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4 years ago

What is 0.4 divided by 7.29

Mathematics

2 answers:

telo118 [61]4 years ago

6 0

Answer:

0.054 or 40/729

denis-greek [22]4 years ago

4 0

The answer should be 0.05486

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∆ABC is reflected about the line y = -x to give ∆A'B'C' with vertices A'(-1, 1), B'(-2, -1), C(-1, 0). What are the vertices of

blsea [12.9K]

Hi there!
Reflections across the line y = -x always go by the rule (-y, -x). We can use this rule to get our answer here. We are given the aftermath of the reflection coordinates, which are <span>A'(-1, 1), B'(-2, -1), and C'(-1, 0). All we have to do now is switch up the coordinate values and multiply them by -1. Here is the work -
A'(-1, 1) => (1, -1) => x -1 => A(-1, 1)
B'(-2, -1) => (-1, -2) => x -1 => B(1, 2)
C'(-1, 0) => (0, -1) => x -1 => C(0, 1)
Therefore, the coordinates of Triangle ABC are A(-1, 1); B(1, 2); C(0,1). Hope this helped and have a phenomenal day!</span>

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4 years ago

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What is the slope of line m?

Svetradugi [14.3K]

Answer:y = MX+b

Step-by-step explanation:

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3 years ago

For a repeated-measures study comparing two treatments with a sample of n = 9 participants, the difference scores have a mean of

Korolek [52]

Answer: Estimated standard error for the sample mean difference would be 1.

Step-by-step explanation:

Since we have given that

Mean of MD = 4.90

So, Sum of difference would be

$4.9\times 9=44.1$

S = 288

n = 9

We need to find the standard error for the sample mean differences.

Estimated standard error for the sampled mean difference would be

$\dfrac{\sqrt{Sum(D^2)-(\dfrac{sum(d)^2}{n})}}{n(n-1)}}\\\\=\dfrac{\sqrt{288-\dfrac{44.1^2}{9}}}{9(9-1)}\\\\=0.99\\\\\approx 1$

Hence, estimated standard error for the sample mean difference would be 1.

6 0

3 years ago

Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) $M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Step-by-step explanation:

Given:

moment generating function of X as:

MX(t) = $\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}$

a) Now

Mean, E(X) = $M_{X}'(t=0)$

Thus,

$M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}$

$M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}$

also,

$E(X^{2})=M_{X}''(t=0)$

Thus,

$M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}$

$M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}$

Therefore,

Mean, E(X) = $M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}$

Mean, E(X) = - 0.5

and

$E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}$

$E(X^{2})$ = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

$M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Therefore,

at x = 0, P(x) = $\frac{1}{2}$

at x= - 4 ,P(x) = $\frac{1}{3}$

at x = 5, P(x) = $\frac{1}{6}$

Thus,

E(X) = $\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})$

E(X) = - 0.5

also, $E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})$

$E(X^{2})$ = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0

4 years ago

Subtract 65 from the product of 20 and 4.

Klio2033 [76]

Answer:

15 is the answer

Step-by-step explanation:

20×4=80

80-65=15

3 0

3 years ago

Read 2 more answers