A coin is tossed 3 times. What is the probability that the number of tails obtained will be between 1 and 3 inclusive?

Answer:

The correct option is option a. an extra variable that could explain result differences between groups.

Step-by-step explanation:

Confusion variables, also called third variables, are variables that the investigator did not control or did not eliminate and that damage the internal validity of an experiment, that is, the degree to which the results are valid.

The dependent and independent variables are the two main variables of any experiment or investigation. The independent is one that changes or is controlled to study its effects on the dependent variable. The dependent is that variable that is investigated and measured.

Confusion variables may cause the investigator to analyze the results incorrectly. The results may show a false correlation between dependent and independent variables.

So the correct option is option a. an extra variable that could explain result differences between groups.

This is because this additional variable that was not controlled or taken into account during the experimental study modifies the result. And the other groups that conducted the experiment and took into account the variable will have different results because they take into account other independent variables.

The linear equation 1/2x + y = 5, we can choose values for x and solve for y. When we plot the points on a coordinate plane, we get a line graph.

To find solutions to the linear equation 1/2x + y = 5, we can arbitrarily choose values for x and solve for y.

Let's choose x = 0:
1/2(0) + y = 5
y = 5
So one solution is (0, 5).

Now let's choose x = 2:
1/2(2) + y = 5
1 + y = 5
y = 4
Another solution is (2, 4).

We can continue this process and find more solutions:
x = 4, y = 3
x = 6, y = 2
x = 8, y = 1
x = 10, y = 0
x = -2, y = 6
x = -4, y = 7
x = -6, y = 8
x = -8, y = 9
x = -10, y = 10
These are ten solutions to the equation.

If we plot these points on a coordinate plane, we will see that they all lie on a straight line.

Therefore the shape of the graph is a line. The equation represents a linear relationship between x and y.

Answer: confidence interval = b. ( 116.42 to 123.58)

Step-by-step explanation:

Given;

Number of samples n = 30

Standard deviation r = 10

Mean x = 120

Confidence interval of 95%

Z' = t(0.025) = 1.96

Confidence interval = x +/- Z'(r/√n)

= 120 +/- 1.96(10/√30)

= 120 +/- 3.58

= ( 116.42, 123.58)

Answer:

32 mi

Step-by-step explanation:

Solve using proportions.

[tex]\frac{\frac{1}{2}in}{8 mi} =\frac{2in}{y}[/tex]

Find the scale factor (how to get from left to right)

To get from left numerator to right numerator, multiply by 4.

(1/2) X 4 = 2

The scale factor is 4.

Multiply the left denominator by the scale factor to get "y".

8 mi X 4 = 32 mi

Therefore 2 inches represent 32 miles.

Answer:

b. 85

Step-by-step explanation:

Average grade (μ) = 75

Standard deviation (σ) = 8

Assuming a normal distribution, the z-score corresponding to the upper 10% of the test grades is z = 1.28.

The minimum grade 'X' within the top 10% is given by:

[tex]z=\frac{X-\mu}{\sigma}\\1.28=\frac{X-75}{8}\\X=85.24[/tex]

Rounding to the nearest percent, the dividing point between an A and a B grade is 85.

Final answer:

The dividing point between an A and B grade for the upper 10% of the test grades is calculated using the normal distribution properties. A Z-score of 1.28 corresponds to the 90th percentile, resulting in a score of 85.24, which is rounded to 85 according to standard rounding rules.

Explanation:

To determine the dividing point between an A and B grade for the upper 10% of the test grades in History 101, we need to refer to the properties of the normal distribution. The mean score is given as 75 with a standard deviation of 8. We are looking for the score that corresponds to the 90th percentile since the instructor wants to award an A to the upper 10%. Z-scores allow us to translate percentile ranks into scores on a given normal distribution.

Using a standard normal distribution table, we find that a Z-score of approximately 1.28 corresponds to the 90th percentile. To find the actual score, we use the formula Score = Mean + (Z-score imes Standard Deviation). Plugging in the values:

Mean = 75

Standard Deviation = 8

Z-score for the 90th percentile = 1.28

Score = 75 + (1.28 imes 8) = 75 + 10.24 = 85.24

When applying the standard rules for rounding, we round 85.24 to the nearest whole number, which is 85. Therefore, the dividing point between an A and a B grade is an 85. Students need to score at this point or higher to be in the top 10% and receive an A grade.

