You roll the same die three times. Consider the possible outcomes if the order of the resultsis not recorded (meaning, e.g., that 1-2-1 and 2-1-1 are considered the same outcome).(a) Order the possible results in lexicographical order and show by direct counting that thenumber of possible outcomes is

Answer:

B. The number of girls is significantly low.

Step-by-step explanation:

Given that 200 births are randomly selected and 6 of the births are girls.

Normally we expect 50% of births to be boys and 50% girls.  If it is not exactly 50% atleast near to 50% we expect

But here out of 200 births, only 6 births are girls.

This means percentage of girls birth= [tex]\frac{6}{200} =0.03[/tex]=3%

while that of boys is 97%

This is a very unbalanced figure posing danger to the future generations.

B. The number of girls is significantly low.

The number of girls out of 200 randomly selected births is significantly low.

The number of girls out of a randomly selected group of 200 births is 6. To determine if this number is significantly high or low, we can use statistical methods. To do this, we can calculate the expected number of girls using probability. Assuming a 50% probability of having a boy or a girl, the expected number of girls would be 200 * 0.5 = 100. Comparing the observed number of girls (6) to the expected number (100), we can see that the number of girls is significantly low.

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Check the picture below.

Answer:

2√10 / 7

Step-by-step explanation:

to solve cos(sin^-1(3/7))

we break it into simpler terms

sin^-1(3/7) ------ these will be taken as an angle when dealing with cos

sin Ф = opposite / hypothenus = 3/7

Using Pythagoras Theorem

Hypothenus ² = opposite² + adjacent ²

7² = 3² + a²

49 = 9 + a²

a² = 49 - 9 = 40

a = √40 = √2x2x10 = 2√10

cos Ф = adjacent / hypothenus = 2√10 / 7  

cos(sin^-1(3/7)) = 2√10 / 7  

The value of the given expression is 2√10 / 7.

Trigonometry is the branch of mathematics that set up a relationship between the sides and angle of the right-angle triangles.

Trigonometric identities are equality conditions in trigonometry that hold for all values of the variables that appear and are defined on both sides of the equivalence. These are identities that, geometrically speaking, involve certain functions of one or more angles.

Solve cos(sin⁻¹(3/7)). Break it into simpler terms sin⁻¹(3/7) these will be taken as an angle when dealing with cos.

sin Ф = opposite / hypothenus = 3/7

Using Pythagoras Theorem

Hypothesis ² = opposite² + adjacent ²

7² = 3² + a²

49 = 9 + a²

a² = 49 - 9 = 40

a = √40 = √2x2x10 = 2√10

cos Ф = adjacent / hypothenus = 2√10 / 7  

cos(sin⁻¹(3/7)) = 2√10 / 7  

Hence, the value of cos(sin⁻¹(3/7)) will be 2√10 / 7.

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Answer:

verifications proved as shown in the attachment

Step-by-step explanation:

The detailed step by step and appropriate derivation is as shown in the attached file.

Final answer:

The problem requires deriving a formula for the area of a disk slice given its width h and verifying specific values for A(h). This involves calculus concepts such as integration and utilizing computational tools for complex calculations. We prove the equalities as shown below.

Explanation:

To derive a formula A(h)  for the area of the slice of the unit disk with width h, we first need to visualize the problem. Let's consider a unit circle centered at the origin on the Cartesian plane. Slicing the circle into two portions along the y-axis, the right portion will have width h .

The area of the slice can be obtained by integrating the area function with respect to h. Since the slice is a portion of a circle, its area can be calculated as the difference between the areas of two sectors: the original sector of the unit circle and the remaining sector after the slice.

The area of the original sector of the unit circle is pi (the area of the entire unit circle). The area of the remaining sector is given by the formula for the area of a sector of a circle:

[tex]\[ A_{\text{remaining}} = \frac{1}{2} r^2 \theta \][/tex]

Now, let's find [tex]\( \theta \)[/tex]. Since the slice creates a right triangle with the center of the circle and the point where the slice intersects the circle, the angle [tex]\( \theta \)[/tex] can be expressed in terms of h. Using trigonometry, we find that:

[tex]\[ \theta = 2 \arccos{\left(\frac{1}{2}\right)} = \frac{\pi}{3} \][/tex]

Thus, the area of the remaining sector is:

[tex]\[ A_{\text{remaining}} = \frac{1}{2} \cdot 1^2 \cdot \frac{\pi}{3} = \frac{\pi}{6} \][/tex]

Therefore, the area of the slice [tex]\( A(h) \)[/tex] is:

[tex]\[ A(h) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \][/tex]

Now, let's verify that \[tex]( A(0) = 0 \) and \( A(2) = \pi \).[/tex]

[tex]For \( A(0) \):[/tex]

[tex]\[ A(0) = \frac{5\pi}{6} \][/tex]

[tex]\[ A(0) = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \][/tex]

For [tex]\( A(2) \):[/tex]

[tex]\[ A(2) = \frac{5\pi}{6} \][/tex]

Thus,[tex]\( A(0) = 0 \) and \( A(2) = \pi \)[/tex], as desired.

