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Galina-37 [17]
3 years ago
10

Please help me. if you can help me answer multiple questions only 3. offering 30 points for it

Mathematics
1 answer:
lubasha [3.4K]3 years ago
7 0

Answer:

(-9, -5)

Step-by-step explanation:

Ok, so when you move an image to the right, you are moving along the x-axis, and when you move up, you are moving up the y-axis. So if the altered image is (x,y) and the values are (-5, -1), you reverse what has been done to the image. In this case, since we moved to the right 4 units, we know that means we added 4 to x, so we subtract 4 to get -9. And then, for the y-value, because we added 4, we do the opposite, and subtract 4 to get -5. So the pre-image should be (-9, -5)

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A multiple-choice examination has 20 questions, each with five possible answers, only one of which is correct. Suppose that one
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Answer:

The probability that the student answers at least seventeen questions correctly is 8.03\times 10^{-10}.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of correctly answered questions.

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Then the probability of selecting the correct option is,

P(X)=p=\frac{1}{5}=0.20

There are <em>n</em> = 20 question in the exam.

It is also provided that a student taking the examination answers each of the questions with an independent random guess.

Then the random variable can be modeled by the Binomial distribution with parameters <em>n</em> = 20 and <em>p</em> = 0.20.

The probability mass function of <em>X</em> is:

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Compute the probability that the student answers at least seventeen questions correctly as follows:

P(X\geq 17)=P (X=17)+P (X=18)+P (X=19)+P (X=20)

=\sum\limits^{20}_{x=17}{{20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x}}\\\\=0.00000000077+0.000000000032+0.00000000000084+0.000000000000042\\\\=0.000000000802882\\\\=8.03\times10^{-10}

Thus, the probability that the student answers at least seventeen questions correctly is 8.03\times 10^{-10}.

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