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3 years ago

What can 0.2 be rounded off to?

2 answers:

8 0

0 because two is less than five which means that your rounding downwards not upwards like if the number was 0.5 you would round that to 1

Nesterboy [21]3 years ago

6 0

The only possible thing is 0 which is being rounded to the nearest whole number.

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Alexis is trying to run a certain number of miles by the end

LuckyWell [14K]

Answer:

30 miles

Step-by-step explanation:

12 is to 40 as X is to 100

So 40X=12X100

40X=1200

Divide 40 from both sides

X=30

3 0

3 years ago

Help me please I'm on a time limit

olga_2 [115]

The original price times the off price to get how much is the discount 3.25*0.25=0.8125 Subtract the discount from the original price 3.25-0.8125=2.4375=2.44 Or 3.25*0.75=2.4375=2.44

7 0

3 years ago

800 divided by 2 ( 100 points & I give brainliest )

nordsb [41]

Answer:

800÷2=400

Step-by-step explanation:

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6 0

3 years ago

Read 2 more answers

A certain circuit board consists of two resistors, green and red. The circuit board manufacturer has two huge bins filled with t

Lesechka [4]

Answer:

(a) 0.675

(b) 0.3

Step-by-step explanation:

$P(R) = 90\% = 0.9$ (Probability of functional red)

$P(\sim R) = 90\% = 1 - 0.9 = 0.1$ (Probability of non-functional red)

$P(G) = 75\% = 0.75$ (Probability of functional green)

$P(\sim G) = 1 - P(G) = 1 - 0.75 = 0.25$ (Probability of non-functional green)

(a) The probability the board is functional is the probability that the red and the green resistors are functional.

$P(\text{board is functional}) = P(R\cap G) = P(R) \times P(G) = 0.9\times0.75 = 0.675$

(b) The probability that exactly one of the resistors chosen is functional is the probability that either red is functional and green is not or green is functional and red is not.

$P(\text{exactly one is functional}) = P((R\cap \sim G) \cup (G\cap \sim R))$

$P(\text{exactly one is functional}) = P(R\cap \sim G) + P(G\cap \sim R) = (P(R) \times P(\sim G)) + (P(G) \times P(\sim R))$

$P(\text{exactly one is functional}) = (0.9 \times 0.25) + (0.75 \times 0.1) = 0.225 + 0.075 = 0.3$

6 0

4 years ago

Read 2 more answers

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