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4 years ago

X + 8 < 10 Please help

Mathematics

1 answer:

taurus [48]4 years ago

8 0

In this question, you're solving the inequality by solving for x.

Solve for x:

x + 8 < 10

subtract 8 from both sides

x < 2

Answer:

x < 2

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liq [111]

Answer:

Integration of I= $\int {x^{n}logx \, dx$ = $[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]$

Step-by-step explanation:

Given integral is I= $\int {x^{n}logx \, dx$

Take logx=t

$x=e^{t}$

$x^{n}=e^{nt}$

$\frac{1}{x} dx=dt$

$dx=xdt$

$dx=e^{t}dt$

I= $\int (e^{nt})(t)(e^{t})\, dt$

I= $\int (e^{(n+1)t})(t)\, dt$

Using integration by part,

$I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]$

Writing in terms of x

I= $[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]$

I= $[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]$

I= $[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]$

I= $[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]$

Thus,

Integration of I= $\int {x^{n}logx \, dx$ = $[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]$

3 0

3 years ago

A set of data has a normal distribution with a mean of 5.1 and standard deviation of 0.9. Find the percent of data between 6.0 a

jolli1 [7]

The percent data between 6.0 and 6.9 would be 13.59%. Here’s the complete solution for this specific problem.

z = (X - Mean)/SD 
z1 = (6 - 5.1)/09 = + 1 
z2 = (6.9 - 5.1)/09 = + 2 
Required Percent = P(6 < X < 6.9)*100 
= P(1 < z < 2)*100 
= [P(z < 2) - P(z < 1)]*100 
= [0.9772 - 0.8413]*100 
= 0.1359*100 
= 13.59%

I am hoping that this has answered your query.

5 0

4 years ago

Enter the domain and range of the function g(x) = 9x2 − 5 as an inequality, using set notation and using interval notation. Comp

tia_tia [17]

The domain and the range of a function are the set of input and output values, the function can take.

The domain and the range of $g(x) = 9x^2 - 2$ is $(-\infty, \infty)$ .
The parent function $f(x) =x^2$ is vertically compressed by 9, then shifted down by 5 units to get $g(x) = 9x^2 - 2$