Answer:

True

Step-by-step explanation:

The given questions can be solved scientifically by framing hypothesis and then testing the hypothesis. The best mode is to assume a hypothesis.  

After this a survey questionnaire can be prepared to get the inputs of a mass in terms of likeliness towards a particular brand on a numerical scale ranging from 1 to 5 where 5 represents higher likeliness towards a brand.  

Once the survey is completed, the obtained result can be analyzed and test statistically for higher probable solution.  

Hence, the given statement is true

Answer:

a) We can use the following excel code to find it:"=NORM.DIST(-2.34,0,1,TRUE)"

[tex] P(Z \leq -2.34)=0.0096[/tex]

b) [tex] P(Z \geq -2.34)= 1-P(X \leq -2.34)= 1-0.0096=0.9904[/tex]

c) We can use the following excel code: "=1-NORM.DIST(1.74,0,1,TRUE)"

[tex] P(Z > 1.74)= 1-P(X \leq 1.74)=0.0409[/tex]

d) We can use the following excel code: "=NORM.DIST(1.74,0,1,TRUE)-NORM.DIST(-2.34,0,1,TRUE)"

[tex] P(-2.34<Z < 1.74)= P(Z<1.74)-P(Z<-2.34)=0.959-0.0096=0.949[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

We want to find this probability:

[tex] P(Z \leq -2.34)[/tex]

And we can use the following excel code to find it:"=NORM.DIST(-2.34,0,1,TRUE)"

[tex] P(Z \leq -2.34)=0.0096[/tex]

Part b

We want to find this probability:

[tex] P(Z \geq -2.34)[/tex]

And for this case we can use the complement rule:

[tex] P(Z \geq -2.34)= 1-P(X \leq -2.34)[/tex]

And using the result from part a we got:

[tex] P(Z \geq -2.34)= 1-P(X \leq -2.34)= 1-0.0096=0.9904[/tex]

Part c

We want to find this probability:

[tex] P(Z > 1.74)[/tex]

And for this case we can use the complement rule:

[tex] P(Z > 1.74)= 1-P(X \leq 1.74)[/tex]

And we can use the following excel code: "=1-NORM.DIST(1.74,0,1,TRUE)"

[tex] P(Z > 1.74)= 1-P(X \leq 1.74)=0.0409[/tex]

Part d

We want to find this probability:

[tex] P(-2.34<Z <1.74)[/tex]

And for this case we can find this probability with this difference:

[tex] P(-2.34<Z < 1.74)= P(Z<1.74)-P(Z<-2.34)[/tex]

And we can use the following excel code: "=NORM.DIST(1.74,0,1,TRUE)-NORM.DIST(-2.34,0,1,TRUE)"

[tex] P(-2.34<Z < 1.74)= P(Z<1.74)-P(Z<-2.34)=0.959-0.0096=0.949[/tex]

The given z-scores correspond to various portions of a standard Normal distribution. Using a standard Normal distribution table, the proportions for the respective regions are derived as: z <= -2.34 is approximately 0.0094, z >= -2.34 is approximately 0.9906, z > 1.74 is approximately 0.0409, -2.34 < z < 1.74 is approximately 0.9497.

To answer your question, let's use the standard Normal distribution table which provides the proportion of observations less or equal to a particular z-score (for the values of z <= 0 and z >= 0).

Please remember that these proportions represent the areas under the curve (or the total probability) that the z-score will fall within the outlined regions.

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Answer:

1)

[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]

2) z=3.164

3) Critic value z₀=1.282.

4) P=0.00078

5) The correct answer is: "Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0"

Step-by-step explanation:

5.1) Being:

p₁: proportion of men who keep track of the deadlines in their head

p₂: proportion of women who keep track of the deadlines in their head

If we want to test if p₁ is larger than p₂, the null hypothesis and the alternative hypothesis should be:

[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]

In this way, if we reject the null hypothesis, it can be claimed that p₁ is larger than p₂.

5.2) Compute the test statistic for the test.

First, we have to estimate a proportion as if the null hypothesis is true. This means the average of proportion of the samples taken from men and women, weighted by the sample size.

[tex]\bar p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{500*0.54+500*0.44}{500+500}= 0.49[/tex]

Then, we used this average to estimate the standard error

[tex]s=\sqrt{\frac{p(1-p)}{n_1}+{\frac{p(1-p)}{n_2}}}=\sqrt{\frac{0.49(1-0.49)}{500}+{\frac{0.49(1-0.49)}{500}}}=\sqrt{0.0004998+0.0004998}\\\\s= 0.0316[/tex]

Lastly, we calculate the statistic z

[tex]z=\frac{p_1-p_2}{s}=\frac{0.54-0.44}{0.0316}=\frac{0.10}{0.0316}=3.164[/tex]

5.3) Give the rejection region for the test, using α = 0.10

For a one-tailed test with α = 0.10, the z value to limit the rejection region is z=1.282.

