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jeyben [28]
3 years ago
12

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

at all, and decides to randomly guess the answers. Find the probabilities of each of the following events: (a) the first question she gets right is the 3rd question
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

We have 20 category Questions split into 5

Where we would assume that the 3rd question is simply 1/4 + 5/20 chance.

This is not a determiner as Robin is asked more questions.

So it is simply 1/4 and kept to factors that Robin will answer one of the 4 choices upon the 3rd question as 1/4 regardless of other questions being asked. Where Robins first answer would be the third question we subtract 4/5 = 16/20 - 5/20= 11/40. Which can only be rounded down to 2 decimal places 0.275

Step-by-step explanation:

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The exponential function has the form y = a(b^x) where a is the starting value or y-intercept. B is the rate of change and x is the variable typically time. In this situation, y = 4(3^{-x}) has starting value 4, rate 3 and a -x.

Normally an exponential curve glides left to right growing steeper as it goes. But a -x flips this behavior. The curve starts very steep then gradually slows down. This means only graphs B or D are options. Since only graph B of these two has a y-intercept at 4, graph B is the solution.

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explanation

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Not sure how to do these. i need help with 13! needing it to be reduced just like y should be 8
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Annie’s change purse contained 5 quarters 8 dimes and 8 nickels if Annie chose 2 coin randomly without replacing them in her pur
Bogdan [553]
<h3>Answer:</h3>

2/15

<h3>Explanation:</h3>

There are 8C2 = 28 ways to choose 2 dimes from the 8 dimes in Annie's purse. There are 21C2 = 210 ways to choose 2 coins from the 21 coins in Annie's purse.

Of the 210 ways to choose 2 coins, 28 of the choices will result in 2 dimes being chosen. The probability of choosing 2 dimes is 28/210 = 2/15.

_____

<em>Comment on nCk</em>

The number of ways to choose k objects from n, when order does not matter, is ...

... n!/(k!(n -k)!)

For the computations above, we have ...

... 8C2 = 8·7/(2·1) = 28

... 21C2 = 21·20/(2·1) = 210

8 0
3 years ago
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