All categories

3 years ago

Help me answer question 16 and 18. I already know how to do question 17.

Mathematics

1 answer:

irakobra [83]3 years ago

6 0

Bro what is the question can’t answer if I don’t know the question

You might be interested in

(accounting)

Blizzard [7]

Answer:

Step-by-step explanation:

expense is usually done everywhere.

8 0

3 years ago

Help meeeee please you get some brainss

Setler [38]

Hi there!

For number 1 it shows that the last two blanks would be 8/8 and 4/4 since adding 1/8 would be 8/8 and adding 1/4 would be 4/4.

For number 2, it would be 3/6 or a half. It wants to have another equivalent to 3/6 and with dividing 3 for each side, it would be 1/2.

7 0

3 years ago

Read 2 more answers

According the the U.S. Department of Education, full-time graduate students receive an average salary of $15,000 with a standard

Firdavs [7]

Answer:

Step-by-step explanation:

Given that :

population Mean = 15000

standard deviation= 1200

sample size n = 100

sample mean = 16000

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o : \mu = 15000 }\\ \\ \mathtt{H_1 : \mu > 15000}$

Using the standard normal z statistics

$z = \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}$

$z = \dfrac{16000 -15000}{\dfrac{1200 }{\sqrt{100}}}$

$z = \dfrac{1000}{\dfrac{1200 }{10}}$

$z = \dfrac{1000\times 10}{1200}$

z = 8.333

degree of freedom = n - 1 = 100 - 1 = 99

level of significance ∝ = 0.05

P - value from the z score = 0.00003

Decision Rule: since the p value is lesser than the level of significance, we reject the null hypothesis

Conclusion: There is sufficient evidence that the Dean claim for his graduate students earn more than average salary of $15,000

5 0

3 years ago

Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.

jek_recluse [69]

Answer:

$\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

Step-by-step explanation:

Let $X$ be a set of all students in your class. The set $X$ is the domain. Denote

$C(x) - ' \text{$x $ has a cat}'\\D(x) - ' \text{$x$ has a dog}'\\F(x) - ' \text{$x$ has a ferret}'$

$\mathbf{a)}$

Consider the statement 'A student in your class has a cat, a dog, and a ferret'. This means that $\exists x \in X$ so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)$

$\mathbf{b)}$

Consider the statement 'All students in your class have a cat, a dog, or a ferret.' This means that $\forall x \in X$ at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)$

$\mathbf{c)}$

Consider the statement 'Some student in your class has a cat and a ferret, but not a dog.' This means that $\exists x \in X$ so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)$

$\mathbf{d)}$

Consider the statement 'No student in your class has a cat, a dog, and a ferret..' This means that $\forall x \in X$ none of the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

$\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)$

$\mathbf{e)}$

Consider the statement ' For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.'

This means that for each of the statements C, F and D there is an element from the domain $X$ so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

$\left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)$

5 0

4 years ago

Help? Please. I really need help

tia_tia [17]

Answer:

Step-by-step explanation:

Any other product in Quadrant IV will be negative also, so Quadrant IV is part of my answer. The points (x, y), with xy < 0, lie in Quadrants II and IV.

3 0

3 years ago