Answer:

[tex]\oint_cF.dr=0\\[/tex]

Step-by-step explanation:

Given that a circle C of radius 7

[tex]x^{2} +y^{2} =49---(1)[/tex]

To find:

[tex]\oint_{C}F.dr[/tex]

As NO function is given so we suppose it to be:

[tex]F=<x,y>[/tex]

Parametric equations:

[tex]x=rcos\theta=7cos\theta\\y=rsin\theta=7sin\theta[/tex]

Each point on circle can be then found as

[tex]r(\theta)=<7cos\theta,7sin\theta>---(2)[/tex]

From (2) dr can be found as:

[tex]dr=<-7sin(\theta),7cos(\theta)>d\theta---(3)[/tex]

From (2) and (3)

[tex]\oint_cF.dr=\int_{0}^{2\pi}{<7cos\theta,7sin\theta><-7sin(\theta),7cos(\theta)>}\,d\theta\\\\\oint_cF.dr=\int_{0}^{2\pi}{<(-7cos\theta)(7sin\theta),(7sin(\theta))(7cos(\theta))>}\,d\theta\\\\\\\oint_cF.dr=\int_{0}^{2\pi}{-49cos\theta sin\theta+49sin(\theta)cos(\theta)}\,d\theta\\\\\oint_cF.dr=0\\[/tex]

When the sample size is small (< 30) and the population standard deviation is unknown , then we use t-test.

The confidence interval for population mean will be :

[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]   (1)

, where [tex]\overline{x}[/tex] = sample mean

t* = Critical value (based on degree of freedom and significance level).

s= sample standard deviation

n= sample size.

As per given we have

n= 9

Degree of freedom = n-1 = 8

[tex]\overline{x}=2.4[/tex]

s= 0.75

Significance level =[tex]\alpha=1-0.80=0.20[/tex]

Using students' t distribution table ,

Critical value : [tex]t^*=t_{\alpha/2,df}=t_{0.10,8}=1.3304[/tex]

We assume that the population is approximately normal.

Then, a 80% confidence interval for the mean waste recycled per person per day for the population of Florida will be :

[tex]2.4\pm (1.3304)\dfrac{0.75}{\sqrt{8}}[/tex]   (Substitute the values in (1))

[tex]2.4\pm (1.3304)\dfrac{0.75}{2.82842712475}[/tex]

[tex]2.4\pm (1.3304)(0.265165042945)[/tex]

[tex]2.4\pm 0.352775573134\approx2.4\pm0.353=(2.4-0.353,\ 2.4+0.353)=(2.047,\ 2.753)[/tex]

Hence, the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. = (2.047, 2.753)

Answer:

(a,b) → (a,-b) : Reflection about x axis.

(a,b) → (a, b+5) : Translation of the point by 5 units up.

(a,b) → (b,-a) : Rotation by 90 degree clockwise.

Step-by-step explanation:

Given:

The transformation of points are given as:

(a,b) → (a,-b)

(a,b) → (a, b+5)

(a,b) → (b,-a)

Now, let us consider each transformation one by one.

(1) (a,b) → (a,-b)

Here, the order of the coordinates has not changed. But, the y coordinate of the point has changed. The y coordinate was 'b' and it has changed only its sign but not value. So, it is a transformation related to reflection.

In reflection, only the sign changes. Since, the 'y' coordinate sing is reversed, so, it is a reflection about x axis.

(2) (a,b) → (a, b+5)

Here, the 'y' coordinate of the point has changed. The change is from 'b' to 'b+5'. So, 5 is added to the y coordinate. As per transformation rules, if a positive number 'C' is added to the y coordinate, then the point shifts vertically up by 'C' units. Hence, there is a translation of 5 units up here.

(3) (a,b) → (b,-a)

Here, the 'x' and 'y' coordinates interchange their values and also the new y coordinate has its sign reversed. This happens in rotation.

