According to one pollster, 43 % of people are afraid of flying. Suppose that a sample of size 26 is drawn. Find the value of standard error , the standard deviation of the distribution of sample proportions.

Answer with Step-by-step explanation:

1.Let p={Aaron , bob ,phill,john,chad}

Number of elements in set p=5

Formula : Number of subset of the set which contain n elements

[tex]2^n[/tex]

Total number of subset =[tex]2^5=32[/tex]

The subsets which  contain neither  Aaron nor  bob are

{phil},{john},{chad}[tex],\phi[/tex],{phil,john},{phil,chad},{john,chad},{phil,john,chad}

There are eight subsets which do not contain neither Aaron nor bob .

Answer given by the method [tex]2^{5-2}=8[/tex]

Where 5= Total number of elements

2= Number of elements in subsets which do not contain

Therefore,answer is correct.

2.A={1,2,3,4,5,6,7}

There are total seven elements

Therefore, total number of subsets =[tex]2^7[/tex]

We have to find the number of subsets of A which include 2,4 and 6.

The number of subsets which contain {2,4, 6}

{2,4,6},{2,4,6,1},{2,4,6,3},{2,4,6,5},{2,4,6,7},{1,2,3,4,6},{1,2,4,5,6},{1,2,4,6,7},{2,4,5,6,7},{2,3,4,5,6},{2,3,4,6,7},{1,2,3,4,5,6},{2,3,4,5,6,7},{1,3,4,5,6,7},{1,2,4,5,6,7},{1,2,3,5,6,7},{1,2,3,4,5,7},{1,2,3,4,5,6,7}

Total number of subsets  which contain {2,4,6} are eighteen.

Answer  given by [tex]2^{7-3}=2^4=16[/tex]

The answer is false.

We can not use these method for calculating number of subsets which contain{2,4,6}.

Answer:

The probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.

Step-by-step explanation:

Let X be the number of defective bulbs.

[tex]X\sim B(n,p)[/tex]

Where n is sample size and p is probability of success.

According to the given information,

[tex]n=100[/tex]

[tex]p=0.02[/tex]

[tex]q=1-p=1-0.02=0.98[/tex]

According to binomial distribution the probability of exactly r success from n is

[tex]P(X=r)=^nC_rp^rq^{n-r}[/tex]

The probability that exactly two are defective is

[tex]P(X=2)=^{100}C_2(0.02)^2(0.98)^{98}\approx 0.2724[/tex]

The probability that the number of defective bulbs is two or fewer is

[tex]P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)[/tex]

[tex]P(X\leq 2)=^{100}C_0(0.02)^0(0.98)^{100}+^{100}C_1(0.02)^1(0.98)^{99}+^{100}C_2(0.02)^2(0.98)^{98}[/tex]

[tex]P(X\leq 2)=0.1326+0.2707+0.2734[/tex]

[tex]P(X\leq 2)=0.6767[/tex]

We have to find the probability that exactly two are defective given that the number of defective bulbs is two or fewer.

[tex]P(x=2|x\leq 2)[/tex]

According to the conditional probability

[tex]P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}[/tex]

[tex]P(x=2|x\leq 2)=\frac{P(x=2\cap x\leq 2)}{P(x\leq 2)}[/tex]

[tex]P(x=2|x\leq 2)=\frac{P(x=2)}{P(x\leq 2)}[/tex]

[tex]P(x=2|x\leq 2)=\frac{0.2734}{0.6767}=0.4040[/tex]

Therefore the probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.

