Here we show that function defined on an interval value property cannot have (a; b) and satisfying the intermediate removable or (a) jump discontinuity. Suppose has & jump discontinuity at Xo € (a,b) and lim f (x) lim f (x) xx0 {ix0 Choose 0 such that lim f (x) < 0 < lim f (x) and 0 + f(xo) xI*o Xx0 In Exercise & we showed there is interval [xo 0,.Xo) such that f(x) < 0 if Xe [xo 6,xo): Likewise, there an interval (xo, Xo + 6] such that f(x) > 0 if xe(xo, Xo + 6]. Conclude that does not satisly the intermediate value property on [xo 6,xXo + 6]. (6) Suppose has a removable discontinuity at Xo € (a,b) and a = lim f(x) < f(xo) Show that there is an interval [xo = 6,Xo) such that f(x)< a+[f(xo) - &] if x e[xo 6,Xo]: Conclude that f does not satisfy the intermediate value property

Answers

Answer 1

f cannot have a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex] and  [tex]$$ \lim _{x \uparrow x_0} f(x) < \lim _{x \mid x_0} f(x) .$$[/tex]

f cannot have a removable discontinuity at [tex]$$x_0 \in(a, b) $$[/tex] and [tex]\alpha=\lim _{x \rightarrow x_0} f(x) < f\left(x_0\right)[/tex]

Let f be a function defined on (a, b) satisfies intermediate value property.

Claim: f ca not have removable on jump discontinuity.

Suppose f has a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex]

We take [tex]$\theta$[/tex] such that

[tex]$$\lim _{x \rightarrow x_0} f(x) < \theta < \lim _{x \downarrow x_0} f(x) \text { and } \theta \neq f\left(x_0\right)$$[/tex]

Now there exist [tex]$\delta > 0$[/tex] such that [tex]$f(x) < \theta$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right)$[/tex] and [tex]$f(x) > \theta$[/tex] for all [tex]$x \in\left(x_0, x_0+\delta\right]$[/tex]

Now [tex]$f\left(x_0-\delta\right)[/tex][tex]< \theta < f\left(x_0+\delta\right)$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0+\delta\right] \backslash\left\{x_2\right\}$[/tex] and [tex]$f\left(x_0\right) \neq \theta$[/tex].

Therefore the point [tex]$\theta$[/tex] has no preimage under f

that is, there does not exists [tex]$y \in\left[x_0-\delta, x_0+\delta\right][/tex] for which

[tex]$$f(y)=\theta[/tex] because [tex]\left\{\begin{array}{l}y=x_0 \Rightarrow f(y) \neq \theta \\y > x_0 \Rightarrow f(y) > \theta \\y < x_0 \Rightarrow f(y) < \theta\end{array}\right.$$[/tex]

Therefore f does not satisfies intermediate value property on [tex]$\left[x_0-\delta, x_0+\delta\right]$[/tex],

Hence f does not satisfies IVP on (a, b) which is not possible because we assume f satisfies IVP on (a, b),

Therefore f can not have a jump discontinuity.

Suppose f has a removable point of discontinuity at [tex]$x_0 \in(a, b)$[/tex],

Let [tex]$\alpha=\lim _{\alpha \rightarrow x_0} f(x)$[/tex],

Let [tex]\alpha < f\left(x_0\right)$[/tex] so [tex]$f\left(x_0\right)-\alpha > 0$[/tex].

Now [tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex] then [tex]\exists$ \delta > 0$[/tex] such that

[tex]$$\begin{aligned}& |f(x)-\alpha| < \frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left\{x_0-\delta, x_0-\alpha\right]-\left\{x_0\right\} \\& \Rightarrow \quad f(x) < \alpha+\frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left[x_0-\delta, x_0+\delta\right]-\left\{x_0\right\}\end{aligned}$$[/tex]

So [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2}$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right]-\left\{x_0\right\}$[/tex]

Now [tex]$f\left(x_0\right) > \alpha$[/tex].

And  [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2} < f\left(x_0\right)$[/tex] for all [tex]$x \in\left[\left(x_0 \delta, x_0\right)\right.$[/tex]

Let [tex]$\mu=\frac{f\left(x_0\right)+\alpha}{2}$[/tex].

Then there does not exist [tex]$e \in\left[x_0-\delta, c\right]$[/tex] such that [tex]$f(c)=\mu$[/tex]

Because for [tex]$e=x_0 \quad f(e) > \mu$[/tex]

                for [tex]$c < x_0 \quad f(c) < \mu$[/tex].

Therefore f does not satisfy IVP on [tex]$\left[x_0-\delta_1 x_0\right]$[/tex] which contradict our hypothesis,

therefore [tex]$\alpha \geqslant f\left(x_0\right)$[/tex]

Let [tex]$\alpha > f\left(x_0\right)$[/tex]. so [tex]$\alpha-f\left(x_0\right) > 0$[/tex]

[tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex]

Then [tex]\exists $ \varepsilon > 0$[/tex] such that

[tex]$|f(x)-\alpha| < \frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$\left.x \in\left[x_0-\varepsilon_0 x_0+\varepsilon\right]\right\}\left\{x_i\right\}$[/tex]

[tex]$\Rightarrow f(x) > \alpha-\frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0+\varepsilon\right] \backslash\left\{x_0\right\}$[/tex]

[tex]$\Rightarrow f(x) > \frac{\alpha+f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0\right)$[/tex]

Now [tex]$f\left(x_0\right) < \alpha$[/tex]

The [tex]$f(x) > \frac{f\left(x_0\right)+\alpha}{2} > f\left(x_0\right)$[/tex].

