Construct a​ stem-and-leaf plot of the test scores:67, 72, 86, 75, 89, 89, 87, 90, 99, 100.67, 72, 86, 75, 89, 89, 87, 90, 99, 100. How does the​ stem-and-leaf plot show the distribution of these​ data?

Answer:

There are 252 possible outcomes.

Step-by-step explanation:

For each economic course, the student can select nine mathematic courses.

For each mathematic couse, the student can select four computer courses.

There are 7 economic courses.

So in all, there are 9*4*7 = 252 possible outcomes, that is, the number of different ways which the student can select his courses.

In order to find the percentage of bulbs with a certain lifespan, one must calculate the z-scores for the given values and find the probabilities using the standard normal distribution, converting these into percentages.

The student's question involves using the properties of the normal distribution to determine the probability of light bulb life spans within certain intervals. To solve these problems, the z-score formula is used, which is (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation.

Remember that the answer will be in the form of a percentage representing the likelihood that any given bulb falls within the specified hour range.

Answer:

B

Step-by-step explanation:

617/500 = 1234/1000

Therefore option B is the answer.

Answer:

See explanation below.

Step-by-step explanation:

Definition

The cross product is a binary operation between two vectors defined as following:

Let two vectors [tex] a = (a_1 ,a_2,a_3) , b=(b_1, b_2, b_3)[/tex]

The cross product is defined as:

[tex] a x b = (a_2 b_3 -a_3 b_2, a_3 b_1 -a_1 b_3 ,a_1 b_2 -a_2 b_1)[/tex]

The last one is the math definition but we have a geometric interpretation as well.

We define the angle between two vectors a and b [tex]\theta[/tex] and we assume that [tex] 0\leq \theta \leq \pi[/tex] and we have the following equation:

[tex] |axb| = |a| |b| sin(\theta)[/tex]

And then we conclude that the cross product is orthogonal to both of the original vectors.

Some properties

Let a and b vectors

If two vectors a and b are parallel that implies [tex] |axb| =0[/tex]

If [tex] axb \neq 0[/tex] then [tex]axb[/tex] is orthogonal to both a and b.

Let u,v,w vectors and c a scalar we have:

[tex] uxv =-v xu[/tex]

[tex] ux (v+w) = uxv + uxw[/tex] (Distributive property)

[tex] (cu)xv = ux(cv) =c (uxv)[/tex]

[tex] u. (vxw) = (uxv).w[/tex]

Other application of the cross product are related to find the area of a parallelogram for two dimensions where:

[tex] A = |axb|[/tex]

And when we want to find the volume of a parallelepiped in 3 dimensions:

[tex] V= |a. (bxc)|[/tex]

Answer:

a) [tex] E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+[/tex]

b) [tex]E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2[/tex]

c) [tex] E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2[/tex]

d) [tex]E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6[/tex]

e) [tex] E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1[/tex]

f) [tex] E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3[/tex]

g) [tex]E(u) = E(v) +E(X) = 0+1=1[/tex]

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

[tex] Y = 1+X +u[/tex]

[tex] u = v+X[/tex]

with both Y and u random variables, we also know that:

[tex] [tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2[/tex]

And we want to calculate this:

Part a

[tex] E(u|X=1)= E(v+X|X=1)[/tex]

Using properties for the conditional expected value we have this:

[tex] E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1[/tex]

Because we assume that v and X are independent

Part b

[tex]E(Y| X=1) = E(1+X+u|X=1)[/tex]

If we distribute the expected value we got:

[tex]E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2[/tex]

Part c

[tex] E(u|X=2)= E(v+X|X=2)[/tex]

Using properties for the conditional expected value we have this:

[tex] E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2[/tex]

Because we assume that v and X are independent

Part d

[tex]E(Y| X=2) = E(1+X+u|X=2)[/tex]

If we distribute the expected value we got:

[tex]E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6[/tex]

Part e

[tex] E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1[/tex]

Part f

[tex] E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3[/tex]

Part g

[tex]E(u) = E(v) +E(X) = 0+1=1[/tex]

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

To evaluate the triple integral ∭Tx²dV, we need to integrate over the solid tetrahedron T defined by the vertices (0,0,0), (3,0,0), (0,3,0), and (0,0,3).

