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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Which question is statistical?
A.How many Monarch butterflies live in the stake park?
B.How many of each species of butterflies did you see on your camping trip?
C.How many different species of butterflies are there in the world?
D.How many Monarch butterflies migrated south this year?
Answer:
i think c
Step-by-step explanation:
Find Angel C !!!
Thank you
Answer:
c = 24
Step-by-step explanation:
Angle c is supplementary to 146 and their sum is 180
c + 146 = 180 subtract 146 from both sides
c = 34
Find the value of x
Help plz
someone please anwser at this point im sad and listening to nirvana
Answer:
6) 9 3/5
7) 1/50
8) 28/45
Step-by-step explanation:
6) 8÷5/6=8x6/5=48/5=9 3/5
7) 1/5÷10=1/50
8) 7/5÷9/4=7/5x4/9=28/45
What is the value? 3/4 + 7/12 - (-4)
Answer: 3/4 + 7/12 - (-4) = 3/4 + 7/12 + 4 (two minuses make a plus)
3/4 + 7/12 + 4/1 = 9/12 + 7/12 + 48/12 (4 = 4/1 or 4 divided by 1)
(to add the fractions the denominators need to be the same. Multiply both the denominator and numerator by the same number)
(9+7+48)/12 = 64/12 = 5 and 1/3 or 16/3
We can also get this answer by leaving the 4 as a whole number:
9/12 + 7/12 +4 = 16/12 + 4 = 1 and 1/3 + 4 = 5 and 1/3
If you put the sum in a calclator you also get 5 and 1/3.
Hope this helps :)
How many times does the digit 6 appear in the numbers between 1 to 99 I’m suffering
Answer:
21 times
Step-by-step explanation:
Answer:
20 Times
Step-by-step explanation:
The length of a rectangle is 4 inches less than its width and the area is 21 square inches. Find
the smallest dimension of the rectangle.
which of the following sequences of transformation maps MNO onto ABC?
Answer:
MNO no se estoy en 6th y no se pero esta es mi respuesta espero y me la saco bien
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).
T(2, –4) maps M"(-3,0) to M'(-1,-4), N"(-1,2) to N"(1,-2), and O"(1,1) to O"(3,-3).
The resulting coordinates M'(-1,-4), N'(1,-2), and O'(3,-3) are the coordinates of triangle M"N"O".
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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(39-7) divided by 4^2 = ?
Answer:
Its 2
Step-by-step explanation:
(39-7) / 4^2
32 / 4^2
32 / 2^4
2^5 / 2^4 divide which gets you
2
find which fraction is larger 3/12 ,2/6 ,1/24
Answer:
3/12
Step-by-step explanation:
Answer:
2/6
Step-by-step explanation:
it is equal to 1/3. The others are equal to 1/4 and 1/24 so 1/3 is the largest
I don't understand this question like for example what do you do with the $28.00 if it's not in the graph?
Answer:
Choice C
Step-by-step explanation:
You find the ratio, so every if 4 McDonalds is $28 thats 4:28
You find the unit rate 1:x
divide each side by 4 and the unit rate is 7. Graph c shows cordinate (1,7)
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.
The length is _____________ ft
The width is ______________ ft
Answer:
length = 8 ft
width = 5 ft
Step-by-step explanation:
width = w
length = w + 3
Area of rectangular garden = 40 square feet
length *width = 40
(w + 3) *w = 40
w*w + 3*w = 40
w² + 3w = 40
w² + 3w - 40 = 0
w² + 8w - 5w - 8*5 = 0
w(w + 8) - 5 (w +8) =0
(w + 8) (w - 5) = 0
w - 5 = 0 {Ignore w + 8 =0, as measurement will not be in -ve}
w = 5 ft
l = 5 + 3
l = 8 ft
giving out brainliest answer !!!!! help me out ASAP !!!
Answer:
0.7880
Step-by-step explanation:
0.7883
Hope it helps!
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.
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.
Leonard wrote the sequence of numbers below. 9, 15, 21, 27,... Which expression did be use to form the sequence, where n is the term number?
A 3n + 3
B 3n - 3
C 6n+3
D 6n-3
Answer:
I think they took 3 * 3 is equals to 9
3 * 5 equals to 15
3 * 7 equals to 21
and 3 * 9 equals to 27.
Step-by-step explanation:
can can you realise that they skipped 3 * 4 3 * 6 and 3 * 8 and the difference between 9 and 15 is 6 meaning there's a gap of 6 at each sequence so I guess my answer is 3 and + 3 if it's not if this is not the answer please tell me if it is not matching with your book tell me so that i can ask the people around me. answer 3n + 3
please help will give brainliest!
Answer:
1. x = 14
2. c = 12
3. d = 4
4. w = 8
5. z = 54
6. d = 20
7. n = 26.4
8. k = 15
Step-by-step explanation:
how do u calculate the area of a circle
The area of a circle is pi times the radius squared (A = π r²).
Determine the length of UT.
What is z in this equation
Answer:
I got you rn
Step-by-step explanation:
its 42 because opposite angles are congruent
Answer:
Step-by-step explanation:
z = 53 + 39 {Exterior angle property of triangle}
z = 92°
bing + us ????? need help ples asap
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
The sales tax rate is 5%. If Carmen buys a fountain priced at $329.97, how muchztax will
she pay?
Round your answer to the nearest cent: $
Answer:
$16.50
Step-by-step explanation:
Take the total price, convert the percentage to a decimal and multiply the two together.
329.97*0.05=16.498
Round to the nearest cent.
$16.50
Answer:
346.47
Step-by-step explanation:
the equation is: 329.97 * 5/100 + 329.97
= 16.4985 +329.97=346.4685
rounding to the nearest cent we get 346.47
Which number has a greater value than |-2.6|?
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:
|-2.6| = 2.6
So any number higher than that work for this question :-)
Last year, scientists claimed there were 10,000 different species of ants around the world. Since then, the number of different species have decreased by 5.1%. How many different species of ants are estimated currently? Round to the nearest whole number.
help..
Answer:
9490 species
Step-by-step explanation:
Hope this helps! You take 5.1% of 10,000 which is 510. Then you take 10,000-510= 9490
Answer: there are more than 12,500 species of ants today like right now
Step-by-step explanation:
To ship a package, a shipping company charges $1.68 for each pound. How much would it cost to ship a 5.5 pound package?
Answer:
$9.24
Step-by-step explanation:
1.68÷2=0.84
1.68×5+0.84=9.24
5. 14,5.13) 1. Amelia has 16 ounces of perfume to divide equally among 3 bottles. How many ounces of perfume will she pour into each bottle?
Answer:
-16/3 = -5.3333333333
Step-by-step explanation:
The 25th term of a sequence is 121. The 80th term is 506. What is the first term
Answer:
-47
Step-by-step explanation:
that's the correct answer
ILL GIVE BRAINLEST, tell whether the angles are adjacent or vertical. then find the value of x.
Answer:
The angles are adjacent and x=100
Select all the factor pairs. What are the factor pairs of -144 that add to 0.
If L 1and"2 are parallel lines and l1 has a slope of -2 what is the slope of l2
Answer:
-2
Step-by-step explanation:
If these two lines are parallel, they have the same slope, which is -2.
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.
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.