For every statistic larger than 1.282, the null hypothesis should be rejected.

5.4) Find the p-value for the test.

The p-value for a z=3.164 is P=0.00078 (corresponding to the area ot the standard normal distribution for a z larger than 3.164).

5.5) Choose the correct answer below.

The correct answer is: "Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0"

The difference between the proportions is big enough to be statistically significant and enough evidence to reject the null hypothesis.

Answer:

5

Step-by-step explanation:

The sample space is considered to be the total number of possibilities in a given sample or study. Here we are told that the bag contains two red marbles, two blue marbles, and one yellow marble. So the sample space is 5, the total number of marbles available and possible in a selection

Final answer:

The sample space for selecting two marbles without replacement from a bag with two red, two blue, and one yellow marble consists of the pairs RR, RB, RY, BR, BB, BY, YR, YB.

Explanation:

When we are picking two marbles without replacement from a bag containing two red marbles, two blue marbles, and one yellow marble, we are dealing with a probabilistic experiment. To find the sample space of this experiment, we need to list all possible pairs of marbles that could result from this process.

Here are the possible combinations without replacement:

Yellow and Blue (YB)

Note that combinations like Red and Blue (RB) and Blue and Red (BR) are distinct since the marbles are drawn one after the other. With this comprehensive listing, we have fulfilled the task of defining the sample space, which consists of the following pairs: RR, RB, RY, BR, BB, BY, YR, YB.

Answer:

Step-by-step explanation:

When data points are plotted between two variables, seeing the scatter plot we can form idea about the curve which is close to all points.

In other words, using least squares method, the curve of best fit would be the curve which has squares of deviations form the observed points a minimum

Thus conditions are

i) The scatter plot suggests a linear relationship

ii) Pearson correlation coefficient has value near to 1 or -1 suggesting linearlity

iii) There seems to be constant increase of y for unit increase in x

Under the above linear line fits

In all the other cases, quadratic or cubic function fits well.

Answer:

[tex] 1.05-2.65 \frac{0.23}{\sqrt{65}} \approx 0.98[/tex]

[tex] 1.05+2.65 \frac{0.23}{\sqrt{65}} \approx 1.12[/tex]

So on this case the 99% confidence interval for the mean would be given by (0.98;1.12)

And the best option is:

C..98 to 1.12

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=1.05[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=0.23 represent the sample standard deviation

n=65 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=65-1=64[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,64)".And we see that [tex]t_{\alpha/2}=2.65[/tex]

Now we have everything in order to replace into formula (1):

[tex] 1.05-2.65 \frac{0.23}{\sqrt{65}} \approx 0.98[/tex]

[tex] 1.05+2.65 \frac{0.23}{\sqrt{65}} \approx 1.12[/tex]

So on this case the 99% confidence interval for the mean would be given by (0.98;1.12)

And the best option is:

C..98 to 1.12

Answer:

t = 45 cubic light years to find a star with this certainty.

Step-by-step explanation:

The Poisson random probability equation is given by:

[tex]P(k events in interval t)=\frac{(\lambda t)^{k}e^{-\lambda t}}{k!}[/tex]

We can use the next equation to quantify how many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94.

[tex]P(k\ge 1) \ge 0.94[/tex]

[tex]1-f(0)=1-\frac{(\frac{1}{16}*t)^{0}e^{-\frac{1}{16}*t}}{0!}=1-e^{-\frac{1}{16}*t}} \ge 0.94[/tex]

So, here we just need to solve it for t:

[tex]1-e^{-\frac{1}{16}*t}} \ge 0.94[/tex]

[tex]e^{-\frac{1}{16}*t}} \ge 0.06[/tex]

[tex]ln(e^{-\frac{1}{16}*t}}) \ge ln(0.06)[/tex]

[tex]-\frac{1}{16}*t \ge -2.8[/tex]

[tex]t \ge 44.8[/tex]

Therefore t = 45 cubic light years to find a star with this certainty.

I hope it helps you!

To find the volume of space where the probability of encountering one or more stars exceeds 0.94 in the Milky Way, use the inverse of the cumulative distribution function of the Poisson distribution, with the given star density of one per 16 cubic light years, and solve for the volume V such that the probability of zero stars is below 0.06.

In order to solve the problem of how many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94, we need to use the properties of the Poisson distribution. The Poisson distribution is often used for modeling the number of events in fixed intervals of time or space under certain conditions. If we let λ denote the average number of stars in a given volume, then for our Milky Way Galaxy in the vicinity of our solar system where the density is one star per 16 cubic light years, λ is 1/16 per cubic light year. The probability of finding no stars in a volume V is given by e^{-λV}. To find when the probability of one or more stars exceeds 0.94, we need to find V such that the probability of finding no stars is less than 0.06 (which is 1 - 0.94).

Let's calculate the volume V:

We first solve the inequality e^{-λV} < 0.06 for V.

Taking the natural logarithm of both sides gives us -λV < ln(0.06).

We then solve for V: V > ln(0.06) / -λ.