To solve this problem, we need to use the concept of conditional probability and the binomial distribution. The binomial distribution is a discrete probability distribution that models the number of successes in a sequence of independent experiments.
Let \( X \) be the random variable representing the number of defective light bulbs. Since each light bulb is either defective or not with a certain probability, and the events are independent, \( X \) follows a binomial distribution with parameters \( n = 100 \) and \( p = 0.02 \), where \( n \) is the number of trials and \( p \) is the probability of success on an individual trial.
1. Calculate \( P(X = 2) \), the probability of exactly two defective bulbs.
The probability mass function (PMF) for a binomial distribution is given by
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
For \( k = 2 \), \( n = 100 \), \( p = 0.02 \), the PMF becomes
\[ P(X = 2) = \binom{100}{2} (0.02)^2 (1-0.02)^{98} \]
2. Calculate \( P(X \leq 2) \), the probability of two or fewer defective bulbs.
This is the cumulative distribution function (CDF) for \( k = 0, 1, 2 \). We sum the probabilities for 0, 1, and 2 defective bulbs:
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) = \binom{100}{0} (0.02)^0 (1-0.02)^{100} + \binom{100}{1} (0.02)^1 (1-0.02)^{99} + \binom{100}{2} (0.02)^2 (1-0.02)^{98} \]
3. Compute the conditional probability \( P(X = 2 | X \leq 2) \), which is the probability of exactly two defective bulbs given that there are two or fewer defective bulbs.
By definition, the conditional probability is
\[ P(X = 2 | X \leq 2) = \frac{P(X = 2 \cap X \leq 2)}{P(X \leq 2)} \]
Since \( P(X = 2 \cap X \leq 2) = P(X = 2) \), we can simplify this to
\[ P(X = 2 | X \leq 2) = \frac{P(X = 2)}{P(X \leq 2)} \]
We have already calculated both \( P(X = 2) \) and \( P(X \leq 2) \), so we just need to substitute the values:
\[ P(X = 2 | X \leq 2) = \frac{\binom{100}{2} (0.02)^2 (1-0.02)^{98}}{\binom{100}{0} (0.02)^0 (1-0.02)^{100} + \binom{100}{1} (0.02)^1 (1-0.02)^{99} + \binom{100}{2} (0.02)^2 (1-0.02)^{98}} \]
To finalize, you would calculate the above expression using a calculator or software capable of performing binomial calculations to get the numerical probability. The solution will be the conditional probability that exactly two light bulbs are defective given that two or fewer are defective.

Both the teams are held at their place by the friction between the ground and their feet. If the pulling force as mentioned above exceeds this friction, the losing team experiences a net force and accelerates.

If object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on the object.

When a team pulls the rope, the rope pulls the team.

This was Newton's third law.

When both teams pull in opposite directions, they often don't exert equal forces; the stronger team exerts a larger force.

So the two forces don't balance each other and the rope accelerates towards the stronger team.

The weaker team experiences a net force pulling them.

This force is the sum of (i)The reaction by the rope as mentioned above and (ii) The friction between the rope and the team members' hands.

Both the teams are held at their place by the friction between the ground and their feet. If the pulling force as mentioned above exceeds this friction, the losing team experiences a net force and accelerates.

To know more about Newton's third law click the link given below.

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Newton's Third Law explains the forces exerted in a tug-of-war, but the winning team is often determined by factors like friction and the angle of pulling. Greater friction helps a team resist the opposing force, while a lower angle of pulling may provide an advantage in distributing the tug's force.

Even though Newton's Third Law of Motion states that for every action there is an equal and opposite reaction, this does not mean that all forces will cancel each other out. In the context of a tug-of-war, friction and the angle in which the tug is exerted can influence outcomes.

The tug-of-war is not only a battle of forces but also a battle against friction. Each team is pulling as hard as they can in opposite directions, establishing the action-reaction pair. However, their ability to maintain their ground or pull the other team relies significantly on the friction between their feet and the ground. If either team can increase this friction (e.g., by wearing shoes with better grip, or planting their feet more effectively), they can resist the pulling force of the other team, leading to their victory.

Moreover, the angle at which a team pulls the rope can also determine the advantage in a tug of war. Teams that manage to keep the rope lower to the ground distribute the force more efficiently and therefore have a greater chance of winning.

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Answer:

(-∞, -1 )∪(3, ∞)

Step-by-step explanation:

Given function,

[tex]f(x) = x^3 - 3x^2 - 9x + 4[/tex]

Differentiating with respect to x,

[tex]f'(x)=3x^2-6x-9[/tex]

For increasing or decreasing,

f'(x) = 0

[tex]\implies 3x^2-6x-9=0[/tex]

By quadratic formula,

[tex]x=\frac{-(-6)\pm \sqrt{(-6)^2-4\times 3\times -9}}{6}[/tex]

[tex]=\frac{ 6\pm \sqrt{36+108}}{6}[/tex]

[tex]=\frac{6\pm \sqrt{144}}{6}[/tex]

[tex]\frac{6\pm 12}{6}[/tex]

[tex]\implies x = 3\text{ or } x = -1[/tex]

In interval (-∞, -1 ), f'(x) = positive,

⇒ f(x) is increasing on (-∞, -1 ),

In interval (-1, 3), f'(x) = negative,

⇒ f(x) is decreasing on (-1, 3),

In interval (3, ∞), f'(x) = positive,

⇒ f(x) is increasing on (3, ∞),

To find the interval where a function is increasing, find the derivative, then set it greater than 0 and solve for x to get the critical points. These points split the domain into intervals which are classified as increasing or decreasing by plugging into the derivative. Increasing intervals are then written in interval notation.

