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3 years ago

8

Find the product of 6372×75

2 answers:

sweet-ann [11.9K]3 years ago

6 0

The answer is 477900

BabaBlast [244]3 years ago

6 0

the answer is 477900

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Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have:

$\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)$

And for the next step we have:

$\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}$

And with this we have the requiered proof.

And since $b=1-p$ we have that:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

4 0

3 years ago

Represent the relationship between the number of copies made and time in seconds in the table

alexdok [17]

Answer:

you need to put the table

Step-by-step explanation:

7 0

3 years ago

McAllister et al. (2012) compared varsity football and hockey players with varsity athletes from noncontact sports to determine

jonny [76]

Answer:

$t _{critical} = 1.760$

t = 2.2450

d. 0.264

Step-by-step explanation:

The null hypothesis is:

$H_o: \mu_1 - \mu_2 = 0$

Alternative hypothesis;

$H_a : \mu_1 - \mu_2 > 0\\$

The pooled variance t-Test would have been determined if the population variance are the same.

$S_p^2 = \dfrac{(n_1-1)S_1^2+(n_2-1)S^2_2}{(n_1-1)+(n_2-1)}$

$S_p^2 = \dfrac{(8-1)2.507^2+(8-1)2.8282^2}{(8-1)+(8-1)}$

$S_p^2 = 7.14$

The t-test statistics can be computed as:

$t= \dfrac{(x_1-x_2)-(\mu_1 - \mu_2)}{\sqrt{Sp^2 ( \dfrac{1}{n} +\dfrac{1}{n_2})}}$

$t= \dfrac{(9-6)-0}{\sqrt{7.14 ( \dfrac{1}{8} +\dfrac{1}{8})}}$

$t= \dfrac{3}{1.336}$

t = 2.2450

Degree of freedom $df = (n_1 -1) + ( n_2 +1 )$

df = (8-1)+(8-1)

df = 7 + 7

df = 14

At df = 14 and ∝ = 0.05;

$t _{critical} = 1.760$

Decision Rule: To reject the null hypothesis if the t-test is greater than the critical value.

Conclusion: We reject $H_o$ and there is sufficient evidence to conclude that the test scores for contact address s less than Noncontact athletes.

To calculate r²

The percentage of the variance is;

$r^2 = \dfrac{t^2}{t^2 + df}$

$r^2 = \dfrac{2.2450^2}{2.2450^2 + 14}$

$r^2 = \dfrac{5.040025}{5.040025+ 14}$

$r^2 = 0.2647$

7 0

3 years ago

Solve the equation for x. 1 6 (2x − 12x) = 30 A) −12 B) −15 C) −18 D) −24

lubasha [3.4K]

$\frac{1}{6}$ *(2x-12x) = 30

$\frac{1}{6}$ *(-10x) = 30

$\frac{-10}{6}$ x = 30

$\frac{-5}{3}$ x = 30

$\frac{5}{3}$ x = -30

x = -30/ $\frac{5}{3}$ = -30 * $\frac{3}{5}$

x = -18

the correct answer is C

8 0

3 years ago

Read 2 more answers

Please help me i will give brainliest

Rzqust [24]

Answer:

the answer is A. :)

7 0

2 years ago