We know that, (x, y) → (y, –x) is true when there is rotation by 90 degree clockwise.

So, the point (a,b) → (b,-a) is rotated by 90 degree clockwise.

Answer:

Null hypothesis: The favorite sport of 53% of adults in the country is professional football ( with revenue of about $13 billion per year)

Alternate hypothesis: The adults in the country whose favorite sport is not professional football (with revenue of about $13 billion per year) is less than 53%

Step-by-step explanation:

The given claim is that the favorite sport of 53% of adults in the country is professional football (with revenue of about $13 per year) which is the null hypothesis. This means that the favorite sport of the remaining 47%(less than 53%) of adults in the country is not professional football (with revenue of about $13 billion per year) which is the alternate hypothesis

Answer:

Step-by-step explanation:

Linear relations are those which are represented by straight line on a graph i.e. increase or decrease a variable directly effect the other in a linear way

for example

y=2x+3

increase in value of x increase the value of y linearly.

Non-linear relationship are those in which change in one entity does not correspond to change in other quantity.

Real life situation for linear relation is when we apply force on a block it accelerates and it goes on increasing as force is increasing.

For Non-linear relation

increase in temperature vs time spend

Time spent is more to increase the slight amount of temperature.            

Linear and non-linear systems exist in everyday scenarios. A linear example is cost increasing by a fixed amount per item purchased, while a non-linear example might be a room's temperature: heating up quickly initially, then more slowly as it approaches the desired warmth.

In everyday life, both linear and non-linear systems abound. A straightforward example of a linear system is calculating the total cost of goods: if each item costs the same, your total cost represents a linear relationship with the number of items - the cost increases by a set amount per item.

On the other hand, a non-linear system might relate to how the temperature of a heated room changes over time: it may start increasing rapidly, then slowly once the room approaches the desired temperature. This represents a non-linear relationship as the rate of temperature increase isn't constant.

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The subject of the question is Mathematics, relevant to High School students. It involves simplifying and combining radicals, eliminating terms to simplify algebraic expressions, and understanding transcendental numbers and their properties.

Add and Subtract Radicals

When working with radicals, it is essential to simplify each radical by removing perfect square roots in order to combine like radicals effectively. For instance, to add \\(\\sqrt{18} + \\sqrt{8}\\), we must first simplify. \\(\\sqrt{18}\\) becomes \\(\\sqrt{9*2}\\) or \\(\\sqrt{9}\\cdot\\sqrt{2}\\), which simplifies to \\(\\sqrt{2}\\) times 3. Similarly, \\(\\sqrt{8}\\) can be rewritten as \\(\\sqrt{4*2}\\) or \\(\\sqrt{4}\\cdot\\sqrt{2}\\), which simplifies to \\(\\sqrt{2}\\) times 2. Now we have like radicals and can combine them: 3\\(\\sqrt{2}\\) + 2\\(\\sqrt{2}\\) equals 5\\(\\sqrt{2}\\).

Simplify Algebra and Reasonableness

To simplify the algebra, we identify and eliminate terms where possible, checking for reasonableness of the answer at all times. Verify if the operations you have performed are correct and the solution looks reasonable given the original equation or expression.

Transcendental Numbers and Functions

Transcendental numbers, like \\(\\sqrt{2}\\) or \\(\\sqrt{5}\\), are those that are not the root of any non-zero polynomial equation with rational coefficients. In algebra, these play a vital role in understanding real numbers and theorems related to them.

Answer: Option 'a' is correct.

Step-by-step explanation:

Since we have given that

P(A) = 0.8

P(B|A) = 0.4

Since A and B are independent events.

Since P(B|A) is given, we will use the formula for "conditional probability":

So, [tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\0.4=\dfrac{P(A\cap B)}{0.8}\\\\0.4\times 0.8=P(A\cap B)}\\\\0.32=P(A\cap B)[/tex]

Hence, it is true.

Therefore, Option 'a' is correct.

Answer:

The answer is c): They are less expensive and lower-risk.

Step-by-step explanation:

Line extensions are more common than new products because they do not require an overhaul of old or existing products or marketing strategies: old/existing products only have to be changed, and this requires lesser money, and is less risky. On the other hand, creating new products is more expensive and risky than line extensions.

