corroborate

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corroborate
▪ I. corroborate, ppl. a. Obs. or arch. (kəˈrɒbərət) [ad. L. corrōborāt-us, pa. pple. of corrōborāre: see next.] Strengthened, confirmed, corroborated. a. as pa. pple. arch.1532 Thynne Chaucer's Wks. Ded., Maynteyned, corroborate, and defended. c 1534 tr. Pol. Verg. Eng. Hist. (Camden) I. 112 That..... Oxford English Dictionary
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corroborate
corroborate/kəˈrɔbəreɪt; kə`rɑbəˌret/ v[Tn]confirm or give support to (a statement, belief, theory, etc) 证实, 支持(某种说法、 信仰、 理论等) Experiments have corroborated her predictions. 实验证实了她的预言. 牛津英汉双解词典
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Hacked Records Corroborate Claims in Hydroxychloroquine ... - The Intercept
Aug 21, 2023AFLDS records, provided to The Intercept by an anonymous hacker in September 2021, corroborate parts of Hatfield's account. Culver is included in the list of 225 AFLDS physicians who prescribed ...
theintercept.com 0.0 0.90000004 0.0
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NBA 'Could Not Corroborate' Claim that Ja Morant's ...
Feb 6, 2023 — The NBA said they 'could not corroborate' a claim that members of Ja Morant's entourage pointed a gun laser at people from the Indiana ...
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Analysts corroborate rumors that upcoming MacBook Pro will ... - 9to5Mac
Chance is an editor for the entire 9to5 network and covers the latest Apple news for 9to5Mac. Tips, questions, typos to chance@9to5mac.com ...
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True or false: For all subsets $ A $ and $ B $ of $ X $, if $ A \subseteq B $, then $ f[A] \subseteq f[B] $. I am trying to determine if the following is true or false: > For all subsets $ A $ and $ B $ of $ X $, if ...
Assume $A \subseteq B$. Let $y \in f(A)$. Then $\exists x \in A : f(x)=y$. But $A \subseteq B$, so if $x \in A$, then $x \in B$. Therefore, $\exists x \in B : f(x) = y$ so that $y \in f(B)$ as required. We conclude that $f(A) \subseteq f(B)$.
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Stumped on a Bayes Theorem Question A certain medical syndrome is usually associated with two overlapping sets of symptoms, A and B. Suppose it is known that: P($A|B$) = 0.8 P($B|A$) = 0.9 P($B'|A'$) = 0.85 Find P...
We know that $$ P(A\mid B) = \frac{P(A \cap B)}{P(B)}, $$ and vice versa, so $$ P(A \cap B) = 0.8 P(B) = 0.9 P(A). $$ On the other hand, $$ P(B' \mid A') = \frac{P(A' \cap B')}{1-P(A)} = \frac{1-P(A \cup B)}{1-P(A)}, $$ and $P(A \cup B) = P(A)+P(B)-P(A \cap B)$ So $$ 1+P(A \cap B) -P(A)-P(B) = 0.85(...
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Question about proof of Heine-Cantor (i.e. compact and continuous implies uniform continuous) If anyone has seen the wikipedia page for the Heine-Cantor theorem, I find something off about the proof it presents. It wo...
That's not what it says: suppose that $d_X(x,y) < \delta$ for two arbitrary points of $X$. It might have been less confusing for you if the proof had said: "let $x,y$ be two arbitrary points of $X$ such that $d_X(x,y) < \delta$" instead: Reason: we want to show $$\forall x,y \in X: d_X(x,y) < \delta...
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What is the minimal polynomial of $T^{-1}$? > Suppose $T$ is an invertible linear operator on a finite dimensional vector space with minimal polynomial $p_T=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$. Is it true that the minima...
Lemma: Given a non-constant polynomial $$ p(x) = x^m+a_{m-1}x^{m-1}+\ldots+a_0,\qquad a_0\neq 0$$ then $\xi\neq 0$ is a root of $p(x)$ iff $\frac{1}{\xi}$ is a root of the reciprocal polynomial $$ q(x) = a_0 x^m + a_1 x^{m-1} +\ldots + 1. $$ Proof: $p(\xi)=0$ implies $\frac{p(\xi)}{\xi^m} = q\left(\...
prophetes.ai 0.0 0.3 0.0
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Simplifying $\log_{\frac{1}{3}}(3^{2x})$ So I have the following problem: $\log_{\frac{1}{3}}(3^{2x})$ How do I solve this? Somewhere I stumbled onto the solution where they find a common exponent and base so they ca...
You have a good idea. But you need to define variables properly. Let $$ y=\log_{\frac{1}{3}}(3^{2x}) $$ Then $$ 3^{-y}=\left(\frac{1}{3}\right)^y = 3^{2x}. $$ The exponents in the previous equation must be the same (since the exponential function is injective). Thus $$ y=-2x. $$
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How extensive is CD47? CD47 aka the "don't eat me" signal has recently been claimed to be expressed on all tumor cells. This doesn't seem to corroborate with other cell-biology experiments. On what other cells is CD47...
I don't know how extensive. Let's run a simple data query and find out: Go to GEO at NCBI. In the "Gene profiles" window, type CD47, and hit enter to launch the query. At the top of the resulting page, use the link labeled "Limits" to restrict the 7000+ results to human by entering the term "human" ...
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Why is "nobody" called the "most privileged user after root"? When I was first learning Linux I recall reading that because it is so widely used, "nobody" is (jokingly?) known as the "second most privileged user, afte...
Sometime Linux distributions incorrectly implement the standards and assign files to the user nobody. In that case programs have access to a bunch of files which are not for them. According to the standards, the user nobody shouldn't own any files and it should only be used to run programs which don...
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Knowing the distribution of $(X,Y)$ implies knowing the distribution of $(X,Y, X-Y)$? Consider a random vector $(X,Y)$ and suppose we know the **joint** probability distribution $P$ of $(X,Y)$. Does this mean that we...
If $f$ is the (continuous) mapping $(x,y)\to(x,y,x-y)$ from $\Bbb R^2$ to $\Bbb R^3$, then the joint distribution of $(X,Y,X-Y)$ is the probability measure $Q$ given by $Q(B)=P(f^{-1}(B))$, for Borel sets $B\subset\Bbb R^3$. This provides a rather abstract YES answer to your question. In your normal...
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Show that $A \in M_n$ and $B \in M_n$ need not be equal if $x^TAx = x^TBx$ for all $x \in \mathbb{C}^n$. Show that $A \in M_n$ and $B \in M_n$ need not be equal if $x^TAx = x^TBx$ for all $x \in \mathbb{C}^n$. My att...
So you need $C\not=0$ such that $x^TCx = 0$ for all $x$. For $2\times2$-matrices this may be achieved with $C=C_2=\pmatrix{0&1\\\\-1&0}$. For $n>2$ you may use $\pmatrix{C_2&0\\\0&0}$.
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Did Amber Heard physically abuse Johnny depp? and admit to it? The Daily Mail has published an article claiming that "Amber Heard admits to 'hitting' ex-husband Johnny Depp and pelting him with pots, pans and vases". ...
Depp is alleging abuse by Heard in **an ongoing lawsuit** filed back in March, and Heard denies this allegation. The alleged admission by Heard is on a 2015 audio recording obtained exclusively by the _Daily Mail_. Unless and until the audio is released, a lot depends on your opinion of the _Mail's_...
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