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3 years ago

Tangent Line h(t) = sec t / t @ (Pi, 1/pi)

1 answer:

7 0

Take derivitive

note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of $\frac{f(x)}{g(x)} = \frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$
so

the derivitive of $\frac{sec(t)}{t} = \frac{sec(t)tan(t)t-(1)(sec(t))}{t^2}$
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
$\frac{(-1)(0)(pi)-(pi)(-1)}{pi^2}= \frac{pi}{pi^2}= \frac{1}{pi}$
slope=1/pi

use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)

y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1

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A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

2 years ago

What is the value of x? (SEE ATTACHMENT)

coldgirl [10]

The answer should be A. 4.5

Because line x is just over half of the length of the line that is 8ft

6 0

3 years ago

The mean of birth weights for individual babies in the population is 3,500 grams. What is the mean birth weight for the sampling

Gala2k [10]

Answer:

The mean birth weight for the sampling distribution is

3,500 grams.

Step-by-step explanation:

The sample mean is the average of the sample values collected divided by the number of the samples, while the population mean is the average or mean of all the values in the population. If the sample is random and the sample size is large enough, then the sample mean would be a good estimator of the population mean. This implies that with a randomly distributed and unbiased sample size, the sample mean and population mean will be equal, according to the central limit theorem. Therefore, the mean of the sample means will always approximate the population mean.

8 0

2 years ago

Find the point below that lies on the line y-5=7(x-6).

MrMuchimi

Answer:

(6,5)

Step-by-step explanation:

The equation is in point slope form: $y-y_1=m(x-x_1)$

'm' is the slope.

(x1, y1) is the point used to create the equation.

The number 6 replaces the x1 spot.

The number 5 replaces the y1 spot.

So a point that lies on the line should be (6,5).

5 0

2 years ago