the pool for subsequent draws. It's a type of 'simple random sampling'.

The sampling technique used when 50 contestants’ names are written on 50 cards, placed in a bag, and three are chosen, is a method called sampling without replacement. In this process, once a name is chosen, it is not returned to the bag, reducing the pool of options for subsequent selections. As each card is drawn, the remaining pool of names shrinks, meaning the probability of drawing any remaining name changes with each draw. This process is often used when the desire is to avoid repetition in the selection.

In contrast, sampling with replacement would mean that after a card is drawn, it is put back into the bag before the next draw, ensuring that the same name could potentially be drawn more than once. Both methods are types of simple random sampling, a fundamental sampling method in statistics.

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Answer:

  The sub rose 121 feet.

Step-by-step explanation:

The sub was less deep at 12 pm than at 10 am, so rose in the water. The amount it rose was ...

  410 -289 = 121 . . . . feet

__

If you like, you can think of the sub's elevation relative to the surface as changing from -410 feet to -289 feet. The difference is then an increase in elevation (rise) of ...

  -289 -(-410) = -289 +410 = 121 . . . feet

_____

12 am is midnight, 14 hours after 10 am. The time 2 hours after 10 am is 12 pm, noon.

The mean, median, and mode of the original set of scores are 54.9, 41.5, and 38 respectively. When the score of 21 is recorded as 2, the mean decreases but the median and mode remain unchanged. When the score of 21 is removed, both the mean and median change but the mode stays the same.

To answer these questions, we first need to find the mean, median, and mode of the original set of scores. The mean is the sum of the values divided by the total number of values. For the given set of numbers, the mean is 54.9. The median is the number that divides the set of values into two equal halves. In the given set, the median is 41.5. The mode is the number that appears most often, which in this case is 38.

When the score of 21 is mistakenly recorded as 2, the mean decreases to 51.3, the median remains the same, and the mode also stays the same as this change does not affect the frequency of the numbers.

When the score of 21 is removed from the set, the mean decreases to 55.3, and the median increases to 43. Still, the mode remains unchanged as removing a number 21 doesn't change the frequency distribution of the scores.

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Answer:

a) positive

b)

[tex]^\circ F\text{ per Minute}[/tex]      

c) Interpretation of f'(20)=2  

Step-by-step explanation:

We are given the following in the question:

[tex]T=f(t)[/tex]

where T is the temperature in degrees Fahrenheit of a cold yam placed in a hot oven and t is the time in minutes since the yam was put in the oven.

a)  sign of f'(t)

f'(t) will represent the rate of change in temperature.

f'(t) will represent the change in temperature of yam when 1 minute has passed since it was kept in oven.

Since the temperature will always increase in oven, f'(t) will have a positive sign.

b) units of f'(20)

Since, f'(t) represent the rate of change in temperature. the unit will be

[tex]\dfrac{\text{degrees Fahrenheit}}{\text{Minute}}[/tex]

That is degrees Fahrenheit per minute.

c)  f'(20)=2

f'(20) will tell the change in temperature when 20 minutes have passed after the yam has been kept in oven.

Thus, the given statement means that 20 minutes after the yam was kept for  in the oven, the temperature of yam was increasing by 2 degree Fahrenheit per minute.

The sign of f'(t) is positive . The units of f'(20) are degrees Fahrenheit per minute. The statement f'(20)=2 means that at 20 minutes, the temperature of the yam is increasing at a rate of 2°F per minute.

The temperature, T, of a cold yam placed in a hot oven, as a function of time t is given by T=f(t), where t is the time in minutes since the yam was put in the oven.

The sign of f'(t) is positive. This is because, as time t increases, the temperature of the cold yam increases due to the hot environment of the oven.

The units of f'(20) are degrees Fahrenheit per minute (°F/min). This is because f'(t) represents the rate of change of temperature with respect to time.

The practical meaning of the statement f'(20)=2 is that at t=20 minutes, the temperature of the yam is increasing at a rate of 2 degrees Fahrenheit per minute.