The subject of the question is about finding the interval on which the function f(x) = x^3 − 3x^2 − 9x + 4 is increasing. This involves the concept of determining increasing and decreasing intervals in calculus, which analyzes the slope of the function's graph.

To find the intervals, we first find the derivative of the function, f'(x) which gives us the slopes of the tangent lines at any point x. The function f(x) = x^3 - 3x^2 - 9x + 4 becomes f'(x) = 3x^2 - 6x - 9. Set f'(x) > 0 to figure out when the function is increasing. Solve for x gives the critical points, which splits the domain of f into intervals. Testing these intervals in the derivative equation helps us classify them as increasing or decreasing.

The final step is writing the increasing intervals in the proper notation, usually in the form of (a, b) where 'a' and 'b' are the endpoints of the interval where the function is increasing. If an interval goes to infinity, you would use a notation like (a, ∞).

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Answer:

Explanation:

The little red square indicates that the angle C measures 90°, which means that ABC is a right triangle.

The opposite side to angle C is the hypotenuse of the triangle. Then, the hypotenuse measures 9.8 in.

The requested measure, m ∠ ABC, is the measure of the angle B.

With respect to angle B you have:

Thus, the trigonometric ratios for angle B are:

Finally, to find the measures of angle B, you just need to find the inverse functions, which are:

Answer:A&E

Step-by-step explanation:

JUST TOOK TEST

Answer: 0.05

Step-by-step explanation:

Given : Interval for uniform distribution : [46.0 minutes, 56.0 minutes]

The probability density function will be :-

[tex]f(x)=\dfrac{1}{56-46}=\dfrac{1}{10}=0.1\ \ , 46<x<56[/tex]

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-

[tex]P(50.75<x<51.25)=\int^{51.25}_{50.75}f(x)\ dx\\\\=(0.1)[x]^{51.25}_{50.75}\\\\=(0.1)(51.25-50.75)=0.05[/tex]

Hence, the probability that a given class period runs between 50.75 and 51.25 minutes = 0.05

Answer:

The correct answer here is option c) 0.05 mg(milligram)

Step-by-step explanation:

The first step here to calculate the milligram of active ingredient in a 5% solution per milliliter is to assume that Z milligram of active ingredient is present in the 5% solution per milliliter.

   NOTE* -    milligram ( mg) and milliliter (ml)

so therefore, the equation would become,

5\% = \frac{Z mg}{1 ml}

.05 = \frac{Z mg}{1 ml}

.05 x 1 = Z mg

therefore there is .05 mg of active ingredient in the 5% solution per ml.

The correct option is A. 50 mg.  This is calculated by knowing that 5% means 5 grams or 5000 milligrams in 100 millilitres, which divides down to 50 milligrams per millilitre.

To determine how many milligrams of active ingredients are in a 5% solution per millilitres, we need to understand what a 5% solution means. A 5% solution means that there are 5 grams of the active ingredient in every 100 millilitres of the solution.

Let's break it down step-by-step:

→ 5% solution means 5 grams of active ingredient per 100 millilitres (mL) of solution.

→ Convert grams to milligrams: 5 grams = 5000 milligrams.

→ Therefore, 100 mL of solution contains 5000 milligrams of active ingredient.

To find out how many milligrams are in 1 mL, we divide the total milligrams by the total millilitres:

→ 5000 milligrams / 100 mL = 50 milligrams per mL

Thus, the correct answer is A. 50 mg.

Answer:

16 in.

Step-by-step explanation:

We have the ratio

[tex]\frac{in.}{ft}: \frac{\frac{2}{3} }{1}[/tex]

How about let's make this easier.  Easier is better, right?  Let's get rid of the fraction 2/3.  We will do that by multiplying 2/3 by 3 and 1 by 3 to get the equivalent ratio of

[tex]\frac{in.}{ft}:\frac{2}{3}[/tex]

Now we need to know how many inches there would be if the number of feet is 24:

[tex]\frac{in.}{ft}:\frac{2}{3}  =\frac{x}{24}[/tex]

Cross multiply to get

3x = 48 so

x = 16 in.

 Length of the wall on blueprint will be 16 inches.

  Scale used by an architect on a blueprint,

                 [tex]\frac{2}{3}\text{ inch}= 1\text{ foot}[/tex]

Scale represents the ratio of the length of the wall on blueprint and actual length,

[tex]\frac{\text{Length on the blueprint}}{\text{Actual length}} =\frac{\frac{2}{3}\text{ inches}}{1\text{ feet}}[/tex]

[tex]\frac{\text{Length on the blueprint}}{\text{Actual length}} =\frac{2}{3}[/tex]

If actual length of the east wall of the building = 24 feet

Substitute the value in the expression representing the ratio,

[tex]\frac{\text{Length on the blueprint}}{24} =\frac{2}{3}[/tex]

Length of the blueprint = [tex]\frac{2}{3}\times 24[/tex]

                                       = [tex]16[/tex] inches

    Therefore, length of the wall on blueprint will be 16 inches.