To evaluate the triple integral ∭Tx²dV, we need to integrate over the solid tetrahedron T defined by the vertices (0,0,0), (3,0,0), (0,3,0), and (0,0,3). Since T is a three-dimensional shape, we need to perform a triple integral.

The limits of integration for each variable are as follows:

x: 0 to 3-y-z

y: 0 to 3-z

z: 0 to 3

Substituting the limits into the integrand Tx², we can then evaluate the triple integral by integrating with respect to x, y, and z in the given limits.

The data, the number of days traveled by randomly selected employees, is classified as quantitative, discrete data with a ratio level of measurement.

The data in question, namely, the number of days traveled last month by 100100 randomly selected employees, is considered quantitative data. This is because it deals with numbers that can be quantitatively analyzed. In terms of whether the data is discrete or continuous, it is discrete. The number of days traveled can be counted in whole numbers (you can't travel 2.5 days for example); thus, it is a countable set of data. Lastly, considering the level of measurement, the data falls under the ratio level as it not only makes sense to say that someone traveled more days than someone else (therefore an ordered relationship), but it also makes sense to say someone traveled twice as many days as someone else (giving us a proportion and a well-defined zero point).

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Answer:

Enthalpy of Vaporization of substance X = 489.15KJ/mol

Step-by-step explanation:

The Concept of Clausius Clapeyron equation is applied. This equation allows us to calculate the vapor pressure of a liquid over a some range of temperatures. The Clausius Clapeyron equation make use of the assumption that the heat of vaporization does not change as the temperature changes.

The question has data attached to it, I have added the other details and a step by step derivation and application of the Clausius Clapeyron equation was done.

The parametric equations of the position of a particle with constant velocity moving along a straight line path from points P(7,2) to Q(2,7) are x(t) = 7 - 5t and y(t) = 2 + 5t.

In this context, the movement of a particle can be represented on a plane using a usual 2D Cartesian coordinate system. The constant velocity of the particle dictates it will always move along a straight line. The straight line path can be found by determining the slope between points P(7,2) and Q(2,7).

The slope m of the line is given by:

m = (y2 - y1) / (x2 - x1)

Where P = (x1, y1) and Q= (x2, y2). Applying these coordinates gives us:

m = (7 - 2) / (2 - 7) = -1

So, the line equation we have is something like y - y1 = m(x - x1), and substituting in all the values gives:

y - 2 = -1 * (x - 7) which simplifies to y = -x + 9

To get the parametric equations, we can consider the particle moving along the straight line path from P to Q in time t = 1 second. The parametric equations of the position of the particle is therefore:

x(t) = 7 - 5t and y(t) = 2 + 5t

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One angle and its supplement give 180 when summed:

[tex]\alpha+\beta = 180[/tex]

This implies that

[tex]\alpha = 180-\beta[/tex]

we also know that

[tex]\alpha = 2\beta+102[/tex]

So, we wrote [tex]\alpha[/tex] as [tex]180-\beta[/tex], but also as [tex]2\beta+102[/tex]. So, the two expressions must equal each other, because they both equal [tex]\alpha[/tex]:

[tex]180-\beta = 2\beta+102 \iff 78=3\beta \iff \beta = 26[/tex]

This implies that [tex]\alpha[/tex] must complete [tex]\beta[/tex] so that they reach 180 together:

[tex]\alpha = 180-\beta = 180-26 = 154[/tex]

The concept of supplementary angles is used to solve the given problem. By forming a pair of linear equations from the problem and solving them, the measures of the two angles are found to be 154 degrees and 26 degrees respectively.

The subject in question here involves the concept of supplementary angles. Supplementary angles are two angles that add up to 180 degrees. In this case, the problem states that one angle is twice its supplement increased by 102 degrees. This forms a pair of linear equations that we can solve for.

Let's denote the unknown angle as x and its supplement as y. According to the problem, we have the following system of equations:

By substitifying the second equation into the first, we receive: 2y + 102 + y = 180. Solving this equation gives y = 26 degrees. Substituting y back into the first equation gives x = 180 - 26 = 154 degrees.