Answer:

The answer to your question is See the picture below

Step-by-step explanation:

Analysis of the graphs

a) In the first graph, we observe that Natasha is running after 1 h and 1/4, the comeback for to her origin, finally she continues running forward. This option is incorrect.

b) In the second graph, Natasha runs for 1 1/4 hours, after this time, she comes back to her origin, takes a rest and continues running forward. This answer is incorrect.

c) In the third graph, Natasha runs for 1 1/4 hour, takes a rest and comes back to the origin. This option is incorrect.

d) In the forth graph, Natasha runs for 1 1/4 hour, takes a rest and continues running upwards. This is the right option.

Answer:

the last one to the far right at the bottom

Step-by-step explanation:

Answer: x = 5.0, y = 4.0

We multiply the second equation by -2.

-2(0.50 x + 0.20 y)= -2(3.30)

-x - 0.40y = -6.60

Then we add it to x + y = 9.0.

So we get: -x - 0.40y + x + y = -6.60 + 9.0

or, 0.60y = 2.4

or, y = 4.0

From x + y = 9.0 we get:

x + 4.0 = 9.0

x = 5.0

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To solve the system of linear equations, we used the substitution method, resulting in the solution x = 5 and y = 4.

The system of linear equations provided is:

To find the values of x and y, we can use the method of substitution or elimination. In this case, we'll use substitution:

Therefore, the solution to the system of equations is x = 5 and y = 4.

To solve the given differential equation, we employ the method of integrating factors, multiplying the equation by an integrating factor to simplify it. After manipulation, it involves integrating sec(x) to find the general solution, which includes trigonometric and possibly logarithmic functions.

The differential equation given is cos(x) dy/dx + (sin(x))y = 1. To find its general solution, we will use the method of integrating factors. This method involves multiplying the entire equation by a function, called the integrating factor, that will allow the left side of the equation to be expressed as the derivative of a product of functions.

Firstly, the differential equation can be rewritten in the form dy/dx + (sin(x)/cos(x))y = 1/cos(x). Notice that the integrating factor, μ(x), can be found using the formula μ(x) = e^(∫ P(x) dx), where P(x) = sin(x)/cos(x). This evaluates to μ(x) = e^(ln|sec(x)|), which simplifies to μ(x) = sec(x).

Multiplying the entire original differential equation by sec(x) leads to sec(x)cos(x) dy/dx + sec(x)sin(x)y = sec(x). This simplifies to dy/dx + tan(x)y = sec(x). Integrating both sides with respect to x, we find the general solution involves an integration of sec(x), which may lead to a solution involving logarithmic and trigonometric functions, depending on the specifics of the integration.

Thus, the general solution to the differential equation involves integrating sec(x) on the right side after applying the integrating factor, leading to a solution of the form y(x) = C*sec(x) + f(x), where C is a constant and f(x) is an integral involving sec(x).

Answer:

Statistical Inference.

Step-by-step explanation:

Statistical Inference:

Thus,

The branch of Statistics concerned with using the sample data to make an inference about a large set of data is called statistical inference.

The correct answer is: False.

As the sample size increases, the variation of the sample mean from the population mean typically becomes smaller. This is known as the Central Limit Theorem, which states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution with a mean equal to the population mean and a standard deviation that decreases as the sample size increases. Therefore, with larger sample sizes, the sample mean tends to be a more accurate estimate of the population mean, and the variation around the population mean decreases.

Answer:

Standard error

Step-by-step explanation:

Standard error: In statistics, the term standard error is also denoted as SE, and is reffered to as the estimate of a particular parameter.

It is defined as the standard deviation in statistics related to its sampling distribution or else the the standard deviation's estimate.

Therefore, if a particular parameter or statistic is considered as a mean then it would be denoted as the standard error of that mean.

The standard error of an estimate may also be defined as the square root of the estimated error variance √[tex]\sigma^{2}[/tex] of the quantity,

[tex]S_{e}[/tex] =√[tex]\sigma^{2}[/tex]

In the given question, the appropriate answer would be standard error.

Answer:

[tex]C(d,m) = 20 + 45d + 0.15m[/tex]

Step-by-step explanation:

In this problem, we have that the cost of renting a car C is a function of the number of days d, and the number of miles driven, m.

There is also a fixed cost F.

So the equation for the cost of renting a car is:

[tex]C(d,m) = F + a*d + b*m[/tex]

In which a is the daily cost and b is the cost per mile.

