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4 years ago

7

Can someone pls help me with proving that (f)x^2-2|x| is increasing over [1;infinity] and decreasing over [0;1] then deduce that

(f) admits a minimum to be determined

2 answers:

Zinaida [17]4 years ago

7 0

Hello :
f(x) = x²-2x if x ≥ 0 ( <span>|x | = x)
f'(x) = 2x-2
f'(x) = 0 ... x=1
f'(x) </span>< 0 .... x in : <span>]0 ; 1[<span>
</span></span> f'(x) > 0 ..... x in : ]1 ; + ∞[
f(1) is a <span> minimum : f(1) = 1² -2(1)= -1</span>

dedylja [7]4 years ago

4 0

Derivatives galore. Don't forget you might need to split the function because of the absolute value.

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Can someone please help me with this? I don't really understand it and I would really appreciate if you can show me step by step

7nadin3 [17]

Answer:

225

I used

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3 0

3 years ago

A researcher is concerned about the impact of students working while they are enrolled in classes, and she likes to know if stud

8_murik_8 [283]

Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(c) Margin of error = 0.800

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

$\bar x = 7.10$ -- sample mean

$\sigma=5$ --- sample standard deviation

$n = 150$ --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

$CI = 90\%$ --- the confidence interval

Calculate the $\alpha$ level

$\alpha = 1 - CI$

$\alpha = 1 - 90\%$

$\alpha = 1 - 0.90$

$\alpha = 0.10$

Divide by 2

$\frac{\alpha}{2} = 0.10/2$

$\frac{\alpha}{2} = 0.05$

Subtract from 1

$1 - \frac{\alpha}{2} = 1 - 0.05$

$1 - \frac{\alpha}{2} = 0.95$

From the z table. the critical value for $1 - \frac{\alpha}{2} = 0.95$ is:

$z = 1.960$

Solving (c): Margin of error

This is calculated as:

$E = z * \frac{\sigma}{\sqrt n}$

$E = 1.960 * \frac{5}{\sqrt {150}}$

$E = 1.960 * \frac{5}{12.25}$

$E = \frac{1.960 *5}{12.25}$

$E = \frac{9.80}{12.25}$

$E = 0.800$

Solving (d): The confidence interval

This is calculated as:

$CI = (\bar x - E, \bar x + E)$

$CI = (7.10 - 0.800, 7.10 + 0.800)$

$CI = (6.3, 7.9)$

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

6 0

3 years ago

Question 8.am I correct I got B

Lyrx [107]

According to the options given yes

4 0

3 years ago

What equation represents the graphed function

Marina86 [1]

Answer: $y=\dfrac{-1}{3}x+3$

Step-by-step explanation:

The equation of ab line that passes through two points (a,b) and (c,d) is given by :-

$(y-b)=\dfrac{d-c}{b-a}(x-a)$

From the given picture , the line is passing through points (0,3) and (3,2).

Then, the equation of the line will be :-

$(y-3)=\dfrac{2-3}{3-0}(x-0)$

$\Rightarrow y-3=\dfrac{-1}{3}x$

$\Rightarrow y=\dfrac{-1}{3}x+3$

Hence, the required equation of the line that represents the graphed function : $y=\dfrac{-1}{3}x+3$

7 0

4 years ago

Read 2 more answers

Given that P = (-5, 5) and Q = (-13, 10), find the component form and magnitude of 2 vector PQ.

Brut [27]

Define unit vectors along the x-axis and the y-axis as $\hat{i} , \, \hat{j}$ respectively.

Then the vector from P to Q is
$\vec{PQ} = (-13+5)\hat{i} + (10-5)\hat{j} = -8\hat{i} + 5\hat{j}$
In component form, the vector PQ is (-8,5).

The magnitude of vector PQ is
√[(-8)² + 5²] = √(89) = 9.434

Answer:
The vector PQ is (-8, 5) and its magnitude is √89 (or 9.434).

4 0

3 years ago

Read 2 more answers