$f$ cts on $[a,b]$ and $f(x)\ne 0$ for all $x\in [a,b]$ implies that $f(x)$ is either always positive or negative on $[a,b]$.

Owen Mathis
2022-11-18
Answered

$f$ cts on $[a,b]$ and $f(x)\ne 0$ for all $x\in [a,b]$ implies that $f(x)$ is either always positive or negative on $[a,b]$.

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Brooklyn Mcintyre

Answered 2022-11-19
Author has **18** answers

Intermediate value theorem.

asked 2022-10-13

A virus has been spread around a population. The prevalence of this virus is 84%. A diagnostic test, with a specificity of 94% and sensitivity of 15%, has been introduced. If a patient is drawn randomly from the population, what is the probability that:

a) a person has the virus, given that they tested positive?

b) a person has the virus, given that they tested negative?

a) a person has the virus, given that they tested positive?

b) a person has the virus, given that they tested negative?

asked 2022-05-09

True or false: if ${a}_{n}$ is any decreasing sequence of positive real numbers and ${b}_{n}$ is any sequence of real numbers converges to $0$, then $\frac{{a}_{n}}{{b}_{n}}$ diverges.

asked 2022-11-09

"The population, for a disease D, has a true rate of T%"

"Some Test ST, has false positive rate of FP% and a false negative rate of FN%."

"T, FP, and FN are elements of the set of real numbers."

Was the number T determined by some 100% accurate and possibly expensive test?

"Some Test ST, has false positive rate of FP% and a false negative rate of FN%."

"T, FP, and FN are elements of the set of real numbers."

Was the number T determined by some 100% accurate and possibly expensive test?

asked 2022-10-15

True or False: If $x\notin \mathbb{Q}$ then $\sum _{m\ge 0}m{x}^{m-1}\notin \mathbb{Q},$, where $|x|<1.$

Considered the contra-positive of the above statement:

If $\sum _{m\ge 0}m{x}^{m-1}\in \mathbb{Q}$ then $x\in \mathbb{Q}.$

Now if the contra-positive is true/false then the statement is true/false.

Thus, if

$$\sum _{m\ge 0}m{x}^{m-1}\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}1+2x+3{x}^{2}+...\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{1}^{\prime}+{x}^{\prime}+({x}^{2}{)}^{\prime}+({x}^{3}{)}^{\prime}+...\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(1+x+{x}^{2}+{x}^{3}+...{)}^{\prime}\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(\frac{1}{1-x}{)}^{\prime}\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{-1}{(1-x{)}^{2}}\in \mathbb{Q}$$

Now from the last step we can conclude that $x\in \mathbb{Q}$, since $|x|<1.$ Therefore the contrapositive is true, so the given statement is also true.

Is my analysis correct?

Considered the contra-positive of the above statement:

If $\sum _{m\ge 0}m{x}^{m-1}\in \mathbb{Q}$ then $x\in \mathbb{Q}.$

Now if the contra-positive is true/false then the statement is true/false.

Thus, if

$$\sum _{m\ge 0}m{x}^{m-1}\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}1+2x+3{x}^{2}+...\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{1}^{\prime}+{x}^{\prime}+({x}^{2}{)}^{\prime}+({x}^{3}{)}^{\prime}+...\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(1+x+{x}^{2}+{x}^{3}+...{)}^{\prime}\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(\frac{1}{1-x}{)}^{\prime}\in \mathbb{Q}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{-1}{(1-x{)}^{2}}\in \mathbb{Q}$$

Now from the last step we can conclude that $x\in \mathbb{Q}$, since $|x|<1.$ Therefore the contrapositive is true, so the given statement is also true.

Is my analysis correct?

asked 2022-05-10

in order to remember stuff i need to understand their reason. Right now i cannot remember what is type 1 error and what is type 2 error why is the reason type 1 is false positive?

asked 2022-07-11

Simplification step by step (exponential and logarithm) - regarding BLOOM FILTER - FALSE POSITIVE probability

$(1-\frac{1}{m}{)}^{kn}\approx {e}^{-kn/m}$

I don't get how to reach ${e}^{-kn/m}$ from $(1-\frac{1}{m}{)}^{kn}$

I have reviewed logarithm and exponent rules but I always get stuck, here is what I have tried; Starting from $(1-\frac{1}{m}{)}^{kn}$, I can write:

- ${e}^{(ln(1-\frac{1}{m}{)}^{kn})}$

- ${e}^{(kn\times ln(1-\frac{1}{m}))}$

Or the other way around:

$ln({e}^{(1-\frac{1}{m}{)}^{kn}})$

$ln({e}^{(1-\frac{1}{m})\times kn})$

$ln({e}^{(1-\frac{1}{m})})+ln(kn)$

$(1-\frac{1}{m})+ln(kn)$

After few steps I fall in an unsolvable loop hole playing with $ln$ and $e$ trying to reach ${e}^{-kn/m}$.

$(1-\frac{1}{m}{)}^{kn}\approx {e}^{-kn/m}$

I don't get how to reach ${e}^{-kn/m}$ from $(1-\frac{1}{m}{)}^{kn}$

I have reviewed logarithm and exponent rules but I always get stuck, here is what I have tried; Starting from $(1-\frac{1}{m}{)}^{kn}$, I can write:

- ${e}^{(ln(1-\frac{1}{m}{)}^{kn})}$

- ${e}^{(kn\times ln(1-\frac{1}{m}))}$

Or the other way around:

$ln({e}^{(1-\frac{1}{m}{)}^{kn}})$

$ln({e}^{(1-\frac{1}{m})\times kn})$

$ln({e}^{(1-\frac{1}{m})})+ln(kn)$

$(1-\frac{1}{m})+ln(kn)$

After few steps I fall in an unsolvable loop hole playing with $ln$ and $e$ trying to reach ${e}^{-kn/m}$.

asked 2022-04-07

A pregnancy test kit is 98.5% accurate for true positive result, i.e. the result is positive when the tester is actually pregnant. If she is not pregnant, however, it may yield a 0.8% false positive. Suppose a woman using this pregnancy kit is 60% at risk of being pregnant.

Not sure about her first test which turned out to be negative, the woman decides to take the test again. This second test, however, turns out to be positive. Assuming the two test are independent, find the probability that she is actually pregnant.

Now she is so confused whether or not she is pregnant. So she take the tests n more times and the results for these n more tests are all positive. Find the minimum value for n so that she can be at least 99.99% sure of pregnancy, assuming all test are independent.

Not sure about her first test which turned out to be negative, the woman decides to take the test again. This second test, however, turns out to be positive. Assuming the two test are independent, find the probability that she is actually pregnant.

Now she is so confused whether or not she is pregnant. So she take the tests n more times and the results for these n more tests are all positive. Find the minimum value for n so that she can be at least 99.99% sure of pregnancy, assuming all test are independent.