Substituting λ (1/16) gives us V > ln(0.06) / -(1/16).

Finally, calculate the numeric value and round to the nearest integer to find the minimum volume V that meets the requirement.

By performing these calculations, you can find how many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.94.

Answer:

Step-by-step explanation:

Prime numbers are numbers that are divisible by itself and '1'. This means prime numbers are numbers with just two factors while composite numbers are numbers that are divisible by more than two numbers, that is they have more than 2 factors.

Example of prime numbers are 5,7,11 etc. We see that these numbers are only divisible by themselves and by '1' while examples of composite numbers are 10, 15, 20, etc. We clearly see that these numbers have more than 2factors.

Prime numbers are divisible by only 2 numbers: 1 and themselves.

Composite numbers have 3 or more factors.

Hope this helps!

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[tex]GraceRosalia[/tex]

~Just a joyful teen

The vector [4,7] represents the direction from the origin to the point (4,7). It has the same numerical values for both x and y components.

In mathematics,

A point represents a specific location in space, while a vector represents both magnitude and direction.

However, when the components of a vector match the coordinates of a point, we can say that the vector points from the origin to that specific point.

So in this case,

The vector [4,7] can be thought of as pointing from the origin (0,0) to the point (4,7).

It's like an arrow starting from the origin and ending at the point (4,7).

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The point (4,7) marks a specific location in the Cartesian coordinate system, while the vector [4,7] represents a direction and magnitude from the origin to the point. In a graphical representation, a vector is drawn as an arrow from the origin to the point, showing displacement, while a point is just a dot marking a location.

Answer:

a) m = 0.78

b) Average annual earnings by Males

c)  F = $ 30,548.685

Step-by-step explanation:

Given:

- The relationship between the average annual earning of Females F is expressed as a function of earnings by males M as:

                                          F = 0.78*M - 1.315

Find:

(a) Viewing F as a function of M, what is the slope of the graph of this function?

(b) what is M?

(c) When the average annual earnings for males reach $65,000, what does the equation predict for the average annual earnings for females?

Solution:

a)

- The given expression can be compared with a linear equation of the form:

                                            y = m*x + c

Where,

y is the dependent variable

m is the slope of the graph

x is the independent variable

c is the intercept when x = 0

- So for our case , y = F , m = 0.78 , x = M , c = -1.315

b)

- The variable M is the independent variable with the domain [ 0 , infinity ] denotes the average annual earnings of Males in dollars ($).

c)

- When average annual earnings by males is $ 65,000 i.e M = 65,000, We compute F by the given expression:

                                          F = 0.47 * 65,000 - 1.315

                                          F = $ 30,548.685

- The average annual earning by females is $ 30,549 when earning by males reach $65,000.

Answer:

[tex]\Omega=\{1,2,3,4,5,6\}[/tex]

[tex]A=\{1,2,3,4\}[/tex]

Step-by-step explanation:

Sample Space

The sample space of a random experience is a set of all the possible outcomes of that experience. It's usually denoted by the letter [tex]\Omega[/tex].

We have a number cube with all faces labeled from 1 to 6. That cube is to be rolled once. The visible number shown in the cube is recorded as the outcome. The possible outcomes are listed as the sample space below:

[tex]\Omega=\{1,2,3,4,5,6\}[/tex]

Now we are required to give the outcomes for the event of rolling a number less than 5. Let's call A to such event. The set of possible outcomes for A has all the numbers from 1 to 4 as follows

[tex]A=\{1,2,3,4\}[/tex]

Possible outcomes of rolling a cube numbered from 1 to 6 are 1, 2, 3, 4, 5, 6. Outcomes of rolling a number less than 5 are 1, 2, 3, 4.

When a number cube with faces labeled from 1 to 6 is being rolled once, the possible values that can be observed (i.e., the sample space) are 1, 2, 3, 4, 5, and 6. These are all the possible outcomes of this particular event.

When we want to find the outcomes for the event of rolling a number less than 5, we need to look at our sample space and identify which numbers meet these criteria. These numbers would be 1, 2, 3, and 4.

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Answer:

P= 0.0039

Step-by-step explanation:

From Exercise we have 12 women and 4 men.

We calculate the number of combinations of 16 people, to be divided into 4 teams of 4 people each.

{16}_C_{4} · {12}_C_{4} · {8}_C_{4} =

=\frac{16!}{4!·(16-4)!} · \frac{12!}{4!·(12-4)!}· \frac{8!}{4!·(8-4)!}

=1820 · 495 ·  105

=94594500

The number of favorable combinations is

{12}_C_{3} · {9}_C_{3} · {6}_C_{3} =

=\frac{12!}{3!·(12-3)!} · \frac{9!}{3!·(9-3)!}· \frac{6!}{3!·(6-3)!}

=220 · 84 ·  20

=369600

The  probability is 369600/94594500=16/4095=0.0039

P= 0.0039

Answer: a) {x | x ∈ N , 1 ≤x≤7} b ) {10 x  | x ∈ Z ,  0≤x≤4 }

Step-by-step explanation:

A set builder form is the the mathematical way to represent aset by using its particular property of characteristic.

