(x) = arcsec(x) − 8x
f'(x) = d/dx( arcsec(x) −
8x )
<span> 1/xsqrt( x^2 - 1) - 8</span>
f'(x) = 0
1/xsqrt( x^2 - 1) - 8 = 0
8 x sqrt (x^2-1) = 1
<span> ( 8 x sqrt (x^2-1) )^2 = 1</span>
64 x^2 ( x^2 - 1) = 1
64 x^4 - 64 x^2 =1
64 x^4 - 64 x^2 - 1 = 0
x = 1.00766 , - 1.00766
<span> x = - 1.00766</span>
f(- 1.00766) = arcsec(-
1.00766) − 8( - 1.00766)
f( - 1.00766 ) = 11.07949
x = 1.00766
f(1.00766) =
arcsec(1.00766) − 8( 1.00766)
f(1.00766 ) = -7.93790
relative maximum (x, y) =
(- 1.00766 , 11.07949 ) relative minimum (x, y) = ( 1.00766 ,
-7.93790 )
X - 9 + 2wx = y Add 9 to both sides
x + 2wx = y + 9 Factor out the x
x (1 + 2w) = y + 9 Divide both sides by (1 + 2w)
x = (y + 9) / (1 + 2w)
omg theres a chipmunk who entered our school gym and its running around
everywhere its currently under the bleachers
The margin of error of a given statistic is an amount that is allowed for in case of miscalculation or change of circumstances.
It is usually the radius or half of the width of the confidence interval of that statistic.
Given that a<span>
survey of the students in Lance’s school found that 58% of the
respondents want the school year lengthened, while 42% think it should
remain the same. The margin of error of the survey is ±10%.
This means that 58% </span><span>± 10% of the </span>respondents want the school year lengthened, while 42% <span><span>± 10% think it should
remain the same.</span>
Thus, from 48% to 68% </span><span><span>of the respondents want the school year lengthened, while from 32% to 52% <span>think it should
remain the same.</span> </span>
Therefore, according to
the survey data, at least 32% of students want the duration of the school
year to remain unchanged, and at least 48% want the school year to be
lengthened.</span>
Answer: C. Significant at 0.036
Step-by-step explanation:
Given:
Number of selected samples Ns= 500
Number of airplane that arrive on time Na = 482.
Number of airplane that arrive late Nl = 500 - 482 = 18
The probability that an airplane arrive late:
P(L) = Nl/Ns
P(L) = 18/500
P(L) = 0.036
Interpret an event as significant if its probability is less than or equal to 0.05.
Since P(L) < 0.05
P(L) = Significant at 0.036