The estimated cost to feed a family for a week with a household income of $12,000 would be $9,341.33.

To find out how much money would be needed to feed a family for a week whose household income is $12,000, we need to use the regression equation provided. The equation is Food/wk = 101.33 + 0.77HIncome. We substitute the value of HIncome with $12,000 and solve for Food/wk.

Food/wk = 101.33 + 0.77(12,000)

Food/wk = 101.33 + 9240

Food/wk = $9,341.33

Therefore, the estimated cost to feed a family for a week with a household income of $12,000 would be $9,341.33.

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Answer: 17-7i

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given that the Products produced by a machine has a 3% defective rate.

Each product is independent of the other with a constant prob of being defective as 0.03

X - no of defects is binomial with p =0.03

a)  the probability that the first defective occurs in the fifth item inspected

=Prob for first 4 non defective and 5th defective

=[tex](0.97)^4(0.03)^1\\=0.0266[/tex]

(b) the probability that the first defective occurs in the first five inspections

=P(X=1) in binomial with n=5

= 0.9915

c) the expected number of inspections before the first defective occurs

Expected defects in n trials = np

Expected number of inspection before the first defect = 1/p

= 33.33

=34

(c) What is the expected number of inspections before the first defective occurs?

Final answer:

The probability that the first defective product occurs on the fifth item is approximately 0.0895. The probability that a defective product occurs within the first five inspections is approximately 0.1426. The expected number of inspections before a defective product occurs is around 33 items.

Explanation:

To find the probability that the first defective occurs on the fifth item inspected (part a), we consider the first four items to be non-defective and the fifth one to be defective. The probability for one non-defective item is 97%, or 0.97. The probability we are looking for is thus the product of these probabilities:

P(non-defective, non-defective, non-defective, non-defective, defective) = (0.97)⁴* 0.03 ≈ 0.0895

For part b, we want to find the probability that a defective is found at any point in the first five inspections. This is the sum of the probabilities of finding the first defective on the first, second, third, fourth, or fifth inspection:

P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th) = 0.03 + (0.97 * 0.03) + (0.97² * 0.03) + (0.97³ * 0.03) + (0.97⁴ * 0.03) ≈ 0.1426

For part c, the expected number of inspections before finding the first defective item is given by the formula E(X) = 1/p, where p is the probability of finding a defective item. Substituting the given probability:

E(X) = 1/0.03 ≈ 33.33

Therefore, we expect to inspect approximately 33 items before finding the first defective.

Answer:

0.40

Step-by-step explanation:

Given that at the local racetrack, the favorite in a race has odds 3:2 of losing

Here instead of probability odds are given.

Odds of losing = 3/2

Hence Probability of losing = [tex]\frac{3}{3+2} \\=\frac{3}{5} \\=0.6[/tex]

Probability that the favourite wins the race will be the probability for the event which is complement of losing the game.

Hence

Probability that the favourite wins the race will be the probability

= 1- 0.6

=0.40

Probability that favorite wins the race is 0.6

Favorite in a race has odds = 3:2

Probability that favorite wins the race

Probability that favorite wins the race = 3 / [3 + 2]

Probability that favorite wins the race = 3 / 5

Probability that favorite wins the race = 0.6

Option "C" is the correct answer to the following question.

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Answer: 0.77

Step-by-step explanation:

Given :  Probability that a student chooses a turkey sandwich is

P(Turkey )= 0.10

Probability that a student chooses an egg salad sandwich is

P(egg salad)=0.67

Also, it is impossible for a student have both lunches.

∴ P(Turkey and egg salad) =0

Now , the probability that a student chooses a turkey or egg salad sandwich will be

P(Turkey or egg salad) =  P(Turkey )+ P(egg salad)- P(Turkey and egg salad)

=  0.10+ 0.67-0 = 0.77

Hence, the probability that a student chooses a turkey or egg salad sandwich= 0.77

Final answer:

The probability that a student picks either a turkey or an egg salad sandwich for their school field trip is 0.77 or 77%.

Explanation:

To calculate the probability that a student chooses either a turkey or egg salad sandwich for lunch, we use the formula for the probability of an 'or' event.