Learn more about the use of scale to calculate the distances on map.

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[tex]\bf \textit{Pythagorean Identities} \\\\ sin^2(\theta)+cos^2(\theta)=1 \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]\bf tan(\theta )+cot(\theta )=sec(\theta )csc(\theta ) \\\\[-0.35em] ~\dotfill\\\\ tan(\theta )+cot(\theta )\implies \cfrac{sin(\theta )}{cos(\theta )}+\cfrac{cos(\theta )}{sin(\theta )}\implies \cfrac{sin(\theta ) sin(\theta )+cos(\theta )cos(\theta )}{cos(\theta )sin(\theta )} \\\\\\ \cfrac{sin^2(\theta )+cos^2(\theta )}{cos(\theta )sin(\theta )}\implies \cfrac{1}{cos(\theta )sin(\theta )}\implies \cfrac{1}{cos(\theta )}\cdot \cfrac{1}{sin(\theta )}\implies sec(\theta )csc(\theta )[/tex]

Answer:

P=0.05 or 5%

Step-by-step explanation:

The formula to calculate the probability in a continuous uniform distribution is:

[tex]P=\frac{d-c}{b-a}[/tex]

Where a is the lower value of the limits of the distribution (a=47.0), b is the upper value of the limit of the distribution (b=57.0), c It is the lowest value in the range from which probability is calculated (c=50.75) and d It is the highest value in the range from which probability is calculated (d=51.25).

[tex]P=\frac{51.25-50.75}{57.0-47.0} =0.05[/tex]

Final answer:

The probability that a class period runs between 50.75 and 51.25 minutes, based on uniform distribution, is 5%.

Explanation:

The probability that a given class period runs between 50.75 and 51.25 minutes can be found using the principles of uniform distribution. For a uniform distribution, the probability of an event occurring between two points (a and b) is given by the formula \((b - a) / (maximum - minimum)\). Here, the class duration minimum is 47.0 minutes and the maximum is 57.0 minutes. So the probability is \((51.25 - 50.75) / (57.0 - 47.0)\), which equals to 0.5 / 10 or 0.05. Thus, the probability is 5%.

Answer:55.227

Step-by-step explanation:

Given data

mass [tex]\left ( m\right )[/tex]=13lb

spring stretches by [tex]\left ( x\right )[/tex]=4.5in=0.375 ft

g=[tex]32ft/s^2[/tex]

Now Spring constant K

mg=kx

k=[tex]\frac{13\times 32}{y}=58.667lb/ft[/tex]

For critical damping [tex]\zeta =1[/tex]

[tex]2\times \zeta \times \omega_n =\frac{c}{m}[/tex]

and [tex]\omega_n=\sqrt {\frac{k}{m}}[/tex]

[tex]\omega _n=2.1241rad/s[/tex]

substituting values

[tex]2\times 1 \times 2.124 =\frac{c}{m}[/tex]

c=55.227 lb.sec/ft

The directional derivative of [tex]f[/tex] at the point [tex]\vec p[/tex] in the direction of [tex]\vec v[/tex] is

[tex]D_{\vec v}f(\vec p)=\nabla f(\vec p)\cdot\dfrac{\vec v}{\|\vec v\|}[/tex]

We have

[tex]f(x,y,z)=xe^y+ye^z+ze^x\implies\nabla f(x,y,z)=\langle e^y+ze^x,xe^y+e^z,ye^z+e^x\rangle[/tex]

With [tex]\vec p=\langle0,0,0\rangle[/tex],

[tex]\nabla f(0,0,0)=\langle1,1,1\rangle[/tex]

[tex]\vec v[/tex] has magnitude

[tex]\|\vec v\|=\sqrt{5^2+2^2+(-2)^2}=\sqrt{33}[/tex]

and so the directional derivative is

[tex]\langle1,1,1\rangle\cdot\dfrac{\langle5,2,-2\rangle}{\sqrt{33}}=\boxed{\dfrac5{\sqrt{33}}}[/tex]

The directional derivative of the function [tex]\( f(x, y, z) = x e^y + y e^z + z e^x \)[/tex] at the point [tex]\( (0, 0, 0) \)[/tex] in the direction of the vector [tex]\( \mathbf{v} = (5, 2, -2) \) is \( \frac{3}{\sqrt{33}} \).[/tex]