So [tex]$f\left(x_0\right) < \frac{f\left(x_0\right)+\alpha}{2} < f(x)$[/tex] for all [tex]$x \in\left[x_0 \varepsilon, \varepsilon_0\right)$[/tex]

Let [tex]$\eta=\frac{f\left(x_e\right)+\alpha}{2}$[/tex]

Then there does not exist [tex]$d \in\left[x_0-\varepsilon, x_0\right]$[/tex] such that [tex]$f(d)=\xi$[/tex].

Because if [tex]$d=x_0, f(d)=f\left(x_0\right) < \eta$[/tex] if [tex]$d E\left[x_0-\varepsilon, x_0\right)$[/tex]

Then [tex]$f(d) > \eta$[/tex]

Therefore f does not satisfies IVP on [tex]$\left[x_0-\varepsilon, x_0\right]$[/tex] which contradict olio hypothesis.

Therefore [tex]$\alpha \leq f\left(x_0\right)$[/tex] (b) From (a) and (b) it follows [tex]$\alpha=f\left(x_0\right)=\lim _{x \rightarrow x_0} f(x)$[/tex]. Therefore f can not have a removable discontinuous

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Step-by-step explanation:

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The  sequence of transformations T(–2, 4) ry-axis ry-axis T(2, –4) can be used to map triangle MNO onto M"N"O".

What is Statistics?

Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.

The given sequence of transformations is:

T(–2, 4) - translation by (-2, 4)

ry-axis - reflection across the y-axis

ry-axis - reflection across the y-axis

T(2, –4) - translation by (2, -4)

We can check whether this sequence of transformations maps triangle MNO onto M"N"O" by applying each transformation to the coordinates of triangle MNO in order:

T(–2, 4) maps M(5,-4) to M'(3,0), N(3,-2) to N'(1,2), and O(1,-3) to O'(-1,1).

ry-axis reflects M'(-3,0) to M"(-3,0), N'(1,2) to N"(-1,2), and O'(-1,1) to O"(1,1).

ry-axis reflects M"(-3,0) back to M"(-3,0), N"(-1,2) back to N"(-1,2), and O"(1,1) back to O"(1,1).

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Therefore, the given sequence of transformations T(–2, 4) ry-axis ry-axis T(2, –4) can be used to map triangle MNO onto M"N"O".

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Take the total price, convert the percentage to a decimal and multiply the two together.

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The length of rectangular garden is 3 feet longer than the width. If the area of the garden is 40 square feet, find the length and width of the garden.

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Answer:

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Step-by-step explanation:

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Answers

Answer: I'm assuming that there were multiple choices provided for this question.. but just to help you, when a number is within the absolute value, it changes any number to a positive:

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Answer:

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Step-by-step explanation:

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A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of 138 cars owned by students had an average age of 5.13 years. A sample of 111 cars owned by faculty had an average age of 7.75 years. Assume that the population standard deviation for cars owned by students is 3.45 years, while the population standard deviation for cars owned by faculty is 2.08 years. Determine the 95% confidence interval for the difference between the true mean ages for cars owned by students and faculty. Step 1 of 3: Find the point estimate for the true difference between the population means.

Answers

Answer:

The point estimate for the true difference between the population means is of -2.62 years.

The 95% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -3.4 years and -1.84 years.

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem, and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the mean, while the standard deviation is the square root of the sum of variances.

A sample of 138 cars owned by students had an average age of 5.13 years. The population standard deviation for cars owned by students is 3.45 years.

This means that:

[tex]\mu_{s} = 5.13, \sigma_{s} = 3.45, n = 138, s_s = \frac{3.45}{\sqrt{138}} = 0.2937[/tex]

A sample of 111 cars owned by faculty had an average age of 7.75 years.  The population standard deviation for cars owned by faculty is 2.08 years.

This means that [tex]\mu_{f} = 7.75, \sigma_{f} = 2.08, n = 111, s_f = \frac{2.08}{\sqrt{111}} = 0.2658[/tex]

Difference between the true mean ages for cars owned by students and faculty.

s - f

Mean:

[tex]\mu = \mu_s - \mu_f = 5.13 - 7.75 = -2.62[/tex]

This is the point estimate for the true difference between the population means.

Standard deviation:

[tex]s = \sqrt{s_s^2+s_f^2} = \sqrt{0.2937^2+0.2658^2} = 0.3961[/tex]

Confidence interval:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = zs = 1.96*0.3961 = 0.78[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is -2.62 - 0.78 = -3.4

The upper end of the interval is the sample mean added to M. So it is -2.62 + 0.78 = -1.84

The 95% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -3.4 years and -1.84 years.

bing + us ????? need help ples asap

Answers

Bing + us = Bingus. That's all

Answer:

it's very hard lol

Bing+us

=Bingus

Step-by-step explanation:

oh my gawd... hope it's ryt

thanks lol

A sequence of Bernoulli trials consists of choosing components at random from a batch of components. A selected component is either classified as defective or nondefective. If the probability that a selected component is non-defective is 0.8, find the following probabilities: a) Three non-defective components in a batch of seven components. b) 8 non-defective components are drawn before the first defective component is chosen.

Answers

Answer:

a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

Step-by-step explanation:

A sequence of Bernoulli trials composes the binomial distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The probability that a selected component is non-defective is 0.8

This means that [tex]p = 0.8[/tex]

a) Three non-defective components in a batch of seven components.

This is P(X = 3) when n = 7. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287[/tex]

0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 8 non-defective components are drawn before the first defective component is chosen.

Now the order is important, so the we just multiply the probabilities.

8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So

[tex]p = (0.8)^8*0.2 = 0.0336[/tex]

0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

how do u calculate the area of a circle

Answers

The area of a circle is pi times the radius squared (A = π r²).

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