A car rental company charges a one-time application fee of 20 dollars, 45 dollars per day, and 15 cents per mile for its cars.

This means that [tex]F = 20, a = 45, b = 0.15[/tex]

Write a formula for the cost, C, of renting a car as a function of the number of days, d, and the number of miles driven, m.

[tex]C(d,m) = 20 + 45d + 0.15m[/tex]

Answer:

Exponential function if the form [tex]e^{Kx}[/tex] is the constant multiple K of itself if.

[tex]y'=\frac{dy}{dx}=cKe^{Kx} \\y'=Ky[/tex]

Step-by-step explanation:

The exponential function of the form [tex]e^x[/tex] is the function which is the derivative of itself.

Where:

x is independent variable

Now if we talk about constant then again exponential function if the form [tex]e^{Kx}[/tex] is the constant multiple K of itself if we take the dervative.

Mathmatical Prove:

Consider the general equation

let [tex]y=ce^{Kx}[/tex]

Where:

K is a constant

c is the cofficient(could be any number)

Now:

Taking derivative of above equation w.r.t x:

[tex]y'=\frac{dy}{dx}=cKe^{Kx} \\y'=Ky[/tex]

Hence proved exponential function is a constant multiple K of it self.

The function is [tex]f(x) = Ae^kx[/tex]. This function is a solution of the differential equation y' = ky.

We know that the derivative of the exponential function is the function itself. Consider the function [tex]f(x) = Ae^kx[/tex], where A is a constant. Then, take the derivative of [tex]f(x) = Ae^kx[/tex], we get:

[tex]f'(x) = Ake^{kx} = k (Ae^{kx}) = k f(x)[/tex]

Thus, a function whose first derivative is a constant multiple k of itself is[tex]f(x) = Ae^{kx}[/tex], where A is a constant.

Since [tex]f'(x) = k f(x)[/tex] for the function [tex]f(x) = Ae^{kx}[/tex], this function must be the solution of the differential equation y' = ky.

Complete question:

What function do you know from calculus is such that its first derivative is a constant multiple k of itself? (Do not use the function f(x)=0.)

f(x)=____

The above function is a solution of which of the following differential equations?

[tex]y'=y^k[/tex]

[tex]y'=y+k[/tex]

[tex]y'=ky[/tex]

[tex]y'=e^{ky}[/tex]

[tex]y'=k[/tex]

Answer:

99

Step-by-step explanation:

Since the caterpillar crawled through the origin, her movement can be described by a straight line equation modeled with the points (-2.5; -5.5) and (0; 0).

The slope of a linear equation is given by:

[tex]m=\frac{y-y_0}{x-x_0}\\m=\frac{-5.5-0}{-2.5-0}\\m=2.2[/tex]

For x = 45, the value of y is:

[tex]y= 2.2x\\y= 2.2*45\\y=99[/tex]

The value of the y-axis is 99.

Answer:

a. Quantitative

b. Quantitative

c. Qualitative

d. Qualitative

e. Qualitative

Step-by-step explanation:

a. The temperature readings can be quantify and can be represented numerically so, it is quantitative variable.

b. The scores in English test can be quantify and can be represented numerically so, it is quantitative variable.

c. The blood type can be categorized as O plus, O negative, A plus, A negative, AB plus, Ab negative, B plus, B negative and cannot be represented numerically so, it is qualitative variable.

d. The last four digit of social security number take numerical values, yet they are qualitative because these are used as identifiers and mathematical operations of social security don't have meaningful interpretation.

e. The number on the jerseys of 53 football players also takes numerical values yet they are qualitative because these are used as identifiers for players and mathematical operations of number on the jerseys of football players don't have meaningful interpretation.

The measures in question are classified as either quantitative or qualitative.

a. The 30 high-temperature readings of the last 30 days - Quantitative. These temperatures are numerical values that can be measured and compared.

b. The scores of 40 students on an English test - Quantitative. The scores represent numerical values that can be added, averaged, and compared.

c. The blood types of 120 teachers in a middle school - Qualitative. Blood types are descriptive categories or labels that cannot be measured numerically.

d. The last four digits of social security numbers of all students in a class - Qualitative. The last four digits are specific identifiers that cannot be measured or compared numerically.

e. The numbers on the jerseys of 53 football players on a team - Qualitative. The numbers are labels or identifiers, not measurable quantities.

