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3 years ago

If you ask the first 50 females who enter a restaurant to complete a survey about their experience, who is the sample?

Mathematics

2 answers:

balandron [24]3 years ago

8 0

Answer:

the first girl is the answer

Anettt [7]3 years ago

6 0

Answer:

the first girl

Step-by-step explanation:

because the first girl to take the survey is a sample but the rest that follows are just extras(can i get brainliest pls)

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R = -5, s = 2, t = 7, u = -1 <br><br><br> 5r - 4

Doss [256]

5r-4 and r=-5 so it would be 5(-5)=-20, then -4 , so is is -24

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3 years ago

Which inequality is true?

Vladimir79 [104]

Answer:

C) 3pi>9

Step-by-step explanation:

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3 years ago

How to find the perimeter of a triangle with vertices

marissa [1.9K]

Answer:

P = A + B + C

Step-by-step explanation:

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Hit brainliest if this was helpful :)

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3 years ago

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Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

that $b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$ ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

$\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)$

which tells us that $\phi$ is a homorphism. The kernel of $\phi$

is the set of elements g in G such that $g^{n}=1$ . However,

$\phi$ is not necessarily 1-1 or onto, if $G=\mathbb{Z}_6$ and

n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

$kern(\phi) \neq \{1\}$ the mapping is not 1-1, also if we take a negative

real number, it is not in the image of $\phi$ , which tells us that

$\phi$ is not surjective.

(d) Remember that $e^{ix}=\cos(x)+i\sin(x)$ , using this, it holds that

$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

that $kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}$ and

$Im(\phi)$ is the unit circle, hence $\phi$ is neither injective nor

surjective.

7 0

3 years ago

You have a recipe that indicates to use 2 parts of flour for every 5 parts of milk. You use 8 cups of flour. How much milk do yo

Drupady [299]

Answer:

20, thats what is in your little box so that should be the answer

3 0

3 years ago

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