So the two supplementary angles are 154 degrees and 26 degrees.

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Answer:

Sample mean for U=21.5

Sample mean for F=8.4

Step-by-step explanation:

[tex]Sample mean of u=xbar_{u} =\frac{sum(xi)}{n}[/tex]

Where xi are the observations in the urban homes sample and n is the number of observations in the urban homes sample

[tex]sample mean of u=xbar_{u} =\frac{6+5+11+33+4+5+80+18+35+17+23}{11}[/tex]

[tex]sample mean of u=xbar_{u} =\frac{237}{11}=21.545[/tex]

Rounding it to one decimal places

[tex]sample mean of u=xbar_{u}=21.5[/tex]

Now for second sample

[tex]Sample mean of F=xbar_{F} \frac{sumxi}{n}[/tex]

Where xi are the observations in the farm homes sample and n is the number of observations in the farm homes sample

[tex]Sample mean of F=xbar_{F} =\frac{2+15+12+8+8+7+6+19+3+9.8+22+9.6+2+2+0.5)}{15}[/tex]

[tex]Sample mean of F=xbar_{F} =\frac{125.9}{15} =8.393[/tex]

Rounding it to one decimal places

[tex]Sample mean of F=xbar_{F} =8.4[/tex]

The sample mean for urban homes (U) is 21.54 and the sample mean for farm homes (F) is 7.73.

To find the sample mean, you sum up all the data points and then divide by the number of data points. For the Urban homes (U), we first add up all the data points: 6.0 + 5.0 + 11.0 + 33.0 + 4.0 + 5.0 + 80.0 + 18.0 + 35.0 + 17.0 + 23.0 = 237.0. The number of data points is 11, so the sample mean for U is 237.0 / 11 = 21.54 (rounded to two decimal places).

For Farm homes (F), add up all the data points: 2.0 + 15.0 + 12.0 + 8.0 + 8.0 + 7.0 + 6.0 + 19.0 + 3.0 + 9.8 + 22.0 + 9.6 + 2.0 + 2.0 + 0.5 = 115.9. The number of data points is 15, so the sample mean for F is 115.9 / 15 = 7.73 (rounded to two decimal places).

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Answer:

Step-by-step explanation:

The confidence interval for the mean fluoride ion concentration in toothpaste at a 95% confidence level is [tex]$0.14 \pm 0.06\%$[/tex].

To calculate the confidence interval for the mean fluoride ion concentration in toothpaste, we use the formula:

[tex]\[ \text{Confidence interval} = \text{Mean} \pm \left( \text{Critical value} \times \frac{\text{Standard deviation}}{\sqrt{\text{Sample size}}} \right) \][/tex]

Given:

- Mean (sample mean) = 0.14%

- Standard deviation = 0.05%

- Sample size (replicate measurements) = 5

- Confidence level = 95%

We need to find the critical value corresponding to a 95% confidence level. Since the sample size is small (n < 30), we use a t-distribution and degrees of freedom [tex]\(df = n - 1 = 5 - 1 = 4\)[/tex].

From the t-distribution table or a statistical calculator, the critical value for a 95% confidence level with 4 degrees of freedom is approximately 2.776.

Now, we can calculate the confidence interval:

[tex]\[ \text{Confidence interval} = 0.14 \pm \left( 2.776 \times \frac{0.05}{\sqrt{5}} \right) \][/tex]

[tex]\[ \text{Confidence interval} = 0.14 \pm \left( 2.776 \times \frac{0.05}{\sqrt{5}} \right) \]\[ \text{Confidence interval} = 0.14 \pm 0.06 \][/tex]

So, the confidence interval is [tex]$0.14 \pm 0.06\%$[/tex].

Therefore, the correct option is [tex]$0.14( \pm 0.06) \%$[/tex].

Complete Question:

A measurement of fluoride ion in tooth paste from 5 replicate measurements delivers a mean of 0.14 % and a standard deviation of 0.05 %. What is the confidence interval at 95 % for which we assume that it contains the true value?