Since the options are mutually exclusive, meaning a student can only choose one type of sandwich, we simply add the individual probabilities together.

The probability of choosing a turkey sandwich is 0.10 and the probability of choosing an egg salad sandwich is 0.67.

Therefore, we can calculate it as follows:

P(turkey OR egg salad) = P(turkey) + P(egg salad)

P(turkey OR egg salad) = 0.10 + 0.67

P(turkey OR egg salad) = 0.77

So the probability that a student picks either a turkey or an egg salad sandwich is 0.77, or 77%.

Answer:

a) [tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]

b) [tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]

c) [tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]

Step-by-step explanation:

Let X the random variable that represent the number of airline crashes in a country. We know that [tex]X \sim Poisson(\lambda=2)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

And f(x)=0 for other case.

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

(a) What is the probability that there will be at least 3 accidents in the next month?

On this case we are interested on the probability of having at least three accidents in the next month, and using the complement rule we have this:

[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-2} 2^0}{0!}=0.1353[/tex]

[tex]P(X=1)=\frac{e^{-2} 2^1}{1!}=0.2707[/tex]

[tex]P(X=2)=\frac{e^{-2} 2^2}{2!}=0.2707[/tex]

And replacing we have this:

[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]

(b) What is the probability that there will be at least 6 accidents in the next two months?

For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x2= 4[/tex] accidents per 2 months.

[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-4} 4^0}{0!}=0.0183[/tex]

[tex]P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733[/tex]

[tex]P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465[/tex]

[tex]P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954[/tex]

[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]

[tex]P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563[/tex]

Replacing we got:

[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]

(c) What is the probability that there will be at most 6 accidents in the next three months?

For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x3= 6[/tex] accidents per 3 months.

[tex]P(X\leq 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)[/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-6} 6^0}{0!}=0.00248[/tex]

[tex]P(X=1)=\frac{e^{-6} 6^1}{1!}=0.0149[/tex]

[tex]P(X=2)=\frac{e^{-6} 6^2}{2!}=0.0446[/tex]

[tex]P(X=3)=\frac{e^{-6} 6^3}{3!}=0.0892[/tex]

[tex]P(X=4)=\frac{e^{-6} 6^4}{4!}=0.1339[/tex]

[tex]P(X=5)=\frac{e^{-6} 6^5}{5!}=0.1606[/tex]

[tex]P(X=6)=\frac{e^{-6} 6^6}{6!}=0.1606[/tex]

[tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]

Answer:

we can conclude two things that:

Step-by-step explanation:

Why either move creates a new equation with the same solutions as the original equation?

If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.

When we multiple the two sides of any given equation by the same number, it would keep the two sides of that particular equation equal. So, whatever the  the solution the first equation may get, will still work for the second equation.

Determining Lin's first move i.e. to multiply the first equation by 3.

Let us consider the equation

x/3 + 2y = 4      .....[1]

x + y = -3           .....[2]

Lin's first move is to multiply the first equation by 3.

3(x/3 + 2y) = 3(4 )

x + 6y = 12         .....[3]

Now subtract the Equation [2] from Equation [3]

x + 6y - x - y = 12 - (-3)

5y = 15

y = 3

Putting y = 3 in [2]

x + (3) = -3

x = -6

So, x = -6 and y = 3

Determining Priya's first move i.e. to multiply the Second equation by 2.

Let us consider the equation

x/3 + 2y = 4      .....[1]

x + y = -3           .....[2]

Priya's first move is to multiply the second equation by 2.

2(x + y) = 2(-3)          

2x + 2y = -6           .....[3]

Now subtract the Equation [2] from [1]

x/3 + 2y - 2x - 2y= 4 - (-6)

x/3 - 2x = 10

x - 6x = 30

x = -6

Putting x = -6 in Equation [2]

x + y = -3

-6 + y = -3

y = 3

So, x = -6 and y = 3

So, from the entire analysis, we can conclude two things that:

Keywords: system of equation, solution, equation

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Answer:

A. approximately a normal distribution.

Step-by-step explanation:

There may be a few differences, but the sampling distribution of the sample mean is still approximately normal.

