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3 years ago

Select the correct answer.

Mathematics

2 answers:

eimsori [14]3 years ago

7 0

Answer:

Step-by-step explanation:

Given that

f(x)=-2x – 1

g(x) = -1/2x+1/2

I The function f( g(x)) = x for all real x.

-2(-1/2X + 1/2) - 1

X - 1 - 1

X - 2

II. The function glAx)) = x for all real x.

- 1/2( -2X - 1 ) + 1/2

X + 1/2 + 1/2

X + 1

III. Functions fand g are inverse functions

Inverse of f(x)

Y = -2x - 1

X = - 2Y - 1

- 2Y = X + 1

Y = - 1/2X - 1/2

Therefore, none of the above answers is correct.

The correct answer is therefore D

Ira Lisetskai [31]3 years ago

4 0

Answer:

In photo below

Explanation:

In photo below

I hope this helps

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a_sh-v [17]

Answer:

Step-by-step explanation:

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8 0

3 years ago

If f(x) = –5x – 4 and g(x) = –3x – 2, find (f – g)(x).

erica [24]

Answer:

C. (f – g)(x) = –2x – 2

Step-by-step explanation:

(f – g)(x) = f(x) – g(x)

= (–5x – 4) – (–3x – 2)

= –2x – 2

3 0

3 years ago

The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.

Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5 {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean MCAT score = 32.2

s = sample standard deviation = 4.2

n = sample of students = 40

So, the test statistics = $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$ ~ $t_3_9$

= 4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

3 0

3 years ago

The age of trees in a forest is known to be approximately normal with an average age of 40 years and a standard deviation of 10

Keith_Richards [23]

Answer: D. -0.6

Step-by-step explanation:

Z score formula= x-mean/ SD

34-40/10= -3/5= -0.6

6 0

3 years ago

What are the turning points of the function f(x)=x^3+9x^2+24x+16?

Lana71 [14]

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3 0

4 years ago