[tex]$0.14( \pm 0.06) \%$[/tex]

[tex]$0.14( \pm 6.2) \%$[/tex]

[tex]$0.14( \pm 0.07) \%$[/tex]

[tex]$0.14( \pm 0.69) \%$[/tex]

Answer:

The correct option is a) Cluster.

Step-by-step explanation:

Consider the provided information.

Types of sampling:

Here, the population is divided into geographical blocks,

Thus, the type of sampling is cluster.

Therefore the correct option is a) Cluster.

Answer:

21

Step-by-step explanation:

7! / (5! x 2!) = 42/2 = 21

other explanation:

TTHHHHH THTHHHH THHTHHH THHHTHH THHHHTH THHHHHT ... 6

HTTHHHH HTHTHHH ........................................................................................... 5

HHTTHHH HHTHTHH ............................................................................................4

HHHTTHH HHHTHTH .............................................................................................3

HHHHTTH HHHHTHT .............................................................................................2

HHHHHTT ...................................................................................................................1

6+5+4+3+2+1 = 21

The number of different ways should be 21.

Since there is  7-letter permutations can be formed from 5 identical H's and two identical T's

So,

[tex]= 7! \div (5! \times 2!) \\\\= 42 \div 2[/tex]

= 21

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Answer:

y(t)= 11/3 e^(-t) - 5/2 e^(-2t) -1/6 e^(-4t)

Step-by-step explanation:

[tex] Y(s)=\frac{s+3}{(s^2+3s+2)(s+4)} + \frac{s+3}{s^2+3s+2} +\frac{1}{s^2+3s+2} [/tex]

We know that [tex] s^2+3s+2=(s+1)(s+2)[/tex], so we have

[tex] Y(s)=\frac{s+3+(s+3)(s+4)+s+4}{(s+1)(s+2)(s+4)}  [/tex]

By using the method of partial fraction we have:

[tex] Y(s)=\frac{11}{3(s+1)} - \frac{5}{2(s+2)} -\frac{1}{6(s+4)} [/tex]

Now we have:

[tex] y(t)=L^{-1}[Y(s)](t) [/tex]

Using linearity of inverse transform we get:

[tex] y(t)=L^{-1}[\frac{11}{3(s+1)}](t) -L^{-1}[\frac{5}{2(s+2)}](t) -L^{-1}[\frac{1}{6(s+4)}](t) [/tex]

Using the inverse transforms

[tex] L^{-1}[c\frac{1}{s-a}]=ce^{at} [/tex]

we have:

[tex] y(t)=11/3 e^{-t} - 5/2 e^{-2t} -1/6 e^(-4t) [/tex]

Answer:

Step-by-step explanation:

The length of the half marathon is 13.1 miles. it took Grant 7 minutes and 45 seconds to run from mile marker 1 to mile marker 2. Converting 7 minutes and 45 seconds to minutes, it becomes

7 + 45/60 =7.75 minutes

Speed = distance/time

Therefore, his speed from mile marker 1 to mile marker 2 is

1/7.75 = 0.129 miles per minute

He spent 19 minutes and 15 seconds to run from mile marker 2 to mile marker 4. Converting 19 minutes and 15 seconds to minutes, it becomes

19 + 15/60 =19.25 minutes

Therefore, his speed from mile marker 2 to mile marker 4 is

2/19.255 = 0.104 miles per minute

A) his average speed from miles 1 to 4 would be

(0.129 + 0.104)/2 = 0.1165 miles per minute.

B) after running the 9th mile, distance remaining would be

13.1 - 9 = 4.1 miles

Time left to complete the race would be

110 - 69 = 41 minutes

Average speed needed to complete the race would be

4.1/41 = 0.1 miles per minute.

Final answer:

Grant's average speed from mile 1 to mile 4 was approximately 0.1481 miles per minute. To complete the half marathon in his goal time, he needs to run the last 4.1 miles at an average speed of approximately 0.1 miles per minute.