For example :  A = {2,4,5,6,......} is represented as

A= {2x | x ∈ N} , where N is a set of natural numbers.

a)  {1, 2, 3, 4, 5, 6, 7}

Let D = {1, 2, 3, 4, 5, 6, 7}

Then, in set builder form , we have

D = {x | x ∈ N , 1 ≤x≤7}

b) ) {1, 10, 100, 1000, 10000}

Let B = {1, 10, 100, 1000, 10000}

[tex]={10^{0} , 10^1 , 10^2 , 10^3 , 10^4}[/tex]

We can see that each element in set B is a multiple of 10.

So in set builder form , we will have

B = {10 x  | x ∈ Z ,  0≤x≤4 } , where Z is the set of integers.

Set-builder notation is used to describe sets using mathematical notation. For the given sets {1, 2, 3, 4, 5, 6, 7} and {1, 10, 100, 1000, 10000}, we can describe them using set-builder notation.

To describe the set {1, 2, 3, 4, 5, 6, 7} using set-builder notation, we can write it as:

{x | x is an element of the set of natural numbers and 1 ≤ x ≤ 7}

Similarly, to describe the set {1, 10, 100, 1000, 10000}, we can write it as:

{x | x is an element of the set of natural numbers and x is a power of 10}

Answer:

The answer to the question is

Inferential statistics

Step-by-step explanation:

Inferential statistics is used to make informed conclusions about a population that cannot be completely sampled due to the population size.

With Inferential statistics, it is possible to make predictions or inferences from available data. It involves collecting data from a random sample of individuals within the population concerned and make generalizations about the entire population from those samples.

Answer: The answer to the question is

Inferential statistics

Step-by-step explanation:

Inferential statistics is used to make informed conclusions about a population that cannot be completely sampled due to the population size.

With Inferential statistics, it is possible to make predictions or inferences from available data. It involves collecting data from a random sample of individuals within the population concerned and make generalizations about the entire population from those samples.

Answer:

d = 13 / 5 units          

Step-by-step explanation:

Given:

- Point P = (-2 , 3 )

- line : 3x - 4y + 5 = 0

Find:

- Use projection to show the given formula

- use the formula to find the distance from given point and line.

Solution:

- Let line L be defined as:

                                     a*x + b*y + c = 0

- Choose A as fixed point on the line L whose coordinates are ( m , k ).

- Now the coordinates of any point (x , y ) can be given by the line L:

                                    a*( x - m ) + b*( y - k ) + c = 0

- We did that by subtracting the coordinates (x,y) from (m,k).

- The above line can represented by dot product of constants (a , b ) to vector coordinates between points (x,y) to (m,k):

                                   ( a , b ) . ( x - m , y - k) = 0

- For above expression to be true i.e the dot product of two vectors is only zero when the vectors are orthogonal. So vector (a , b ) is always perpendicular to every vector in direction of line L

- The vector:

                                         v =  ( x - m , y - k)

- It may represent the line connecting the points (x_1 , y_1) and (m, k).

Then the distance d from the point P_1: (x_1, y_1) to the line L is given by the magnitude of the scalar projection of v onto a, because the latter is the length of the side of a right  triangle with two of the vertices being P_1 and

( m , k), and the other vertex lying on L So:

                                      d = | component v along x |

                                      d =  | v . ( a , b ) | / | (a , b ) |

                                      d = | (x_1 - m , y_1 - k ) . ( a , b)| / sqrt(a^2 + b^2)

                                      d = | ax_1 + by_1 - (a*m + b*k) | / sqrt(a^2 + b^2)

Where point A (m,k) lies on line hence;

                                      a*m + b*k + c = 0

                                      c = - (a*m + b*k)

Hence,

                                      d = | ax_1 + by_1 + c | / sqrt(a^2 + b^2)

- Given point P (-2,3) and line L : 3x - 4y + 5 = 0

where,

                                    (x_1 , y_1) = (-2 , 3 )

                                     (a , b) = ( 3 , -4 )

                                            c = 5

-Evaluate:

                                    d = | 3*(-2) + -4*3 + 5 | / sqrt(3^2 + 4^2)

                                    d = | - 13 | / 5

                                    d = 13 / 5 units                                    

The distance from the point (-2,3) to the line [tex]\(3x - 4y + 5 = 0\)[/tex] is [tex]\(\frac{|3(-2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}}\)[/tex].

To find the distance from a point [tex]\(P_1(x_1, y_1)\)[/tex] to a line [tex]\(ax + by + c = 0\)[/tex], one can use the formula derived from scalar projection:

[tex]\[ \text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \][/tex]

Here, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the line, and [tex]\(x_1\)[/tex] and [tex]\(y_1\)[/tex] are the coordinates of the point. The numerator represents the absolute value of the expression [tex]\(ax_1 + by_1 + c\)[/tex], which is the scalar projection of the vector from the origin to the point onto the normal vector of the line. The denominator is the magnitude of the normal vector of the line.