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Answer:

The area of the rectangle is [tex]A=5L-L^2[/tex].

Step-by-step explanation:

The perimeter of a rectangle is

[tex]P=2(L+W)[/tex]

where, L is length and W is width.

It is given that perimeter of a rectangle is 10m.

[tex]10=2(L+W)[/tex]

Divide both sides by 2.

[tex]5=L+W[/tex]

Subtract L from both sides.

[tex]5-L=W[/tex]

Area of a rectangle is

[tex]A=L\times W[/tex]

Substitute W=(5-L) in the above formula.

[tex]A=L\times (5-L)[/tex]

[tex]A=5L-L^2[/tex]

Therefore, the area of the rectangle is [tex]A=5L-L^2[/tex].

Final answer:

The area A of a rectangle with a fixed perimeter of 10 meters is expressed as a function of the length L by the formula A(L) = L(5 - L), assuming 0 ≤ L ≤ 5 meters.

Explanation:

The question is asking us to express the area A of a rectangle as a function of the length L, one of its sides, given a fixed perimeter of 10 meters. The perimeter of a rectangle is given by the formula P = 2L + 2W, where L is the length and W is the width. For a rectangle with a perimeter of 10 meters, we have:

2L + 2W = 10

W = (10 - 2L) / 2 = 5 - L

The area A of the rectangle is A = L × W = L(5 - L)

This formula A = L(5 - L) gives the area as a function of the length.A: The rectangle's area is a function of its length, expressed as A(L) = L(5 - L), valid for 0 ≤ L ≤ 5, since the minimum possible width is 0 when the length equals 5, and the maximum possible length is 5 when the width equals 0.

Answer:

ab dc ef is replaced by 12 43 56

Step-by-step explanation:

ab dc ef are replaced by the position they take when writing in an alphabetical order. a is 1, b is 2, c is 3, and so on.

So, ab dc ed is written as

12 43 56

and

56 - 43 - 12 = 1

Answer:false

Step-by-step explanation:

Integer contains only both positive and negative numbers digits but not decimal points.

Answer:False

Step-by-step explanation:The set of integers Z ={....-3,-2,-1,0,1,2,3....}

Answer:

a) True

Step-by-step explanation:

The given statement is true.

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

The standard deviation express the variation of data from the mean and therefore, have the same unit as the data like lbs, inches, dollars, etc.

Coefficient of variation on the other hand is the ratio of standard deviation and mean. It is also a measure of dispersion but since it is a ratio, the units cancel each other.

[tex]C.V = \dfrac{s}{\bar{x}}[/tex]

Thus, coefficient of variation is dimensionless.

Thus, it is a relative measure.

Answer:

Stem-and-leaf plot of the test scores is shown below.

Step-by-step explanation:

The given data set is

67, 72, 86, 75, 89, 89, 87, 90, 99, 100

Stem-and-leaf: Leaf is the last term and stem is other term. If a number is 32, then 3 is stem and 2 is leaf.

Stem-and-leaf plot of the test scores is

Stem            leaf

6                  7

7                  2,5

8                  6,7,9,9

9                  0,9

10                 0

The length of the rows are similar to the heights of bars in a histogram; longer rows of the data correspond to higher frequency.

Answer:

Complete the following statements. In general, 50% of the values in a data set lie at or below the median. 75% of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 500 test scores, of them 0.5*500 = 250 would be at or below the median. If a sample consists of 500 test scores, of them 0.75*500 = 375 would be at or above the first quartile (Q1).

Step-by-step explanation:

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.

The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.

The answer is:

Complete the following statements. In general, 50% of the values in a data set lie at or below the median. 75% of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 500 test scores, of them 0.5*500 = 250 would be at or below the median. If a sample consists of 500 test scores, of them 0.75*500 = 375 would be at or above the first quartile (Q1).

Answer:The percentage of the plants that are grasses is 83%

Step-by-step explanation:

Seventeen percent of the plants in the greenhouse are broad leaf plants and the rest are grasses

The total number of grasses and broad leaf plants in the greenhouse is 100%. Therefore,

The percentage of the plants that are grasses would be

100% - 17% = 83%

Final answer:

By subtracting the percentage of broad-leaf plants (17%) from the total (100%), we find that 83% of the plants in the greenhouse are grasses.