So the correct answer is:

A. approximately a normal distribution.

Answer:

Correct answer is (A) {Normal distribution}

Step-by-step explanation:

sampling distribution of the sample mean is still approximately normal.

Answer:

Matching the vocabulary word with its corresponding example:

1. All children who Ski and Snowboard = population (a group of items, units or subjects which is under reference of study e.g inhabitants of a region, numbers of cars in a city e.t.c)

2. The 92 children who were asked when they took their first lesson = Sample (a part or fraction of a population selected on some basis)

3. The average age that all children take their first to lesson = Parameter (the number that summarizes some characteristics of a population

4. The average age that the 92 children took their first lesson = statistic (a sample characteristic corresponding to the population parameter used when a sample is use to make inference about a population)

5. The age that children take their first listen = variable (anything that has attribute, quality or quantity that varies or a characteristic/attribute that describes a place, person or thing)  

6. The list of the 92 ages that the children from the study took the first listen = data (facts that are collected together for analysis)

Step-by-step explanation:

1. All children who Ski and Snowboard = population (a group of items, units or subjects which is under reference of study , numbers of cars in a city e.t.c)

2. The 92 children who were asked when they took their first lesson = Sample (a part of a population selected on some basis)

3. The average age that all children take their first to lesson = Parameter (the number that summarizes some characteristics of a population

4. The average age that the 92 children took their first lesson = statistic (a sample characteristic corresponding to the population parameter used when a sample is use to make inference about a population)

5. The age that children take their first listen = variable (anything that has attribute, quality or quantity that varies or a characteristic/attribute that describes a place, person or thing)  

6. The list of the 92 ages = data (facts that are collected together for analysis)

Answer:

Step-by-step explanation:

a)  For all x∈ℤ, the number f(x)=x²+1 exists and is unique because f(x) is defined using the operations addition (+) and multiplication (·) on ℤ. Then f is a function. f is not one-to-one: consider -1,1∈ℤ. -1≠1 but f(-1)=f(1)=2- Because two different elements in the domain have the same image under f, f is not 1-1. f is not onto: x²≥0 for all x∈ℤ then f(x)=x²+1≥1>0 for all x∈ℤ. Then 0∈ℕ but for all x∈ℤ f(x)≠0, which means that one element of the codomain doen't have a preimage, so f is not onto.

b) 0∈ℕ, so 0 is an element of the domain of g, but g(0)=1/0 is undefined, therefore g is not a function.

c) Let (z,n)∈ℤ x ℕ. The number h(z,n)=z·1/(n+1) is unique and it's always defined because n+1>0, then h is a function. h is not 1-1: consider the ordered pairs (1,2), (2,5). They are different elements of the domain, but h(2,5)=2/6=1/3=h(1,2). h is onto: any rational number q∈ℚ can be written as q=a/b for some integer a and positive integer b. Then (a,b-1)∈ ℤ x ℕ and h(a,b-1)=a/b=q.

f) For all (x,y)∈ℝ², the pair h(x,y)=(y+1,x+1) is defined and is unique, because the definition of y+1 and x+1 uses the addition operation on ℝ. f is 1-1; suppose that (x,y),(u,v)∈ℝ² are elements of the domain such that h(x,y)=h(u,v). Then (y+1,x+1)=(v+1,u+1), so by equality of ordered pairs y+1=v+1 and x+1=u+1. Thus x=u and y=x, therefore (x,y)=(u,v). f is onto; let (a,b)∈ℝ² be an element of the codomain. Then (b-1,a-1)∈ℝ² is an element of the domain an h(b-1,a-1)=(a-1+1,b-1+1)=(a,b). Because h is 1-1 and onto, then h is a bijection so h has a inverse h^-1 such that for all (x,y)∈ℝ² h(h^-1(x,y))=(x,y) and h^-1(h^(x,y))=(x,y). The previous proof of the surjectivity of h (h onto) suggests that we define h^-1(x,y)=(y-1,x-1). This is the inverse, because h(h^-1(x,y))=h(y-1,x-1)=(x,y) and h^-1(h^(x,y))=h^-1(y+1,x+1)=(x,y).