Given the point [tex]\(P_1(-2, 3)\)[/tex] and the line [tex]\(3x - 4y + 5 = 0\)[/tex], we substitute [tex]\(x_1 = -2\)[/tex] and [tex]\(y_1 = 3\)[/tex] into the formula:

[tex]\[ \text{Distance} = \frac{|3(-2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}} \][/tex]

Now, we calculate the numerator:

[tex]\[ 3(-2) - 4(3) + 5 = -6 - 12 + 5 = -13 + 5 = -8 \][/tex]

The absolute value of [tex]\(-8\)[/tex] is [tex]\(8\)[/tex], so the numerator becomes [tex]\(8\)[/tex].

Next, we calculate the denominator, which is the magnitude of the normal vector of the line:

[tex]\[ \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \][/tex]

Finally, we divide the absolute value we found in the numerator by the denominator to get the distance:

[tex]\[ \text{Distance} = \frac{8}{5} \][/tex]

Therefore, the distance from the point (-2,3) to the line [tex]\(3x - 4y + 5 = 0\)[/tex] is [tex]\(\frac{8}{5}\)[/tex].

Answer:

Yes, the given experiment is a binomial experiment.

Step-by-step explanation:

The conditions of the binomial experiment are

1. The trails are independent.

2. There are two possible outcomes i.e. success or failure.

3. The probability of success p remains constant on each trail.

4. The experiment is repeated fixed number of times n.

When an intruding aircraft enters the scene, the random variable X, the number of radar units that do not detect the plane indicates binomial experiment because

1. The four identical radar units are independent of each other.

2. There are two possible outcomes i.e. radar will detect the plane or radar will not detect the plane.

3. The probability of not detecting the plane p=0.05 remains constant on each trial. Here the probability of success is the probability of not detecting the plane can be calculated as

The probability of not detecting the plane=1- probability of detecting the plane

4. The experiment consists of four radar units i.e. n=4.

Answer:

[tex]z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717[/tex]  

[tex]p_v =P(z<-1.717)=0.0429[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

[tex]\hat p=\frac{46}{115}=0.4[/tex] estimated proportion of people that have traded in their old car

[tex]p_o=0.48[/tex] is the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level

Confidence=90% or 0.9

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.48.:  

Null hypothesis:[tex]p\geq 0.48[/tex]  

Alternative hypothesis:[tex]p < 0.48[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.1[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<-1.717)=0.0429[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

Answer:

dQ/dt = 3 - 2Q/(t+200), Q(0) = 100

Step-by-step explanation:

The rate of change dQ/dt describes the amount of salt in the tank at a given time, this rate of change can be express as rate-in minus the rate-out and those individual rates can be expressed as the rate at which liquid enters or leaves the tank (gal/min) by the amount of salt entering or leaving the tank at a given time (lb/gal).

Collecting everything said:

dQ/dt = r1 - r2 = 3 lb/min - 2 gal/min * Q/(t*(1 gal/min)+200)

Note:

Answer:

z_{calc}=2.57[/tex]

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

[tex]p_v =2*P(z>2.57)=0.0101[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from [tex]\mu_o[/tex] at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X[/tex] represent the sample mean

[tex]\sigma[/tex] represent the population standard deviation

[tex]n[/tex] sample size  

[tex]\mu_o [/tex] represent the value that we want to test  

Confidence = 95% or 0.95

[tex]\alpha=1-0.95=0.05[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to an specified value [tex]\mu_o [/tex], the system of hypothesis would be:  

Null hypothesis:[tex]\mu =\mu_o[/tex]  

Alternative hypothesis:[tex]\mu \neq \mu_o[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) and we got a value calculated let's say [tex] z_{calc}=2.57[/tex]

P-value  

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

[tex]p_v =2*P(z>2.57)=0.0101[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from [tex]\mu_o[/tex] at 5% of signficance.  

For a 95% confidence level in a two-tail hypothesis test, if the computed Z-value is 2.57, you would reject the null hypothesis, as the value lies outside the critical z-scores of ±1.96, indicating statistical significance.

If you use a 95% confidence level in a two-tail hypothesis test, the critical z-scores are ±1.96.

This means that if the computed test statistic falls beyond these values (either less than -1.96 or greater than +1.96), we reject the null hypothesis.

In the scenario where the computed value of the test statistic is Z = 2.57, it is greater than +1.96. Therefore, under these conditions, you will reject the null hypothesis.

The reason for this decision stems from the test statistic lying outside the range of values considered to be consistent with the null hypothesis at the 95% confidence level.

The absolute value of 2.57 is greater than 1.96, indicating that the observed effect is statistically significant, and the likelihood of observing such a result if the null hypothesis were true is less than 5% (which is the alpha level of 0.05 associated with a 95% confidence level).

This indicates that there is sufficient evidence to support the alternative hypothesis over the null hypothesis in a two-tailed test.

