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svp [43]
3 years ago
13

I need help please.

Mathematics
1 answer:
Sidana [21]3 years ago
7 0

Answer:

-1/63y^11

Step-by-step explanation:

because we multiply it

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300 students in a high school freshman class are surveyed about what kinds of pets they have. Of the 300 students, 200 have a do
katrin2010 [14]

This Question is not complete

Complete Question:

300 students in a high school freshman class are surveyed about what kinds of pets they have. Of the 300 students, 200 have a dog, 180 have a cat, and 150 have a cat and a dog. Using this information, answer each of the following questions, Let D be the event that a randomly selected student has a dog and C be the event that a randomly selected student has a cat.

a . What is P(D), the probability that a student in the class has a dog?

b. What is P(C), the probability that a student in the class has a cat?

c. What is P(D and C), the probability that a student in the class has a dog and a cat?

d. What is P(D or C), the probability that a student in the class has a dog or a cat?

Answer:

a. 2/3

b) 3/5

c) 1/2

d) 19/ 15 = 1 4/15

Step-by-step explanation:

The formula for probability =

Number of possible outcomes ÷ Number of events.

Total number of Students = 300

Number of Students that have a dog = 200

Number of Students that have a cat = 180

Number of Students that have a dog and cat = 150

a. What is P(D), the probability that a student in the class has a dog?

In the question, the given data is interpreted as :

Number of possible outcomes = 200

Number of events = 300

P(D) = 200/300

= 2/3

Therefore, the probability that a student has a dog = 2/3

b. What is P(C), the probability that a student in the class has a cat?

In the question, the given data is interpreted as :

Number of possible outcomes = 180

Number of events = 300

P(D) = 180/300

= 3/5

Therefore, the probability that a student has a cat = 3/5

c. What is P(D and C), the probability that a student in the class has a dog and a cat?

In the question, the given data is interpreted as :

Number of possible outcomes = 150

Number of events = 300

P(D) = 150/300

= 1/2

Therefore, the probability that a student has a dog and cat = 1/2

d. What is P(D or C), the probability that a student in the class has a dog or a cat?

The is calculated as

P(D) + P(C)

In the above question,

P(D) is calculated as 2/3

P(C) is calculated as 3/5

Therfore, P(D or C), the probability that a student in the class has a dog or a cat is calculated as:

= 2/3 +3/5

= 19/15 = 1 4/15

7 0
3 years ago
Please someone help me
matrenka [14]

Answer:

B should be correct

Step-by-step explanation:

Please mark me as brainliest :)

5 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS. I suck at age problems
ella [17]

H + 4 = J

4 + 2 =J

J = 6

3 0
3 years ago
Read 2 more answers
3. Write an equation for the line perpendicular to y = 6x - 5 that contains (2, 10)
Artemon [7]

<u>Answer:</u>

y = -\frac{1}{6}x +\frac{31}{3}

<u>Step-by-step explanation:</u>

We are to find the equation for the line which is perpendicular to another line with the equation y = 6x - 5 and passes through a point (2, 10).

Since the slope of a line which is perpendicular to another line is the negative reciprocal of the other line, therefore the slope of our line will be -\frac{1}{6}.

Finding the y-intercept:

y=mx+c

10=-\frac{1}{6} (2)+c

c=\frac{31}{3}

Therefore, the equation of the line will be y = -\frac{1}{6}x +\frac{31}{3}.

6 0
4 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id
dybincka [34]

i) The given function is

f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}

The domain is

(x-5)(x+4)^2\ne 0

(x-5)\ne0,(x+4)^2\ne 0

x\ne5,x\ne -4

ii) For vertical asymptotes, we simplify the function to get;

f(x)=\frac{(2x+1)}{(x+4)^2}

The vertical asymptote occurs at

(x+4)^2=0

x=-4

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

2x+1=0

2x=-1

x=-\frac{1}{2}

iv) To find the y-intercept, we substitute x=0 into the reduced fraction.

f(0)=\frac{(2(0)+1)}{(0+4)^2}

f(0)=\frac{(1)}{(4)^2}

f(0)=\frac{1}{16}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0

The horizontal asymptote is y=0.

vi) The function has a hole at x-5=0.

Thus at x=5.

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

6 0
3 years ago
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