Explanation:

We're given that Grant took 7 minutes and 45 seconds to run from mile marker 1 to mile marker 2, and 19 minutes and 15 seconds to run from mile marker 2 to mile marker 4. To find Grant's average speed from mile marker 1 to mile marker 4, we first convert the time into minutes. He ran 2 miles in 7.75 minutes and then 2 miles in 19.25 minutes. That's a total of 4 miles in 7.75 + 19.25 = 27 minutes, resulting in an average speed of 4 miles / 27 minutes ≈ 0.1481 miles per minute.

Then, we find out how fast Grant needs to run to complete the half marathon in 110 minutes. Grant is at mile marker 9 after 69 minutes, leaving him with 110 - 69 = 41 minutes to complete the remaining 13.1 - 9 = 4.1 miles. The average speed required for this last stretch is 4.1 miles / 41 minutes ≈ 0.1 miles per minute.

Answer:

The random variables in this case are discrete since they have a Multinomial distribution.

The probability mass function for a discrete random variable X is given by:

[tex]P(X=x_{i} )[/tex]

Where are [tex]x_{i}[/tex] are possible values of X.

The joint probability mass function of two discrete random variables X and Y is defined as

P(x,y) =P(X=x,Y=y).

It follows that, The joint probability mass function of [tex]X_{3} , X_{4}[/tex] is :

[tex]P(X_{3}, X_{4} ) = P( X_{3} = x_{3}, X_{4} = x_{4} ) =\frac{1}{8} +\frac{3}{8} =\frac{1}{2}[/tex]

In a Multinomial Distribution, variables are independent. Hence, the joint mass function of a pair (X3 , X4) is the product of their individual mass functions. Their specific joint mass function equals P(X3=x3)P(X4=x4) = (n choose x3)(1/8)^x3(3/8)^x4 for x3+x4 ≤ n and x3, x4 ≥ 0.

This problem relates to the concept of a Multinomial Distribution in probability theory. The Multinomial Distribution describes the probabilities of potential outcomes from a multinomial experiment.

In this particular case, you are given that (X1, X2, X3, X4) follows a Multinomial Distribution with parameters n (number of trials) and 4 categories, with known probabilities 1/6, 1/3, 1/8, and 3/8 respectively.

You are asked to derive the joint mass function of the pair (X3, X4). This is actually very straightforward. Due to the properties of a multinomial distribution, these two variables are independent and the joint mass function is simply the product of the individual mass functions.

So the joint mass function of X3 and X4 would be P(X3=x3, X4=x4) = P(X3=x3)P(X4=x4) = (n choose x3)(1/8)^x3(3/8)^x4 provided x3+x4 ≤ n and each of x3,x4 are nonnegative.

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Answer:

Part A: the a quadratic function that reflects the company's revenue.

R = (800+40x)(80-2x) = 64,000 + 1,600 x - 80 x²

Part B: The price should the company charge to maximize its revenue = $60

Step-by-step explanation:

company that sells 800 phones each week when it charges $80 per phone. It sells 40 more phones per week for each $2 decrease in price

Part A: Find the a quadratic function that reflects the company's revenue.

Let the number of weeks = x, and the revenue R(x)

So, the number of sold phones = 800 + 40x

And the cost of the one phone = 80 - 2x

∴ R = (800+40x)(80-2x)

∴ R = 64,000 + 1,600 x - 80 x²

Part B: What price should the company charge to maximize its revenue?

The equation of the revenue represent a parabola

R = 64,000 + 1,600 x - 80 x²

The maximum point of the parabola will be at the vertex

see the attached figure

As shown, the maximum will be at the point (10, 72000)

Which mean, after 10 weeks

The number of sold phones = 800 + 40*10 = 1,200 phones

The price of the phone = 80 - 2 * 10 = 80 - 20 = $60

So, the price should the company charge to maximize its revenue = $60

In this exercise we have to use the knowledge of quadratic function to calculate the value of the company in this way we can say uqe;

A) [tex]R= 64,000 + 1,600 X - 80 X^2[/tex]

B)[tex]V= \$60[/tex]

A) First, we find the a quadratic function that reflects the company's revenue:

[tex]R = (800+40X)(80-2X)\\ R = 64,000 + 1,600X - 80 X^2[/tex]

B)The equation of the revenue represent a parabola:

