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3 years ago

The function g is given in three equivalent forms. Which form most quickly reveals the vertex?

Mathematics

1 answer:

Nadusha1986 [10]3 years ago

8 0

Answer:

Option A. $g(x)= \frac{1}{2}(x-8)^2-8$

Step-by-step explanation:

<u><em>The complete question is</em></u>

The function g is given in three equivalent forms.

Which form most quickly reveals the vertex?

A)g(x)= 1/2(x-8)^2-8

B)g(x)= 1/2(x-12)(x-4)

C)g(x)= 1/2x^2-8x+24

we know that

The equation of a quadratic function in vertex form is equal to

$y=a(x-h)^2+k$

where

a is the leading coefficient

(h,k) is the vertex of the function

In this problem, the option A is the equation of the function in vertex form

we have

$g(x)= \frac{1}{2}(x-8)^2-8$

where the vertex is the point (8,-8)

therefore

Option A

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Answer:

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Step-by-step explanation:

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3 years ago

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The answer is:

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3 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

3 years ago