Explanation:

If seventeen percent of the plants in the greenhouse are broad-leaf plants, then the remaining percentage of plants must be grasses. Since the total percentage must add up to 100%, we can subtract the percentage of broad-leaf plants from the total to find out the percentage of grasses.

To calculate the percentage of grasses, we use the following steps:

Start with the total percentage which is 100%.

Subtract the percentage of broad-leaf plants which is 17%.

The result is the percentage of grasses.

100% - 17% = 83%

Therefore, 83% of the plants in the greenhouse are grasses.

Answer:

a) [tex]\hat{p} = 0.27[/tex]    

b) All the conditions are met for constructing a confidence interval.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 167

Number of people who had side effects, x = 45

a)  point estimate

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{45}{167} = 0.27[/tex]

The point estimate for the population proportion of patients who received 10-mg doses of a drug daily and reported headache as a side effect is 0.27

b) Conditions for constructing a confidence interval

[tex]n\hat{p}>10\\167\times 0.27 = 45.09 > 10\\n(1-\hat{p})>10\\167(1-0.27) = 121.91 > 10[/tex]

Thus, all the conditions are met for constructing a confidence interval.

Answer:

the correct answer is B. g(x)=3^x +1

Step-by-step explanation:

I just took the test

Hope this helps

The exponential function with a y-intercept of 2 is in the form y = 2b^x, where 'a' represents the y-intercept of the function. To find this, set x to 0 in the function, resulting in y = a(1), and thus y = a. If the function's y-intercept is 2, the value of 'a' is 2.

To determine which exponential function has a y-intercept of 2, you need to recall the standard form of an exponential function, y = abx, where a is the y-intercept of the function. For an exponential function, the y-intercept occurs when x is 0. Thus, when x = 0, the function takes the form y = ab0, and since anything to the power of 0 is 1, the function simplifies to y = a. Therefore, if a function's y-intercept is 2, it means that the value of a must be 2, resulting in the function y = 2bx.

Additionally, understanding the relationship between exponential and logarithmic functions can be helpful. To rewrite a base number b in terms of natural logarithms, you can use the fact that b = eln(b). For example, 2 = eln(2). This is valuable for solving equations involving exponential growth or decay, especially when a y* calculator button is unavailable.

Answer:

Step-by-step explanation:

There are 15 numbers on a list, and the smallest number is changed from 12.9 to 1.29.

a. Is it possible to determine by how much the mean changes? If so, by how much does it change?

Yes.  The total would decrease by 11.61.  so mean would decrease by 11.61/15 =0.774

b. Is it possible to determine the value of the mean after the change? If so, what is the value?

No. new men would be old mean - 0.774

c. Is it possible to determine by how much the median changes? If so, by how much does it changes?

Though the smallest number is changed the position does not change.

Hence median will not change.

d. Is it possible to determine by how much the standard deviation changes? If so, by how much does it change?

The original variance would be average of square of deviations of the old mean.  

Yes, it possible to determine by how much the mean changes. so, 0.7741 change in mean.

No, it is not possible to determine the value of the mean after the change because the value of mean is unknown before the mean

No, it possible to determine by how much the median changes. so, the value of mean will remain same.

No it is impossible to determine how much the standard deviation changes .

Given that,

There are 15 numbers on a list, and the smallest number is changed from 12.9 to 1.29.

We have to find,

All the statements about mean and median are correct or not.

According to the question,

Then,

Mean would decrease by,

[tex]= \dfrac{11.61}{15} \\\\= 0.774[/tex]1

Yes, it possible to determine by how much the mean changes. so, 0.7741 change in mean.

No, it is not possible to determine the value of the mean after the change because the value of mean is unknown before the mean.

Yes, It is possible to determine by how much the median changes because we have 15 numbers in a data set and doesn’t effect the median so the median didn’t change.

Hence median will not change.

No, it possible to determine by how much the median changes. so, the value of mean will remain same.

No it is impossible to determine how much the standard deviation changes .

To know more about Mean click the link given below.

https://brainly.com/question/16776778