1. Find the gradient of the function:

The gradient of a function  is given by the vector:

[tex]\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \][/tex]

For our function [tex]\( f(x, y, z) = x e^y + y e^z + z e^x \)[/tex], the partial derivatives are:

[tex]\[ \frac{\partial f}{\partial x} = e^y + z e^x \][/tex]

[tex]\[ \frac{\partial f}{\partial y} = x e^y + e^z \][/tex]

[tex]\[ \frac{\partial f}{\partial z} = y e^z + e^x \][/tex]

2. Evaluate the gradient at the given point:

Plug in [tex]\( (x, y, z) = (0, 0, 0) \)[/tex] into the partial derivatives:

[tex]\[ \nabla f(0, 0, 0) = (1, 1, 1) \][/tex]

3. Normalize the vector [tex]\( \mathbf{v} \)[/tex]:

To normalize a vector, divide each component by the magnitude of the vector:

[tex]\[ \| \mathbf{v} \| = \sqrt{5^2 + 2^2 + (-2)^2} = \sqrt{33} \][/tex]

[tex]\[ \mathbf{v}_{\text{normalized}} = \left( \frac{5}{\sqrt{33}}, \frac{2}{\sqrt{33}}, -\frac{2}{\sqrt{33}} \right) \][/tex]

4. Find the directional derivative:

The directional derivative of [tex]\( f \)[/tex] in the direction of [tex]\( \mathbf{v} \)[/tex] at the point [tex]\( (0, 0, 0) \)[/tex] is given by the dot product of the gradient and the normalized vector [tex]\( \mathbf{v} \)[/tex]:

[tex]\[ D_{\mathbf{v}} f = \nabla f(0, 0, 0) \cdot \mathbf{v}_{\text{normalized}} = (1, 1, 1) \cdot \left( \frac{5}{\sqrt{33}}, \frac{2}{\sqrt{33}}, -\frac{2}{\sqrt{33}} \right) \][/tex]

 [tex]\[ = \frac{5}{\sqrt{33}} + \frac{2}{\sqrt{33}} - \frac{2}{\sqrt{33}} = \frac{5 + 2 - 2}{\sqrt{33}} = \frac{5}{\sqrt{33}} \][/tex]

So, the directional derivative of the function at the point [tex]\( (0, 0, 0) \)[/tex] in the direction of the vector [tex]\( (5, 2, -2) \)[/tex] is [tex]\( \frac{5}{\sqrt{33}} \).[/tex]

Answer:

She earns that day $103.5

Step-by-step explanation:

Lets explain how to solve the question

- She is paid $12.50 per hour between 8 am and 4 pm

- After 4 pm she is paid $14.25 per hour

- She works one particular day from 10 am to 6 pm

* Lets distribute her time of work into two part according to her

 payment above

# 1st part from 10 am to 4 pm

∵ She is paid $12.50 per hour between 8 am to 4 pm

∵ She works from 10 am to 4 pm

- Calculate how many hours between 10 am and 4 pm

∵ am and pm not like terms in time, so change the time to 24-hours

∵ 4 pm in 24-hour time is 4 + 12 = 16

∴ She works form 10 to 16

∴ She works for 6 hours (16  - 10 = 6)

∴ She is paid = 12.50 × 6 = $75

# 2nd part after 4 pm to 6 pm

∵ She is paid $14.25 per hour after 4 pm

∵ She works from 4 pm to 6 pm

- Calculate how many hours between 4 pm and 6 pm

∵ Both times are pm

∴ She works for 2 hours (6 - 4 = 2)

∴ She is paid = 14.25 × 2 = $28.5

- Add the two answers of the 1st part and the 2nd part

∴ She earns = 75 + 28.5 = $103.5

* She earns that day $103.5

Answer:

$103.50.

Step-by-step explanation:

From 10 am until 4 pm ( 6 hours) she earns 6 * 12.50 = $75.00.

Then from 4 pm to 6 pm she earns 2*14.25 = $28.50.

Total for that day = 75 + 28.50

= $103.50.

Answer:

Proved

Step-by-step explanation:

To prove that every point in the open interval (0,1) is an interior point of S

This we can prove by contradiction method.

Let, if possible c be a point in the interval which is not an interior point.

Then c has a neighbourhood which contains atleast one point not in (0,1)

Let d be the point which is in neighbourhood of c but not in S(0,1)

Then the points between c and d would be either in (0,1) or not in (0,1)

If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.

Then we find that dn is a boundary point of S

But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.

Every point of S=(0, 1) is an interior point of S.

