All categories

2 years ago

The graph of f(x) = x6 – 2x4 – 5x2 + 6 is shown below.

Mathematics

2 answers:

DochEvi [55]2 years ago

5 0

Answer:

There are two rational roots for f(x)

Step-by-step explanation:

We are given a function

$f(x) = x^6-2x^4-5x^2+6$

To find the number of rational roots for f(x).

Let us use remainder theorem that when

f(a) =0, (x-a) is a factor of f(x) or x=a is one solution.

Substitute 1 for x

f(1) = 1-2-5+6=0

Hence x=1 is one solution.

Let us try x=-1

f(-1) = 1-2-5+6 =0

So x =-1 is also a solution and x+1 is a factor

We can write f(x) by trial and error as

$f(x) = (x-1)(x+1)(x^2-3)$

We find that $f(x) (x^2-3)$ factor gives two irrational solutions as

±√3.

Hence number of rational roots are 2.

Assoli18 [71]2 years ago

4 0

Answer:

B. 2

Step-by-step explanation:

got it right on the quiz

You might be interested in

Find the mean, median and mode of the weights of the people shown. 105kg 53kg 76kg 91kg 120kg 61kg 55kg 98kg 61kg

Yuliya22 [10]

First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg

For mean, you add them all up and divide by the amount of numbers (9)
720/9 = 80

For median, you find the middle number (76kg)

For mode, you find the number that appears the most (61kg)

Mean: 80
Median: 76
Mode: 61

5 0

2 years ago

Read 2 more answers

1. An AP Statistics class starts a project to estimate the average number of hours a student at their high school sleeps per nig

Vladimir [108]

Answer:

Option E

Step-by-step explanation:

This survey will not be reliable as it is chosen for convenience; the first 120 students to arrive in school on a particular morning and this collection of individuals may not be a representative of the population. The study may become biased because it does not take into account the latecomers among the students which might have been changed the study from systematically favoring certain outcomes.

5 0

2 years ago

The slope of the line below is -5/7 Write the point-slope equation of the line using the coordinates of the labeled point.

Vesna [10]

Answer:

The answer is C) y - 2 = -5/7(x - 6)

Step-by-step explanation:

Since we have the slope (-5/7) and a point (6, 2), we can just input those in for m and (x1, y1) in point-slope form.

y - y1 = m(x - x1)

y - 2 = -5/7(x - 6)

4 0

2 years ago

Read 2 more answers

the average number of spectators at a football competition for the first five days was 3144.the attendance on the sixth day was

Ksivusya [100]

Answer:

3285

Step-by-step explanation:

5x3144=15720

3990

15720+3990=19710

19710/6=3285

8 0

2 years ago

Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho

oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

7 0

2 years ago