The function f: ℤ → ℕ is neither one-to-one nor onto, g: ℕ → ℚ is one-to-one but not onto, h: ℤ x ℕ → ℚ is neither one-to-one nor onto, and h: ℝ² → ℝ² is a bijective function with an inverse h⁻¹(u,v) = (v-1, u-1).

Explanation:

Let's examine each function individually to determine if they define a function from the domain to the codomain, and if so, whether they are one-to-one or onto, and describe the inverse function for any bijective function.

f: ℤ → ℕ where f is defined by f(x) = x2 + 1. This is indeed a function since each element in the domain ℤ has a unique image in the codomain ℕ. It is not one-to-one because both positive and negative integers will produce the same result when squared. However, it is not onto since no element in ℕ will map to 0, which is not attained by x2 + 1 for any integer x.

g: ℕ → ℚ where g is defined by g(x) = 1/x. This represents a function since each positive integer x will have a unique reciprocal in ℚ. This function is one-to-one, as no two different positive integers have the same reciprocal, but it is not onto because certain rational numbers, like 2/3, cannot be expressed as the reciprocal of a natural number.

h: ℤ x ℕ → ℚ where h is defined by h(z,n) = z/(n+1). This defines a function where each ordered pair of integers and natural numbers corresponds to a unique rational number. However, this function is neither one-to-one nor onto. It is not one-to-one because different integer pairs could result in the same rational number (e.g., h(2,1) = h(-2,-3) = 2/2), and it's not onto as some rational numbers cannot be obtained using this formula, such as 2/3.

h: ℝ2 → ℝ2 where h is defined by h(x,y) = (y+1, x+1). This is a function that maps pairs of real numbers to pairs of real numbers, it is both one-to-one and onto (therefore, bijective), as every pair (x,y) has a unique image and every possible pair (u,v) in ℝ2 is hit. The inverse function is given by h-1(u,v) = (v-1, u-1).

Answer: B. (0.44,0.94)

Step-by-step explanation:

Given : Number of observations : n = 9

Number of successes  : x = 7

Let p be the population proportion of times that the bats would follow the point.

Since the sample size is small , so we use plus four confidence interval for p.

Plus four estimate of p=[tex]\hat{p}=\dfrac{\text{No. of successes}+2}{\text{No. of observations}+4}[/tex]

[tex]=\dfrac{7+2}{9+4}\approx0.69[/tex]

By z-table , the critical value for 95% confidence level : z* = 1.96

Then, the 95% confidence interval for the population proportion of times that the bats would follow the point. will be :

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}[/tex] , where N= 13

[tex]0.69\pm (1.96)\sqrt{\dfrac{0.69(1-0.69)}{13}}[/tex]

[tex]0.69\pm (1.96)\sqrt{0.0163862084615}[/tex]

[tex]0.69\pm (1.96)(0.128008626512)[/tex]

[tex]\approx0.69\pm 0.25=(0.69-0.25,\ 0.69+0.25)[/tex]

[tex](0.44,\ 0.94)[/tex]

Hence, the  95% confidence interval for the population proportion of times that the bats would follow the point = [tex](0.44,\ 0.94)[/tex]

Thus the correct answer is B. (0.44,0.94)

To find the 95% confidence interval for the population proportion, use the formula CI = p ± z * √((p(1-p))/n), where p is the sample proportion, z is the z-score, and n is the sample size. Substituting values, the 95% confidence interval is approximately (0.685, 0.869).

To find the 95% confidence interval for the population proportion, we can use the formula:


CI = p ± z * √((p(1-p))/n)


where p is the sample proportion, z is the z-score for the desired confidence level, and n is the sample size.


In this case, the sample proportion is 7/9 and n is 9. Since we want a 95% confidence interval, the z-score is approximately 1.96.

Substituting these values into the formula:


CI = (7/9) ± 1.96 * √(((7/9)(2/9))/9)


CI = 0.777 ± 1.96 * √(0.123/9)


CI ≈ 0.777 ± 1.96 * 0.047


CI ≈ (0.777 - 0.092, 0.777 + 0.092)


CI ≈ (0.685, 0.869)

https://brainly.com/question/34700241

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The probability of a motorbike rider in this city experiencing carbon monoxide exposure exceeding 20 ppm is 0.3932, while the probability of exceeding 25 ppm is 0.1316, based on the normal distribution with a mean of 18.6 ppm and a standard deviation of 5.9 ppm.

