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2 years ago

Parametric test based on standard error

Mathematics

1 answer:

Delvig [45]2 years ago

8 0

Answer:

The parametric hypothesis tests that we introduced in Chapter 5 were all based upon the means of the sample and population.

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Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

(p+q)^5 do that for 100 points

jonny [76]

Answer:

p^5 + q^5

Step-by-step explanation:

(p + q)^5

p^5 + q^5 Distribute the fifth power to the letters in the paranthesis

5 0

3 years ago

Read 2 more answers

The total weight of three tables is 16.9 pounds. The first table is twice as heavy as the second table and the weight of the thi

Degger [83]

Refer to image for answer

3 0

4 years ago

Identify the type of conic section that has the equation 9x2+ 16y2 = 144 and identify its domain and range.

Kobotan [32]

9 x² + 16 y² = 144 /:144
$\frac{ x^{2} }{16} + \frac{ y^{2} }{9} =1$
General formula of ellipse ( the center is at the origin ): $\frac{ x^{2} }{ a^{2} }+ \frac{ y^{2} }{ b^{2} } =1$
a² = 16, b² = 9
Domain: [-a, a ] = [-4, 4]
Range:[-b, b ]
Answer: B ) ellipse.
Domain: { -4 ≤ x ≤ 4 }
Range: { -3 ≤ y ≤ 3 }

8 0

3 years ago

Which equation is correct regarding the measure of ∠MNP?

Evgen [1.6K]

Answer:

The measure of anle MNP is

m ∠ MNP = (x - y)/2

Explanation:

The image attached shows the figure corresponding to this question.

The <em>angle MNP</em>, wich is also the angle LNP, is formed by the intersection of a secant and a tanget to a circle.

Then, you can use the theorem:

the angle formed by a secant and a tangent to a circle that intersect outside the circle is half the difference of the major arc minus the minor arc.

The major arc formed is identified with the letter x and the minor arc is identified witht he letter y. Thus, the measure of the angle MNP is half the differenc x - y:

m ∠ MNP = (x - y)/2

6 0

3 years ago

Read 2 more answers

Parametric test based on standard error ​

Parametric test based on standard error