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Jet001 [13]
3 years ago
8

Can someone help me with this

Mathematics
1 answer:
timurjin [86]3 years ago
8 0
200 is the answer sir
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What is the separation method of blood?​
natita [175]

Answer:

Blood separation is one of the crucial processes in a clinical lab and is usually conducted via a process called centrifugation.

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The 1998 team payroll of the Baltimore Orioles was $68,988.134. (Source: Major
Gekata [30.6K]
False. This is not a statistic. A statistic would be something more like “Since 1998, the team payroll of the Baltimore Orioles increased 26%” Statistics show things more overtime, rather than that just being a basic fact that the team payroll was that amount.
5 0
3 years ago
Please help me solve numbers 4, 5, and 6 with some type of explanation as to how you got that answer.
Mila [183]

Answer:

4.R=16

D=16+16=32

C=

\pi \times d

C=32π

then do the same thing for other problem

6 0
3 years ago
What is the recursive formula for this sequence . 10,14,18,22,26
Marrrta [24]

add 4 each time

the nth term is a_n or f(n)

the next term after that is a_{n+1} or f(n+1)

so each term is 4 more than previous

basically

a_{n+1}=4+a_n or

f(n+1)=4+f(n)

same thing

4 0
3 years ago
Read 2 more answers
Let S(x, y) denote the statement "x has seen y" and D denote the set of all students in our class and M be the set of all movies
oksano4ka [1.4K]

Answer:

Step-by-step explanation:

a) Recall the quantifiers \forall, \exists.

Then, we can translate the proposition as follows

\forall m \in M \exists x \in D \exists y \in D S(x,m)\land S(y,m)

b) Recall that an expression of the form \exists x P(x) its negation is of the form \forall x \neg P(x) which means that it is not true that for all elements the proposition P holds. Equivalently, we have that the negation of an expression of the form \forall x P(x) is \exists x \neg P(x) which means that there is at least one x such that P doesn't hold. Using this, we get the following

\neg(\forall m \in M \exists x \in D \exists y \in D S(x,m)\land S(y,m))= \exists m \in M \neg (\exists x \in D \exists y \in D S(x,m)\land S(y,m))= \exists m \in M \forall x \in D \neg (\exists y \in D S(x,m)\land S(y,m))= \exists m \in M \forall x \in D \forall y \in D \neg(S(x,m)\land S(y,m))

By De Morgan's law, we have that \neg (A \land B) = \neg A \lor \neg B

So, the final statement is

\exists m \in M \forall x \in D \forall y \in D \neg S(x,m) \lor \neg S(y,m)

c)

This statement means: There is a movie that for every pair of students, at least one of the students hasn't seen the movie yet.

8 0
3 years ago
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