Answer:

C. No, the probability of this outcome at 0.080, would be considered usual, so there is no problem

Step-by-step explanation:

To solve this problem, it is important to know the Central Limit Theorem and the Normal probability distribution.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

Normal Probability Distribution

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.

When a probability is unusual?

A probability is said to be unusual if it is higher than 0.95 or lower than 0.05.

In this problem, we have that:

[tex]\mu = 20, \sigma = 0.11, n = 40, s = \frac{0.11}{\sqrt{15}} = 0.0284[/tex]

Should the machine be reset?

What is the probability of a mean of 20.04 or higher?

This is 1 subtracted by the pvalue of Z when X = 20.04. So

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

[tex]Z = \frac{20.04 - 20}{0.0284}[/tex]

[tex]Z = 1.41[/tex]

[tex]Z = 1.41[/tex] has a pvalue of 0.92.

So the probability of this outcome is 1-0.92 = 0.08 = 8%.

8% is still an usual probability.

So the correct answer is:

C. No, the probability of this outcome at 0.080, would be considered usual, so there is no problem

The sample mean of 20.04 ounces can be calculated using the z-score formula to determine if the machine should be reset. The probability of obtaining this sample mean is extremely close to 1, indicating it is likely to occur by random chance. Therefore, the machine does not need to be reset.

To determine if the machine should be reset, we can calculate the probability of obtaining a sample mean of 20.04 ounces or higher, assuming the machine is still set to the mean of 20 ounces. We can use the z-score formula to standardize the sample mean. The formula is z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size. Plugging in the values, we have z = (20.04 - 20) / (0.11 / sqrt(15)) = 5.45.

Next, we can find the probability associated with this z-score using a z-table or a calculator. From the z-table, we find that the probability is approximately 0.99999999.

Since the probability is extremely close to 1, which means it is very likely to occur by random chance, we can conclude that the sample mean of 20.04 ounces is not significantly different from the mean of 20 ounces. Therefore, there is no need to reset the machine.

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Answer:

The probability that the reporter made no typographical errors for the 3-page article is 0.7165.

Step-by-step explanation:

For Poisson Distribution:

p(k:λ) = [(λ^k)(e^-λ)]/k!

where k is the number of outcomes and λ is the rate at which the outcomes occur.

λ=np

where n=number of errors on one page

           p=probability that an error appears on a given page in the 3 page article

From the question we have n=1 and p=1/3. Computing these values:

λ=np

λ=1 x 1/3

λ=1/3

The question is asking for the probability that no error occurs in the 3-page article i.e. k=0. Using the Poisson formula mentioned above:

P(0:1/3) = (1/3)^0 e^(-1/3)/(0!)

            = (1)(0.7165)/1

P(0:1/3) = 0.7165

The probability that the reporter made no typographical errors for the 3-page article is 0.7165.

Using the Poisson distribution, the probability that a reporter makes no typographical errors in a 3-page article is estimated to be approximately 4.98%.

The probability of a reporter not making any typographical errors in a 3-page article can be calculated using the Poisson distribution. The Poisson distribution is commonly used for estimating the probability of a certain number of rare events occurring in a specified time or space.

The Poisson probability formula is P(x; μ) = (e^-μ) * (μ^x) / x!, where:
x = number of successes that result from the experiment
μ = average rate of success
e = mathematical constant approximately equal to 2.71828.

Plugging the given values into the formula gives us: P(0; 3) = (e^-3) * (3^0) / 0!. After simplifying this equation, we get an answer of approximately 0.0498 or 4.98% (rounded to four decimal places).

Therefore, the probability that the reporter made no typographical errors in the 3-page article is approximately 4.98%.

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Answer:

The confidence interval is (0.444 ± 0.111).

Step-by-step explanation:

The general form of a confidence interval for single proportion is: [tex](\hat p- E<p<p\hat + E)=\hat p \pm E[/tex]

The interval provided is: (0.333 < p < 0.555)

Then

[tex]\hat p-E=0.333...(i)\\\hat p + E = 0.555...(ii)[/tex]

Solving the two equation simultaneously:

Add (i) and (ii),

[tex]2\hat p=0.888\\\hat p=\frac{0.888}{2}\\ =0.444[/tex]

Substitute the value of [tex]\hat p[/tex] in (i) to compute the value of E:

[tex]\hat p-E=0.333\\0.444-E=0.333\\E=0.444-0.333\\=0.111[/tex]

Thus, the confidence interval is,

[tex](0.444-0.111<p<0.444+0.111)=0.444 \pm 0.111[/tex]

Answer:

Volume of the solid generated = pi2/84 cubic unit

Step-by-step explanation:

The deatiled step and appropriate integration is donw as shown in the attached file.

To find the volume of the solid when the region bounded by the curves is revolved about the x-axis, use the disk method to integrate the cross-sectional areas of the disks formed. The limits of integration are determined by finding the intersection points of the curves. The formula for the cross-sectional area is A = πr^2, where r is the distance from the x-axis to the function y(x).

