Please help me ..........

A stick is broken into two pieces, at a uniformly random break point. Find the CDF and average of the length of the longer piece

Oksanka [162]

Answer:

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

Step-by-step explanation:

let assume that stick has length 1.Random variable L that make length of a longer piece and random variable U that mark point .See that L < l means that

U≤ l and 1-U ≤l

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

this means 1-l≤U≤l

so we have

if we have L [1/2,1]

then apply the formula we have E(L)=3/4

5 0

4 years ago

Write a function in any form that would match the graph shown below.

Pani-rosa [81]

Answer:

(x-6)(x+5)^2 - 50

5 0

3 years ago

Find the slope of the line that passes through the points (-6, 2) and (-10, 2).

Alex Ar [27]

Answer:

m = 0

Step-by-step explanation:

$m=\frac{rise}{run}=\frac{2-2}{-10-6}=\frac{0}{-16} =0$

Hope this helps.

4 0

3 years ago

Suppose the national mean SAT score in mathematics was 510. In a random sample of 60 graduates from Stevens High, the mean SAT s

zalisa [80]

Answer:

We accept the alternate hypothesis and conclude that mean SAT score for Stevens High graduates is not the same as the national average.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 510

Sample mean, $\bar{x}$ = 502

Sample size, n = 60

Sample standard deviation, s = 30

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 510\\H_A: \mu \neq 510$

We use Two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }$ Putting all the values, we have

$t_{stat} = \displaystyle\frac{502- 510}{\frac{30}{\sqrt{60}} } = -2.0655$

Now,

$t_{critical} \text{ at 0.05 level of significance, 59 degree of freedom } = \pm 1.671$

Since,

$|t_{stat}| < |t_{critical}|$

We reject the null hypothesis and fail to accept it.

We accept the alternate hypothesis and conclude that mean SAT score for Stevens High graduates is not the same as the national average.

7 0

3 years ago

Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 307. Approximate th

SOVA2 [1]

Answer:

There is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

Step-by-step explanation:

The roll of a six-sided die sum has this parameters:

Mean: 3.5

Standard deviation: 1.7

If the die is rolled 81 times, the distribution of the sum of the 81 rolls will have the following parameters:

$\mu=n*M=81*3.5=283.5\\\\\sigma=\sqrt{n}s=\sqrt{81}*1.7=9*1.7=15.3$

Note: the variance of the sum of random variables is equal to the sum of the variance of the individual variables.

Then, we can calculate the probabilties that the sum of 81 rolls is lower than 307.

$z=\frac{x-\mu}{\sigma}=\frac{307-283.5}{15.3}= \frac{23.5}{15.3}= 1.536\\\\\\P(x$

There is 94% of chances that the sum of 81 rolls is lower than 307, so there is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

5 0

3 years ago

Please help me ..........​

Please help me ..........