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3 years ago

A hypothesis is only tentative: to make sure it's valid, you have to ___________.

Mathematics

1 answer:

GREYUIT [131]3 years ago

4 0

A hypothesis is only tentative: to make sure it's valid, you have to test it.

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Simplify: (7x4 – 7x3 + 7x2 + 2x) + (-5x4 + 7x3 + 5x2 – 9)

ikadub [295]

Answer:

Step-by-step explanation:

8 0

2 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

2 years ago

Can someone please help?

Alecsey [184]

Answer:

73 degrees

Step-by-step explanation:

6 0

2 years ago

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch. Suppose that the speci

Leni [432]

Answer:

0.0002 inch

Step-by-step explanation:

The empirical rule of the normal distribution, the 68-95-99.7 rule, means

if the mean is μ and the standard deviation is σ,

68% of data lies within μ - σ and μ + σ,

95% of data lies within μ - 2σ and μ + 2σ,

99.7% of data lies within μ - 3σ and μ + 3σ.

From the question, μ = 0.002.

The required range is 0.0014 to 0.0026.

With a probability of 0.9963, then 0.9963 × 100% = 99.63% should lie within the range. This approximately corresponds to μ - 3σ and μ + 3σ.

μ - 3σ = 0.0014

0.002 - 3σ = 0.0014

3σ = 0.0006

σ = 0.0002

Hence, the standard deviation is 0.0002 inch

We can check with the other end of the range:

μ + 3σ = 0.0026

3σ = 0.0026 - 0.002

3σ = 0.0006

σ = 0.0002

7 0

3 years ago

What is 35/6 in simplest form

Len [333]

Its already in simplest form u cant simplify it hope this helps!!

3 0

3 years ago