To find the volume of the solid when the region bounded by the curves is revolved about the x-axis, we can use the disk method. The disk method involves integrating the cross-sectional area of each disk formed by rotating a thin vertical strip of the region about the x-axis. The formula for the cross-sectional area of a disk is A = πr^2, where r is the distance from the x-axis to the function y(x).

First, we need to determine the limits of integration. The curves y = cos(21x) and y = 0 intersect at x = 0 and x = π/42. So we integrate from x = 0 to x = π/42.

The distance from the x-axis to the function y(x) is y(x) = cos(21x). Therefore, the cross-sectional area of each disk is A(x) = π[cos(21x)]^2. To find the volume, we integrate A(x) from x = 0 to x = π/42:

V = ∫0π/42π[cos(21x)]^2 dx

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Answer:

0.33 m/s

Step-by-step explanation:

Given,

s = √(63+6t)..................... Equation 1

s' = ds/dt

Where s' = rate of change of the particles position.

Differentiating equation 1,

s = (63+6t)¹/²

s' = 6×1/2(63+6t)⁻¹/²

s' = 3(63+6t)⁻¹/²

s' = 3/√(63+6t)........................ Equation 2

At t = 3 s,

Substitute the value of t into equation 2

s' = 3/√(63+6×3)

s' = 3/√(63+18)

s' = 3/√(81)

s' = 3/9

s' = 0.33 m/s.

Hence the rate of change of the particles position = 0.33 m/s

Answer:

ds/dt = 0.33 m/s

Therefore, the rate of change of the particle's position at t = 3 sec is 0.33 m/s

Step-by-step explanation:

Given;

The position function of the particle.

s(t) = √(63+6t)

The rate of change of the particle's position = ds/dt = s(t)'

Using function of function rule.

Let u = 63+6t

s = √u

ds/dt = du/dt × ds/du

du/dt = 6

ds/du = 0.5u^(-0.5) = 0.5/u^(0.5) = 0.5/(63+6t)^(0.5)

ds/dt = 6 × 0.5/(63+6t)^(0.5)

ds/dt = 3/(63+6t)^(0.5)

At t = 3sec

ds/dt = 3/(63+6(3))^(0.5) = 3/9

ds/dt = 0.33 m/s

Therefore, the rate of change of the particle's position at t = 3 sec is 0.33 m/s

Answer:

From a statistical point of view: A correct decision was made.

so that the extent of the problem may be ascertained.

Step-by-step explanation:

"In statistical hypothesis testing, a type I error is the rejection of a true null hypothesis (also known as a "false positive" finding or conclusion), while a type II error is the non-rejection of a false null hypothesis (also known as a "false negative" finding or conclusion)."

Final answer:

The scenario described indicates that a Type I error was made during the hypothesis test, as the machine was working correctly but was stopped due to the rejection of the null hypothesis.

Explanation:

The question asks about the consequences of a hypothesis test related to the functioning of a machine filling quart size bottles of olive oil. When the hypothesis is rejected and the machine is shut down, but no issue is found upon inspection, this scenario describes a Type I error. This error occurs because the null hypothesis, which would be the assumption that the machine is functioning correctly, is rejected even though it is actually true.

Answer:

4. Slowing down from 100 km/h to 50 km/h

Step-by-step explanation:

The work done by car is given as the change in the kinetic energy of the car. Mathematically,

W = ΔK

where

W = work done by the brakes.

ΔK = Change in kinetic energy.

Kinetic Energy is given as:

[tex]K = \frac{1}{2}mv^{2}[/tex]

Case 1: The car goes from 50 km/h to 0 km/h

ΔK = [tex]K_{f} - K_{i}[/tex]

ΔK =  [tex](\frac{1}{2}*50^{2}*m) - (\frac{1}{2}*0^{2}*m)[/tex]

ΔK = 1250m J

∴ W = 1250m J

Case 2: The car goes from 100 km/h to 50 km/h

ΔK = [tex]K_{f} - K_{i}[/tex]

ΔK = [tex](\frac{1}{2}*100^{2}*m) - (\frac{1}{2}*50^{2}*m)[/tex]

ΔK = 5000m - 1250m

ΔK = 3750m J

∴ W = 3750m J

Note: Mass of the car is constant.

Hence, slowing down from 100 km/h to 50 km/h requires the brakes to do more work, precisely 3 times more work [tex](3750m/1250m = 3)[/tex]

Final answer:

The work done by car brakes is greatest when slowing down from 100 km/h to 50 km/h, as more kinetic energy needs to be dissipated compared to slowing down from 50 km/h to rest.

Explanation:

The question pertains to the work done by brakes when slowing down a car. The work done to slow a car down is directly related to the car's kinetic energy, which depends on the square of its velocity. Therefore, slowing down from a higher speed requires more work, as it involves dissipating a larger amount of kinetic energy. Specifically, the work done by the brakes on a car slowing down from 100 km/h to 50 km/h is greater than the work required for a car slowing down from 50 km/h to rest. This is because the kinetic energy at 100 km/h is four times greater than at 50 km/h due to kinetic energy's dependence on the square of velocity (KE = 1/2 mv²).