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4 years ago

I need an explanation of this pleasee

1 answer:

8 0

Option A is wrong because,

only when p(5)=0 , x-5 can be a factor of p(x)
but the question says P(3)=-2

option B is wrong because,

only when p(2)=0 , x-2 can be a factor of p(x)
but the question says P(3)=-2

option C is wrong because,

only when p(-2)=0 , x+2 can be a factor of p(x)
but the question says P(3)=-2

option D is correct because,

only when p(3)=0 , x-3 can be a factor of p(x)
here p(x) is not equal to 0

therefore, p(x)=0+(-2)=-2
so when p(x) is divided by x-3 it will leave a reminder -2

hope it helps!!

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Explanation:

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3 years ago

Meagan, a contractor, is hired to install cabinets in a new home and has to decide whether to build the cabinets herself or purc

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Meagan should build the cabinets herself only if she has a comparative advantage in the production of cabinets.

<h3>What is comparative advantage?</h3>

Comparative advantage can be defined as ability of an individual or business firm to produce goods and services at a lower opportunity cost than their rivals or trade partners.

The most important benefit of comparative advantage is that, it gives an individual or business firm the ability to sell their finished goods and services at a lower price in comparison with its close rivals (competitors) or trade partners. Thus, comparative advantage results in stronger sales margins for an individual or business firm.

In conclusion, Meagan should build the cabinets herself only if she has a comparative advantage in the production of cabinets.

Read more on comparative advantage here: brainly.com/question/14846093

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2 years ago

Consider a set of cards that has four cards labeled 1, 2, 3, and 4. Suppose you pick two cards, without replacement, to obtain t

Phantasy [73]

Answer:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

Explanation:

Given

$Cards = \{1,2,3,4\}$

Required

The sampling distribution

The possible selection of 2 cards without replacement is as follows:

$S = \{(1,2) (1,3) (1,4) (2,1) (2,3) (2,4) (3,1) (3,2) (3,4) (4,1) (4,2) (4,3)\}$

Calculate the mean

$\begin{array}{cccccccccccc}{Selection} & {(1,2)} & {(1,3)} & {(1,4)} & {(2,1)} & {(2,3)} & {(2,4)}& {(3,1)} & {(3,2)} & {(3,4)} & {(4,1)} & {(4,2)} & {(4,3)} \ \\ {Mean} & {1.5} & {2} & {2.5} & {1.5} & {2.5} & {3} & {2} & {2.5} & {3.5} & {2.5} & {3} & {3.5}\ \end{array}$

List out the mean and the respective frequency

$1.5 \to 2$

$2 \to 2$

$2.5 \to 4$

$3 \to 2$

$3.5\to 2$

$Total \to 12$

Calculate the probability of each mean

$P(1.5) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2.5) \to \frac{4}{12} \to \frac{1}{3}\\$

$P(3) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(3.5) \to \frac{2}{12} \to \frac{1}{6}$

So, the table of sampling distribution is:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

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3 years ago

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MrRa [10]

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2 years ago

Which statement best explains the relationship between lines cd and fg?.

Bezzdna [24]

Answer:

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Explanation:

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3 years ago