Answer:

The reflected across x-axis, stretched vertically by factor 3 and shifted 5 units right and 4 units up.

Step-by-step explanation:

The parent function is

[tex]f(x)=x^2[/tex]

The given function is

[tex]f(x)=-3(x-5)^2+4[/tex]               ... (1)

The transformations of a quadratic function is defined as

[tex]f(x)=k(x+a)^2+b[/tex]                .... (2)

Where, k is vertical stretch or compression factor, a is horizontal shift and b is vertical shift.

If |k|>1, then graph stretch vertically if 0<|k|<1, then graph compressed vertically. If k<0 or negative, then the graph reflected across x-axis.

If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.

If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.

From (1) and (2) it is clear that

[tex]k=-3,a=-5,b=4[/tex]

k=-3<0, it means graph reflected across x-axis. |k|=3>1, so graph stretched vertically by factor 3.

a=-5<0, so the graph shifts 5 units right.

b=4>0, so the graph shifts 4 units up.

Therefore the reflected across x-axis, stretched vertically by factor 3 and shifted 5 units right and 4 units up.

Answer:

The value of g(4) is 2.

Step-by-step explanation:

Given,

The one-to-one functions g is defined as,

g = {(2,4), (4, 2), (5,-1), (8, -9)}

The element of a function is written in the form of order pair where,  the first element is input value and second element is the output value.

Since, under the function g the output value of input value 2 is 4,

That is, we can write,

g(2) = 4

Hence, the value of g(4) is 2.

Answer:  The required equation of the line is [tex]x-y=0.[/tex]

Step-by-step explanation:  We are given to to find the equation of a straight line in the standard form Ax + By = C passing through the points (-4, -4) and (-3, -3).

We know that the slope of a line passing through the points (a, b) and (c. d) is given by

[tex]m=\dfrac{d-b}{c-a}.[/tex]

So, the slope of the given line will be

[tex]m=\dfrac{-3+4}{-3+4}=\dfrac{1}{1}=1.[/tex]

Since the line passes through the point (-3, -3), so its equation will be

[tex]y-(-3)=m(x-(-3))\\\\\Rightarrow y+3=1(x+3)\\\\\Rightarrow y+3=x+3\\\\\Rightarrow y=x\\\\\Rightarrow x-y=0.[/tex]

Thus, the required equation of the line is [tex]x-y=0.[/tex]

Answer:

a) 1.26×10⁻⁴ H

b) 1.26×10⁻⁴ H

Step-by-step explanation:

N₁ = Number of turns in the solenoid = 600

N₂ = Number of turns in the coil = 30

A = Cross sectional area = 7×10⁻³ m²

l = Length of the solenoid = 0.4 m

μ₀ = Permeability of free space = 4π10⁻⁷ T m/A

a) Mutual inductance

[tex]M=\frac{\mu_{0} N_1N_2A}{l}\\\Rightarrow M=\frac{4\times 10^{-7}\times 600\times 30\times 7\times 10^{-3}}{0.4}\\\Rightarrow M=1.26\times 10^{-4}\ H[/tex]

∴ Mutual inductance of this system is 1.26×10⁻⁴ H

b) Mutual inductance does not depend on the cross sectional of the outer coil but depends on the area of the solenoid.

Hence, the mutual inductance remains the same.

Answer:

2063.9 kg

Step-by-step explanation:

Given,

1 kg = 2.2046 pounds

This can also be written as:

1 pound = 1/2.2046 kg

We have to calculate the value of 4550 pounds in kg up to nearest 10th

Thus,

[tex]4550 pounds= \frac{4550}{2.2046} kg[/tex]

Solving the above equation, we get:

4550 pounds = 2063.8664 kg

Rounding the above result to nearest tenth as:

4550 pounds = 2063.9 kg

Final answer:

The probability of picking the double-sided quarter is 1/100. The probability of getting heads on the 8th flip given that the previous 7 flips were all heads is still 1/2.

Explanation:

Let's calculate the probability of picking the double-sided (heads) quarter first.

Out of the 100 quarters, only 1 is double-sided. Therefore, the probability of picking the double-sided quarter is 1/100.

Now, let's calculate the probability of getting heads on the 8th flip given that the previous 7 flips were all heads.

The probability of getting heads on a single flip is 1/2, since there are two possible outcomes (heads or tails) and both are equally likely.

Since the flips are independent, the probability of getting heads on the 8th flip given that the previous 7 flips were all heads is still 1/2.

