WILL MARK BRAINLIEST AMD THANKS

Social Studies

2 answers:

snow_tiger [21]3 years ago

7 0

Answer:

teaching right or wrong, food and clothes

Explanation:

PIT_PIT [208]3 years ago

6 0

I think it would be shelter, clothing, and food

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Who chooses the prime minister

solniwko [45]

The Queen does since she is in charge of the kingdom

4 0

4 years ago

That part of the whole audience a speaker most wants to persuade is called the Group of answer choices specific audience. design

alexira [117]

The target audience is the that part of the whole audience a speaker most wants to persuade.

<h3>Who are the target audience?</h3>

In the analysis of audience, the target audience are those individuals that the speaker actually prepared the speech for. That is, they are the main focus for the speech.

To be able to persuade your target audience the following steps should be followed:

Define your target audience,

Create Useful and Relevant Content, and

Use Targeted Advertisements.

Therefore, the target audience is the that part of the whole audience a speaker most wants to persuade.

Learn more about speech here:

brainly.com/question/26157848

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3 0

2 years ago

26. A standard deck of playing cards consists of 52 cards, half of which are black and half of which are red. If three cards are

nlexa [21]

Answer:

1/4

Explanation:

The probability of getting one black card out of a 52 deck of cards (where 26 are red and 26 are black) is: $\frac{26}{52}$ , <u>(remember that probability of an event = (total amount of favorable outcomes/ total amount of outcomes) </u>by simplifying this fraction we get: $\frac{26}{52} = \frac{13}{26}=\frac{1}{2}$ .

Therefore, the probability of getting one black card out of a deck of 52 cards is 1/2.

Since the way of picking cards is done with replacement, for the <u>second card,</u> if we want to get another black one we would have a probability of <u>1/2.</u>

Again, for the <u>third card we would still have a probability of 1/2 of getting a black one.</u>

We want to get a black one on the first time AND on the second AND on the third one, therefore we have to use the rule of multiplication, which tells us we have to multiply all three probabilities:

P(having 3 black cards) = $(\frac{1}{2}) (\frac{1}{2})(\frac{1}{2})= \frac{1}{8}$

Applying the same way of thinking that we used before but now for the red cards (which are also 26) we have that the probability of getting one red card is 1/2

Thus,

P(having 3 red cards) = $(\frac{1}{2}) (\frac{1}{2})(\frac{1}{2})= \frac{1}{8}$

Since the problem is asking us to get either 3 black cards OR 3 red cards, we are going to use the rule of sum and we're going to sum up both probabilities.

P(having 3 black cards OR 3 red cards) = $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$