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eimsori [14]
3 years ago
7

E^(x/2) /(e^(x/3) +1)

Mathematics
1 answer:
nevsk [136]3 years ago
8 0

\int e^{x/2} /(e^{x/3} +1) dx

start with substituting z=e^x

z=e^x\\dz=e^x dx\\\int \frac{dz}{z^{5/6}+z^{1/2}}

then substitute u = z^(1/6):

u=z^{1/6}\\dz=6 z^{5/6}du\\6 \int \frac{u^2}{u^2+1}du

that integral has a standard form that can be looked up in integral tables, it has the following solution:

6(u - \tan^{-1} u) + \mbox{constant}

substituting back the the variable z and then x you get the final solution:

6 e^{x/6} - 6 \tan^{-1}(e^{x/6}) + \mbox{constant}

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Plot 4/7 on the number line
erik [133]

Answer:

0.57, which is a number between 0 and 1

Step-by-step explanation:

The fraction 4/7 is approximately equal to 0.57 as a decimal, so you would plot 4/7 where 0.57 should go

8 0
3 years ago
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What’s 690 divided by 4???
poizon [28]
172.5! hope this helps
6 0
2 years ago
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Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the
Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million)     Number of companies

25 up to 35                                                    4

35 up to 45                                                    19

45 up to 55                                                    27

55 up to 65                                                    16

65 up to 75                                                     9

TOTAL                                                            75

Let's find the probabilities first:

P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}

For 35 up to 45

P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}

For 45 up to 55

P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}

For 55 up to 65

P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}

For 65 up to 75

P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}

Chi-Square Table can be computed as follows:

Expense   No of   Probabilities(P)  Expe                (O-E)^2   \dfrac{(O-E)^2}{E}

             compa                                 cted E (n*p)

             nies (O)  

25-35           4      0.0612    75*0.0612 = 4.59        0.3481       0.0758

35-45           19     0.2218   75*0.2218 = 16.635     5.5932       0.3362

45-55           27     0.3568   75*0.3568 = 26.76     0.0576      0.021

55-65           16      0.2552   75*0.2552 = 19.14      9.8596      0.5151

65-75           9      0.0839     75*0.0839 = 6.2925   7.331         1.1650

                                                                                           \sum \dfrac{(O-E)^2}{E}= 2.0492                                                                                                      

Using the Chi-square formula:

X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square  \ X^2 = 2.0942

Null hypothesis:

H_o: \text{The population of advertising expenses follows a normal distribution}

Alternative hypothesis:  

H_a: \text{The population of advertising expenses does not follows a normal distribution}

Assume that:

\alpha = 0.02

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from X^2 = 11.667

Decision rule: To reject H_o  \  if \  X^2  test statistics is greater than X^2 tabulated.

Conclusion: Since X^2 = 2.0942 is less than critical value 11.667. Then we fail to reject H_o

6 0
3 years ago
Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2
Arturiano [62]

Answer:

D

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0
3 years ago
Henry buys a softball for 16 dollars plus a 9% tax Ella buys a pair of shoes for 17 dollars plus a 9% tax. Enter the difference
Gwar [14]

Answer:

$1.09

Step-by-step explanation:

16*0.09 = 1.44, 16 + 1.44 = 17.44 (Henry's total cost)

17*0.09 = 1.53, 17 + 1.53 = 18.53 (Ella's total cost)

To find the difference we subtract 18.53-17.44 = 1.09 and so the difference including tax is $1.09

8 0
3 years ago
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