[tex]\vec F[/tex] is conservative if we can find a scalar function [tex]f[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require

[tex]\dfrac{\partial f}{\partial x}=5y^2z^3[/tex]

[tex]\dfrac{\partial f}{\partial y}=10xyz^3[/tex]

[tex]\dfrac{\partial f}{\partial z}=15xy^2z^2[/tex]

Integrate both sides of the first PDE with respect to [tex]x[/tex]:

[tex]f(x,y,z)=5xy^2z^3+g(y,z)[/tex] (*)

Differentiate both sides of (*) with respect to [tex]y[/tex]:

[tex]\dfrac{\partial f}{\partial y}=10xyz^3=10xyz^3+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]

Differentiate both sides of (*) with respect to [tex]z[/tex]:

[tex]\dfrac{\partial f}{\partial z}=15xy^2z^2=15xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]

So we have

[tex]f(x,y,z)=5xy^2z^3+C[/tex]

and so [tex]\vec F[/tex] is indeed conservative.

Answer:

  y = 11(x -1) +6

Step-by-step explanation:

The slope at the given point is ...

  y' = 14x -3x^2 = 14(1) -3(1^2) = 11

Then the point-slope equation for the line can be written as ...

  y = m(x -h) +k . . . . . . for slope m at point (h, k)

  y = 11(x -1) +6

The equation of the tangent line to the curve y = 7x^2 - x^3 at the point (1,6) is y = 11x - 5.

In

Mathematics

, specifically in Calculus, the equation of the tangent line to a curve can be found by first taking the derivative of the function. The derivative of the function y = 7x^2 - x^3 is y'= 14x - 3x^2. To find the slope of the tangent line at the point (1,6), we substitute x = 1 into the derivative to get y'(1) = 14*1 - 3*1^2  = 11. So the slope of the tangent line at the point (1,6) is 11. Using the point-slope formula, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line, we substitute to get the equation of the line: y - 6 = 11(x - 1). Simplifying this, we find the equation of the tangent line to be y = 11x - 5.

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Answer: 0.0089

Step-by-step explanation:

Given : Mean : [tex]\mu=\ 330[/tex]

Standard deviation :[tex]\sigma= 80[/tex]

Sample size : [tex]n=40[/tex]

We know that the standard error of the mean is given by :-

[tex]\text{S.E.}=\dfrac{\sigma}{\sqrt{n}}\\\\\Rightarrow\text{S.E.}=\dfrac{80}{\sqrt{40}}\\\\\Rightarrow\ \text{S.E.}=12.6491106407\approx12.65[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x = 360

[tex]z=\dfrac{360 -330}{\dfrac{80}{\sqrt{40}}}=2.37170824513\approx2.37[/tex]

The p-value = [tex]P(z>2.37)=1-P(z<2.37)= 1- 0.9911059\approx0.0089[/tex]

Hence, the likelihood the sample mean is greater than 360 minutes= 0.0089

The standard error of the mean is 12.68 minutes. The likelihood the sample mean is greater than 360 minutes is approximately 35.1%.

a. The standard error of the mean can be calculated using the formula: standard deviation / square root of sample size. In this example, the standard error of the mean is 80 / sqrt(40) = 12.68 minutes.

b. To find the likelihood that the sample mean is greater than 360 minutes, we need to transform the given information into a z-score and then find the corresponding probability using a standard normal distribution table. First, we calculate the z-score: (360 - 330) / 80 = 0.375. Then, we find the probability associated with a z-score of 0.375, which is approximately 0.649. Therefore, the likelihood the sample mean is greater than 360 minutes is 1 - 0.649 = 0.351, or 35.1%.

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No. The sample size is too small to be reliable. If there were many bags of 500 candies that consistently showed a larger-than-marketed amount of green candy, that will be better proof.

The difference between the observed proportion of green candies (0.12) and candy company's claimed proportion (0.10) does not necessarily mean that the manufacturing process is out of control. It could simply be down to natural variation or chance. A definitive conclusion would require a proper statistical test using more data.

In this scenario, we are looking at a statistic known as the proportion. It is calculated by dividing the number of successful outcomes (green candies) by the total number of outcomes (total candies). The candy company claims a proportion of 10%, which is 0.10. However, the observed proportion in your large bag of candies was 12% of 500, which works out to 0.12.

From a statistical standpoint, a slight variation from the claimed percentage is not unusual, as there will always be some random variability due to chance. While the difference between the observed proportion (0.12) and the expected proportion (0.10) might seem significant at first glance, we cannot definitively say from this one sample that the manufacturing process is out of control without conducting a proper hypothesis test — ideally, using additional data from more bags of candy.

