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igor_vitrenko [27]
3 years ago
12

A point with a positive x-coordinate and a negative y-coordinate is reflected over the y-axis.

Mathematics
2 answers:
icang [17]3 years ago
6 0

Answer:

The x-coordinate is negative, and the y-coordinate is negative.

Step-by-step explanation:

(x, -y)

When you reflect over the y-axis, the x-coordinate becomes negative:

(-x, -y)

goblinko [34]3 years ago
3 0
We need a photo or something
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Need the slope of this line as a fraction
olchik [2.2K]

Answer:

The slope of the line is -5/3.

6 0
3 years ago
Consider an experiment that consists of recording the birthday for each of 20 randomly selected persons. Ignoring leap years, we
8_murik_8 [283]

Answer:

a)  p_{20d} = 0.588

b) 23

c) 47

Step-by-step explanation:

To find a solution for this question we must consider the following:

If we’d like to know the probability of two or more people having the same birthday we can start by analyzing the cases with 1, 2 and 3 people

For n=1 we only have 1 person, so the probability  p_{1} of sharing a birthday is 0 (p_{1}=0)

For n=2 the probability p_{2} can be calculated according to Laplace’s rule. That is, 365 different ways that a person’s birthday coincides, one for every day of the year (favorable result) and 365*365 different ways for the result to happen (possible results), therefore,

p_{2} = \frac{365}{365^{2} } = \frac{1}{365}

For n=3 we may calculate the probability p_{3} that at least two of them share their birthday by using the opposite probability P(A)=1-P(B). That means calculating the probability that all three were born on different days using the probability of the intersection of two events, we have:

p_{3} = 1 - \frac{364}{365}*\frac{363}{365} = 1 - \frac{364*363}{365^{2} }

So, the second person’s birthday might be on any of the 365 days of the year, but it won’t coincide with the first person on 364 days, same for the third person compared with the first and second person (363).

Let’s make it general for every n:

p_{n} = 1 - \frac{364}{365}*\frac{363}{365}*\frac{362}{365}*...*\frac{(365-n+1)}{365}

p_{n} = \frac{364*363*362*...*(365-n+1)}{365^{n-1} }

p_{n} = \frac{365*364*363*...*(365-n+1)}{365^{n} }

p_{n} = \frac{365!}{365^{n}*(365-n)! }

Now, let’s answer the questions!

a) Remember we just calculated the probability for n people having the same birthday by calculating 1 <em>minus the opposite</em>, hence <em>we just need the second part of the first calculation for</em> p_{n}, that is:

p_{20d} = \frac{364}{365}*\frac{363}{365}*\frac{362}{365}*...*\frac{(365-20+1)}{365}

We replace n=20 and we obtain (you’ll need some excel here, try calculating first the quotients then the products):

p_{20d} = 0.588

So, we have a 58% probability that 20 people chosen randomly have different birthdays.

b) and c) Again, remember all the reasoning above, we actually have the answer in the last calculation for pn:

p_{n} = \frac{365!}{365^{n}*(365-n)! }

But here we have to apply some trial and error for 0.50 and 0.95, therefore, use a calculator or Excel to make the calculations replacing n until you find the right n for p_{n}=0.50 and p_{n}=0.95

b) 0.50 = 365!/(365^n)*(365-n)!

n           p_{n}

1              0

2           0,003

3           0,008

….           …

20           0,411

21           0,444

22           0,476

23           0,507

The minimum number of people such that the probability of two or more of them have the same birthday is at least 50% is 23.

c) 0.95 = 365!/(365^n)*(365-n)!

We keep on going with the calculations made for a)

n             p_{n}

…                …

43            0,924

44            0,933

45            0,941

46            0,948

47            0,955

The minimum number of people such that the probability of two or more of them have the same birthday is at least 95% is 47.

And we’re done :)

6 0
4 years ago
What is the quotient of (x3 - 3x2 + 5x – 3) = (x - 1)?
IRINA_888 [86]

Note: Consider "÷" sign instead of "=".

Given:

(x^3-3x^2+5x-3)\div (x-1)

To find:

The quotient.

Solution:

We have,

(x^3-3x^2+5x-3)\div (x-1)

It can be written as

=\dfrac{x^3-3x^2+5x-3}{x-1}

Splitting the middle terms, we get

=\dfrac{x^3-x^2-2x^2+2x+3x-3}{x-1}

=\dfrac{x^2(x-1)-2x(x-1)+3(x-1)}{x-1}

Taking out the common factor (x-1), we get

=\dfrac{(x-1)(x^2-2x+3)}{x-1}

Cancel the common factors.

=x^2-2x+3

The quotient is x^2-2x+3.

Therefore, the correct option is D.

7 0
3 years ago
Read 2 more answers
Find the distance between these points.<br> C(0, 4), T(-6, -3)
beks73 [17]

Answer:

distance=\sqrt{85} =9.22

Step-by-step explanation:

We can use the general distance formula between any two points (x_1,y_1) and (x_2,y_2) on the plane given by:

distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\

We simply identify (x_1,y_1)  with (0, 4)  and  (x_2,y_2)  with  (-6, -3) , thus obtaining:

distance=\sqrt{(-6-0)^2+(-3-4)^2}\\distance=\sqrt{(-6)^2+(-7)^2} \\distance=\sqrt{36+49} \\distance=\sqrt{85} =9.22

3 0
4 years ago
Catherine has a number cube that has 6 faces that are numbered from 1 through 6. She plans on rolling the number cube 60 times.
Leona [35]
Greater than 4 means either 5 OR 6, that means the probability of getting 5 or 6 is 2/6 ==>1/3

Theoretically she needs 3 Rolls to get a number greater than 4 
8 0
3 years ago
Read 2 more answers
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