[tex]R = 64,000 + 1,600X - 80 X^2[/tex]

As shown, the maximum will be at the point 10, which mean, after 10 weeks the number of sold phones

[tex]S = 800 + 40*10 = 1,200 \\P=1,200/200=60[/tex]

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Answer:

1.33 x 10⁻⁸ seconds

Step-by-step explanation:

Assuming that the speed of light is 299,792,458 m/s, and that in order to bounce of Bill's forehead, hit the mirror and return back to his eyes, light must travel 4 meters (distance to the mirror and back) the time that it takes for light to travel is:

[tex]t=\frac{4}{299,792,458} \\t=1.33*10^{-8}[/tex]

It takes 1.33 x 10⁻⁸ seconds.

Answer:

Step-by-step explanation:

Yes, integers like 27,57,87,117,.... and so on gives a remainder of 2 when it is divided by 5 and a remainder of 3 when it is divided by 6.

Yes, there is an integer that has a remainder of 2 when divided by 5 and a remainder of 3 when divided by 6. One example of such an integer is 23.

Let's use variables to rewrite the given statement. We can represent the integer as 'n', and the two remainders as 'r1' and 'r2'.

Therefore, the rewritten statement is: Is there an integer 'n' such that 'n' has a remainder of 2 when divided by 5 and a remainder of 3 when divided by 6?

Yes, such integers exist. One example is 23. When 23 is divided by 5, the remainder is 3, and when it is divided by 6, the remainder is also 3.

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Answer:

a) The 95% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses is (0.4239, 0.5161).

b) The 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses is (0.4094, 0.5306).

c)The margin of error increases as the confidence level increases.

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 = 450, p = 0.47[/tex]

a) Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses.

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}} = 0.47 - 1.96\sqrt{\frac{0.47*0.53}{450}} = 0.4239[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47*0.53}{450}}{119}} = 0.5161[/tex]

The 95% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses is (0.4239, 0.5161).

b. Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition andliving expenses.

95% confidence interval

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 2.575\sqrt{\frac{0.47*0.53}{450}} = 0.4094[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 2.575\sqrt{\frac{0.47*0.53}{450}}{119}} = 0.5306[/tex]

The 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses is (0.4094, 0.5306).

c What happens to the margin of error as the confidence is increased from 95% to 99%?

The margin of error is the subtraction of the upper end by the lower end of the interval, divided by 2. So

95% confidence interval

[tex]M = \frac{(0.5161 - 0.4239)}{2} = 0.0461[/tex]

99% confidence interval

[tex]M = \frac{(0.5306 - 0.4094)}{2} = 0.0606[/tex]

The margin of error increases as the confidence level increases.

Answer:

d. No, you should examine the situation to identify lurking variables that may be influencing both variables

Step-by-step explanation:

Hello!

Finding out that there is a regression between two variables is not enough to claim that there is a causation relationship between the two of them. First you have to test if other factors are affecting the response variable, if so, you have to control them or test how much effect they have. Once you controled all other lurking variables you need to design an experiment, where only the response and explanatory variables are left uncontroled, to learn if there is a regression and its strenght.

If after the experiment, you find that there is a significally strog relationship between the variables, then you can imply causation between the two of them.

I hope it helps!

The ODE has characteristic equation

[tex]r^2+6r+9=(r+3)^2=0[/tex]

with roots [tex]r=-3[/tex], and hence the characteristic solution

[tex]y_c=C_1e^{-3t}+C_2te^{-3t}[/tex]

For the particular solution, assume an ansatz of [tex]y_p=ae^{-t}[/tex], with derivatives

[tex]\dfrac{\mathrm dy_p}{\mathrm dt}=-ae^{-t}[/tex]

[tex]\dfrac{\mathrm d^2y_p}{\mathrm dt^2}=ae^{-t}[/tex]

Substituting these into the ODE gives

[tex]ae^{-t}-6ae^{-t}+9ae^{-t}=4ae^{-t}=4e^{-t}\implies a=1[/tex]

so that the particular solution is

[tex]\boxed{y(t)=C_1e^{-3t}+C_2te^{-3t}+e^{-t}}[/tex]

Answer:

a) Mean = 2.8

b) Median = 3

c) Mode = 4

d) Mid range = 2.5

e) Option C) Only the mode makes sense since the data is nominal.  