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Answer:

Step-by-step explanation:

(A-B)U (A∩B)  [   We know that difference of A and B is A∩B')

A∩B')U(A∩B)    

A∩(B'UB)      (Using Distributive law)

A∩(U)            [ Union of a set and its complement is always a universal set]

A                        Hence proved

Answer:

Option A (134.7mm)

Step-by-step explanation:

Let's find the distance, but first we need to remember that the distance between two points with coordinates (Xa,Ya) and (Xb,Yb) is defined by:

[tex]distance = \sqrt{(Xb-Xa)^{2} + (Yb-Ya)^{2}  }[/tex]

From the situation we notice that:

Xb=31.45 and Xa=65.35, as well as:

Yb=-55.50 and Ya=74.88

Using the previous equation we have:

[tex]distance = \sqrt{(31.45-65.35)^{2} + (-55.50-74.88)^{2}  }[/tex]

[tex]distance = \sqrt{(-33.9)^{2} + (-130.38)^{2}  }[/tex]

[tex]distance = \sqrt{1149.21 + 16998.9444}[/tex]

[tex]distance = \sqrt{18148.1544}[/tex]

[tex]distance = 134.7151mm[/tex]

In conclusion, the distance between points (65.35,74.88) and (31.45,-55.50) is 134.7151mm, which is option A (134.7mm).

Answer:

17 cm

Step-by-step explanation:

In the above figure the question has been explained

the perpendicular from circle on the chord divides chord into two halves

let r be radius of circle in triangle NOY

NO= PO-PN = r-2 and YO=r(radius) YN=8cm

applying pythagorus theorem

[tex]r^{2} =8^{2} +(r-2)^{2}[/tex]

after rearranging and solving the equation we get

r=17 cm

therefore the radius of the triangle is  17 cm

note: on observing we find that the sides of triangle are pythagorian triplet 17, 15 and 8

The radius of the circle is calculated using the Pythagorean theorem for the right triangle formed by the perpendicular bisector, chord, and radius. By finding the length of the radius from the center to the midpoint of the chord and using NP as 2 cm, we calculate the radius to be approximately 8.246 cm.

The student has asked us to calculate the radius of the circle where a perpendicular bisector of a chord XY cuts through the chord at point N and the circle at P, with XY measuring 16 cm and NP measuring 2 cm. To solve this we can apply the theorem stating that in a circle, the radius bisecting an angle at the center is perpendicular to the chord that subtends the angle and bisects this chord. Since NP is half the distance of XY, we can conclude that N is the midpoint of XY, making XD (XD being the length from one end of the chord to its perpendicular bisector) equal to 8 cm. Considering triangle FNP, where F is the center of the circle, FN is the radius, and NP is 2 cm, we can use the Pythagorean theorem to find the radius FN:

FN2 = NP2 + FN2

FN2 = 82 + 22

FN2 = 64 + 4
FN2 = 68
FN = √68
FN ≈ 8.246 cm
Therefore, the radius of the circle is approximately 8.246 cm.

Step-by-step explanation:

Maping is the weak point in fixed point method.

Answer:  g(f(0)) = 2 and  (f ° g)(2) = -3.

Step-by-step explanation:  We are given the following two functions in the form of ordered pairs :

f = {(-2, 3), (-1, 1), (0, 0), (1,-1), (2,-3)}

g = {(-3, 1), (-1, -2), (0, 2), (2, 2), (3, 1)} .

We are to find g(f(0))  and   (f ° g)(2).

We know that, for any two functions p(x) and q(x), the composition of functions is defined as

[tex](p\circ q)(x)=p(q(x)).[/tex]

From the given information, we note that

f(0) = 0,  g(0) = 2,  g(2) = 2  and  f(2) = -3.

So, we get

[tex]g(f(0))=g(0)=2,\\\\(f\circ g)(2)=f(g(2))=f(2)=-3.[/tex]

Thus,  g(f(0)) = 2 and  (f ° g)(2) = -3.

Answer: [tex]H_0:\mu=453[/tex]

[tex]H_1:\mu\neq453[/tex]

Step-by-step explanation:

Given : An operator of chocolate-packaging machine monitors the net weights of the packaged boxes by periodically weighing random samples of boxes.

The operator wishes to see if a random sample of 50 boxes gives evidence that the mean is different than 453 grams.

i.e. Claim : [tex]\mu\neq 453[/tex]

We know that the null hypothesis is a kind of claim which always takes equals sign but alternative hypothesis does not.

Thus , the null and alternative hypotheses for the given situation will be :-

[tex]\text{Null hypothesis}\ H_0:\mu=453\\\text{Alternative hypothesis}\ H_1:\mu\neq453[/tex]