Define the random variable X as the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in this city.

Since the distribution of X is normally distributed with a mean of 18.6 ppm and a standard deviation of 5.9 ppm, we can use the standard normal distribution to calculate probabilities.

To find the probability of X being greater than 20 ppm, we need to find the area to the right of 20 ppm under the standard normal curve.

We can calculate this area using a z-score, which is defined as the number of standard deviations a specific point is away from the mean. In this case, the z-score for 20 ppm is (20 ppm - 18.6 ppm) / 5.9 ppm = 0.271.

Using a standard normal table or calculator, we can find that the area to the right of 0.271 is 0.3932.

Therefore, the probability of someone riding a motorbike for 5 km on a highway in this city experiencing a carbon monoxide exposure of more than 20 ppm is 0.3932.

Follow the same steps as in part a, but use a z-score of (25 ppm - 18.6 ppm) / 5.9 ppm = 1.119.

The area to the right of 1.119 under the standard normal curve is 0.1316.

Therefore, the probability of someone riding a motorbike for 5 km on a highway in this city experiencing a carbon monoxide exposure of more than 25 ppm is 0.1316.

Complete question:

According to a research paper, the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular city is approximately normally distributed with a mean of 18.6 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.9 ppm (a) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monexide exposure of more than 20 ppm? (Round your answer to fou decimal places.) (b) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monoxide exposure of more than 25 ppm?

Answer:

At 0.05 significance level, the p-value is 0.014

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 3 minutes

Sample mean, [tex]\bar{x}[/tex] = 3.11 minutes

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, σ = 0.5 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 3\text{ minutes}\\H_A: \mu > 3\text{ minutes}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

[tex]z_{stat} = 2.2[/tex]

Now, we calculate the p-value from the normal standard z-table.

P-value = 0.014

At 0.05 significance level, the p-value is 0.014

Answer:

[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]

Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140

Step-by-step explanation:

1) 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".  

2)  Solution to the problem

Let X the random variable that represent the IQ scores of the population of interest, and for this case we know the distribution for X is given by:

[tex]X \sim N(100,19)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=19[/tex]

We are interested on this probability

[tex]P(X>140)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula and the complement rule to our probability we got this:

[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]

And we can find this probability with the following excel code:

"=1-NORM.DIST(2.105,0,1,TRUE)"

This number 0.0176 represent the proportion of Americans that present a score higher than 140.

And now since we ar einterested on the approximate number of people in the United States (assuming a total population of 280,000,000) with an IQ higher than 140, we just need to do this:

Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140

Answer:

(a) 0.12

(b) 0.42

(c) 0.46

Step-by-step explanation:

Probability of student A not showing up = 0.3

Probability of student B not showing up = 0.4

(a) Neither will show up for class

[tex]P(A\ and\ B) = 0.3*0.4 = 0.12[/tex]

(b) Both will show up for class

[tex]P (A\ nor\ B) = (1-0.3)*(1-0.4) = 0.42[/tex]

(c) Exactly one will show up for class

[tex]P(A\ or\ B) = 1 -P(A\ and\ B) - P(A\ nor\ B)\\P(A\ or\ B) = 1 -0.12-0.42 = 0.46[/tex]

Answer:

Option A) About ±2.98 minutes

Step-by-step explanation:

We are given the following information in the question:

Sample mean, [tex]\bar{x}[/tex] = 191.3 minutes

Sample size, n = 200

Population standard deviation, σ = 21.5 minutes

Alpha, α = 0.05

The leaders of the organization wish to develop an interval estimate with 95 percent confidence.

[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]

Margin of error =

[tex]z_{\text{critical}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get:

[tex]\pm 1.96\times \displaystyle\frac{21.5}{\sqrt{200}} = \pm 2.9797 \approx \pm 2.98[/tex]

Option A) About ±2.98 minutes