Step-by-step explanation:

We are given the following data set in the question:

1, 4, 4, 4, 2, 1, 4, 3, 1, 4, 4, 3, 3, 1

a) Mean

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

[tex]Mean =\displaystyle\frac{39}{14} = 2.78 \approx 2.8[/tex]

b) Median

[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]

Sorted data:

1, 1, 1, 1, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4

[tex]\text{Median} = \dfrac{7^{th}+8^{th}}{2} = \dfrac{3+3}{2} = 3[/tex]

c) Mode

Mode is the observation with highest frequency. Since 4 appeared maximum time  

Mode = 4

d) Mid range

It is the average of the smallest and largest observation of data.

[tex]\text{Mid Range} = \dfrac{1+4}{2} = 2.5[/tex]

e) Measure of center

Option C) Only the mode makes sense since the data is nominal.

A single taxi fare is more likely to fall between $21 & $24 or be over $24 compared to the mean of a sample of 10 taxi fares.

The question revolves around understanding the distribution of taxi fares and comparing the probabilities associated with the means of samples versus individual observations from a normally distributed population.

(A) Probability of falling between $21 & $24

A single taxi fare has greater variability and thus a greater probability of falling within the range of $21 & $24 compared to the mean of a random sample of 10 taxi fares. This is due to the Central Limit Theorem, which states that the distribution of the sample means will have a smaller standard deviation than that of individual observations, also known as the standard error. For a sample size of 10, the standard error is the population standard deviation divided by the square root of the sample size, which leads to a narrower distribution for the sample means compared to the distribution of individual fares.

(B) Probability of being over $24

The probability of a single randomly selected taxi fare being over $24 is greater than that of the mean of 10 randomly selected taxi fares. This is because individual observations are more spread out, as indicated by the standard deviation of the population, whereas the distribution of sample means is more concentrated around the mean due to the reduced standard error.

(A) The mean of a random sample of 10 taxi fares should have the greater probability of falling between $21 and $24.

(B) The amount of a single randomly selected taxi fare should have a greater probability of being over $24.

The Central Limit Theorem states that the distribution of the sample means will approach a normal distribution as the sample size increases, with the mean of the sample means being equal to the population mean and the standard deviation of the sample means (also known as the standard error) being equal to the population standard deviation divided by the square root of the sample size.

For a single taxi fare, the probability of falling between $21 and $24 can be calculated using the standard normal distribution. We first find the Z-scores corresponding to $21 and $24:

Z-score for $21: [tex]\( Z = \frac{X - \mu}{\sigma} = \frac{21 - 22.27}{2.20} = -0.58 \)[/tex]

Z-score for $24: [tex]\( Z = \frac{X - \mu}{\sigma} = \frac{24 - 22.27}{2.20} = 0.80 \)[/tex]

Using a standard normal table or calculator, we can find the probabilities corresponding to these Z-scores:

P(-0.58 < Z < 0.80) = P(Z < 0.80) - P(Z < -0.58) ≈ 0.788 - 0.278 ≈ 0.510

For the mean of a random sample of 10 taxi fares, the standard error (SE) is:

[tex]\( SE = \frac{\sigma}{\sqrt{n}} = \frac{2.20}{\sqrt{10}} \approx 0.70 \)[/tex]

Now we calculate the Z-scores for $21 and $24 using the standard error:

[tex]Z-score for $21: \( Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{21 - 22.27}{0.70} \approx -1.75 \)[/tex]

[tex]Z-score for $24: \( Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{24 - 22.27}{0.70} \approx 2.53 \)[/tex]

The probability that the sample mean falls between $21 and $24 is then:

P(-1.75 < Z < 2.53) = P(Z < 2.53) - P(Z < -1.75) ≈ 0.994 - 0.040 ≈ 0.954

Comparing the two probabilities, 0.954 for the sample mean is greater than 0.510 for a single fare.

