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nexus9112 [7]
4 years ago
11

Integrate Sec (4x - 1) Tan (4x - 1) Dx ​

Mathematics
1 answer:
Delicious77 [7]4 years ago
5 0

\bf \displaystyle\int~sec(4x-1)tan(4x-1)dx \\\\[-0.35em] ~\dotfill\\\\ sec(4x-1)tan(4x-1)\implies \cfrac{1}{cos(4x-1)}\cdot \cfrac{sin(4x-1)}{cos(4x-1)}\implies \cfrac{sin(4x-1)}{cos^2(4x-1)} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int~\cfrac{sin(4x-1)}{cos^2(4x-1)}dx \\\\[-0.35em] ~\dotfill\\\\ u=cos(4x-1)\implies \cfrac{du}{dx}=-sin(4x-1)\cdot 4\implies \cfrac{du}{-4sin(4x-1)}=dx \\\\[-0.35em] ~\dotfill

\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C

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D) 4x + 5y = -30 = -4/5

Step-by-step explanation:

A) 5x + 4y = 28 slope = -5/4

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2 years ago
You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
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Answer:

a) No

b) 42%

c) 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c)   P(win first game)  = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

Standard deviation \sigma=\sqrt{variance} = \sqrt{0.3844}=0.62

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The dependent t-test (also known as the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference between these two means.

We use the paired t-test if we have two measurements on the same item, person or thing. We should also use this test if we have two items that are being measured with a unique condition.

For instance, an experimenter tests the effect of a medicine on a group of patients before and after giving the doses.

In this case, Kristen assesses the students' knowledge of French vocabulary at the start of the semester and then again at the end of the semester.

So, she collects data before and after the semester for vocabulary learning in introductory-level French class.

Thus, she is using a paired <em>t</em>-test to analyze whether the French students significantly increased their knowledge of French vocabulary.

The hypothesis is defined as:

<em>H₀</em>: There is no difference between the two means, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a significant difference between the two means, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The test statistic is:

t =  Sample mean difference ÷ (Standard deviation of difference/ Sample size)

Thus, Kristen should use the paired <em>t</em>-test.

7 0
3 years ago
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