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svetoff [14.1K]
3 years ago
8

DEFORE SIGN INSTALLATION

Mathematics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

wat

Step-by-step explanation:

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A sample contains 10 pairs of values. Find the critical value for the linear correlation coefficient from Table A-6 correspondin
NikAS [45]

Answer: did you ever get the answer? I’m STRUGGLING

Step-by-step explanation:

3 0
3 years ago
Let l(x) be the statement "x has an Internet connection" and C(x, y) be the statement "x and y have chatted over the Internet,"
Bas_tet [7]

The question is:

Let l(x) be the statement "x has an Internet connection" and C(x, y) be the statement "x and y have chatted over the Internet," where the domain for the variables x and y consists of all students in your class. Use quantifiers to express each of these statements.

a. Jerry does not have an Internet connection.

b. Rachel has not chatted over the Internet with Chelsea.

c. Jan and Sharon have never chatted over the Internet.

d. No one in the class has chatted with Bob.

e. Sanjay has chatted with everyone except Joseph.

f. Someone in your class does not have an Internet connection.

g. Not everyone in your class has an Internet connection.

h. Exactly one student in your class has an internet connection.

i. Everyone except one student in your class has an Internet connection.

j. Everyone in your class with an Internet connection has chatted over the Internet with at least one other student in your class.

k. Someone in your class has an Internet connection but has not chatted with anyone else in your class.

l. There are two students in your class who have not chatted with each other over the Internet.

m. There is a student in your class who has chatted with everyone in your class over the Internet.

n. There are at least two students in your class who have not chatted with the same person in your class.

o. There are two students in the class who between them have chatted with everyone else in the class.

Step-by-step explanation:

Note that: Because the arguments are variables, we write

Ix = I(x)

Cxy = C(x,y)

a. Jerry does not have an Internet connection.

~I(Jerry).

b. Rachel has not chatted over the internet with Chelsea.

~C(Rachel,Chelsea).

c. Jan and Sharon have never chatted over the internet.

~C(Jan,Sharon).

d. No one in the class has chatted with Bob.

~∃x C(x, Bob).

e. Sanjay has chatted with everyone except Joseph.

∀y (y ≠ Joseph → C(Sanjay, y)).

f. Someone in your class does not have an internet connection.

∃x ~Ix.

g. Not everyone in your class has an internet connection.

~∀x Ix.

h. Exactly one student in your class has an internet connection.

∃x (Ix ∧ ∀y (Iy → y = x)).

i. Everyone except one student in your class has an internet connection. This means exactly one student does not have a connection.

∃x (~Ix∧ ∀y (Iy → y = x)).

j. Everyone in your class with an internet connection has chatted over the internet with at least one other student in your class.

∀x (Ix → ∃y (Cxy ∧ x ≠ y)).

Assuming nobody can chat with him or herself, then

∀x (Ix → ∃y (Cxy))

k. Someone in your class has an internet connection but has not chatted with anyone else in your class.

∃x (Ix ∧ ∀y ~Cxy).

l. There are two students in your class who have not chatted with each other over the internet.

∃x ∃y (x ≠ y∧~Cxy).

m. There is a student in your class who has chatted with everyone in your class over the internet.

∃x ∀y Cxy.

n. There are at least two students in your class who have not chatted with the same person in your class. ∃x ∃y (x ≠ y ∧ ∃z (~Cxz ∧ ~Cyz)).

o. There are two students in the class who between them have chatted with everyone else in the class.

∃x ∃y (x ≠ y ∧ ∀z (Cxz ∨ Cyz)).

5 0
4 years ago
How much will a person pay for 8.2 pounds of bananas at a price of $2.46 per pound?
kati45 [8]

Answer:

$19.99

Step-by-step explanation:

1 lb = 16 oz

8.2 lb = (16 x 8) +2= 130

\frac{16}{2.46} =\frac{1}{x}  ---> x = $0.154 per oz

0.154 x 130 =  $19.99

7 0
3 years ago
Read 2 more answers
100 Points!!!
inessss [21]

Answer:

An object moving along the x-axis is said to exhibit simple harmonic motion if its position as a function of time varies as

x(t) = x0 + A cos(ωt + φ).

The object oscillates about the equilibrium position x0.  If we choose the origin of our coordinate system such that x0 = 0, then the displacement x from the equilibrium position as a function of time is given by

x(t) = A cos(ωt + φ).

A is the amplitude of the oscillation, i.e. the maximum displacement of the object from equilibrium, either in the positive or negative x-direction.  Simple harmonic motion is repetitive.  The period T is the time it takes the object to complete one oscillation and return to the starting position.  The angular frequency ω is given by ω = 2π/T.  The angular frequency is measured in radians per second.  The inverse of the period is the frequency f = 1/T.  The frequency f = 1/T = ω/2π of the motion gives the number of complete oscillations per unit time.  It is measured in units of Hertz, (1 Hz = 1/s).

The velocity of the object as a function of time is given by

v(t) = dx(t)/dt = -ω A sin(ωt + φ),

and the acceleration is given by

a(t) = dv(t)/dt = -ω2A cos(ωt + φ) = -ω2x.

4 0
3 years ago
A 27​% solution ​( 27mg per 100 mL of​ solution) is given intravenously. Suppose a total of 1,36 L of the solution is given over
a_sh-v [17]

Answer:

Step-by-step explanation:

a.

(1.36 L)/(10 hr) = (0.136 L)/(hr)

Flow rate = (0.136 L)/(hr) × (1000 mL)/L = (136 mL)/(hr)

136 mL × (27 mg)/(100 mL) = 36.72 mg

Delivery rate = (36.72 mg)/(hr)

b.

(136 mL)/(hr) × (13 gtt)/(mL) = (1868 gtt)/(hr)

c.

10 hr × (36.72 mg)/)hr) = 367.2 mg

6 0
3 years ago
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