Explanation for (B):

For a single taxi fare, we already calculated the Z-score for $24, which is 0.80. The probability of a single fare being over $24 is:

P(Z > 0.80) = 1 - P(Z < 0.80) ≈ 1 - 0.788 ≈ 0.212

For the mean of a random sample of 10 taxi fares, we calculated the Z-score for $24 as 2.53. The probability of the sample mean being over $24 is:

P(Z > 2.53) = 1 - P(Z < 2.53) ≈ 1 - 0.994 ≈ 0.006

Comparing the two probabilities, 0.212 for a single fare is greater than 0.006 for the sample mean. Therefore, the amount of a single randomly selected taxi fare has a greater probability of being over $24.

Answer:

cos 38° = 17/c

Step-by-step explanation:

in the triangle shown

the sum of angles in a triangle is 180°

its a right angle triangle meaning one of the angles is 90°

the other part is 52°

the third part is described as x

90° + 52° + x = 180° ( sum of angles )

142° + x = 180°

x = 180 - 142 = 38°

cos 38° = adjacent/hypothenus = 17/c

cos 38° = 17/c

Answer:

There is enough evidence to support the claim that  the true proportion is in fact different from 0.40  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 120

p = 0.4

Alpha, α = 0.05

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.4\\H_A: p \neq 0.4[/tex]

This is a two-tailed test.  

Formula:

[tex]\hat{p} = 0.3[/tex]

[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

Putting the values, we get,

[tex]z = \displaystyle\frac{0.3-0.4}{\sqrt{\frac{0.4(1-0.4)}{120}}} = -2.236[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

Since,

The calculated z-statistic does not lies in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there is enough evidence to support the claim that  the true proportion is in fact different from 0.40

Answer:

The growth factor b is 1.04

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

[tex]y=a(b^x)[/tex]

where

y ---> is the number of students enrolled at a college

x ----> the number of years

a is the initial value or y-intercept

b is the growth factor

b=(1+r)

r is the rate of change

we have

[tex]a=15,000\ students[/tex]

[tex]r=4\%=4/100=0.04[/tex]

[tex]b=1+0.04=1.04[/tex]

therefore

The exponential function is equal to

[tex]y=15,000(1.04^x)[/tex]

Answer:

26.76% probability that a randomly chosen golfer's score is above 70 strokes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 68.2, \sigma = 2.91[/tex]

What is the probability that a randomly chosen golfer's score is above 70 strokes?

This is 1 subtracted by the pvalue of Z when X = 70. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 68.2}{2.91}[/tex]

[tex]Z = 0.62[/tex]

[tex]Z = 0.62[/tex] has a pvalue of 0.7324.

So there is a 1-0.7324 = 0.2676 = 26.76% probability that a randomly chosen golfer's score is above 70 strokes.

Final answer:

To find the probability that a golfer's score is above 70, calculate the Z-score using the formula Z = (X - μ) / σ, where X is 70, the mean (μ) is 68.2, and the standard deviation (σ) is 2.91. The result, approximately 0.62, corresponds to a cumulative probability that must be subtracted from 1 to find the probability of scoring above 70. This process estimates there's about a 26.8% chance a golfer scores above 70.

Explanation:

The question asks about the probability that a randomly chosen golfer on the PGA tour has a score above 70 strokes, given that the scores are normally distributed with a mean of 68.2 and a standard deviation of 2.91. To find this probability, we use the Z-score formula, which is Z = (X - μ) / σ, where X is the score of interest, μ (mu) is the mean, and σ (sigma) is the standard deviation.

Calculating the Z-score for a score of 70:
Z = (70 - 68.2) / 2.91 ≈ 0.62.

Next, we consult a Z-table or use a calculator to find the probability corresponding to a Z-score of 0.62, which tells us the probability of a score being less than 70. To find the probability of a score being above 70, we subtract this value from 1.

Note: Specific values from the Z-table or calculator are not provided here. Generally, the process would involve looking up the cumulative probability for a Z-score of 0.62, which might be around 0.732. Therefore, the probability of a score above 70 would be 1 - 0.732 = 0.268. This means there's approximately a 26.8% chance that a randomly chosen golfer's score is above 70.