The sum of two consecutive odd integers is 188. list the numbers from smallest to largest:

Tennis Replay In the year that this exercise was written, there were 879 challenges made to referee calls in professional tennis

tiny-mole [99]

Answer:

a. 0.0209 = 2.09% probability that among the 879 challenges, the number of overturned calls is exactly 231.

b. 231 is less than 2.5 standard deviations above the mean, which means that 231 overturned calls among 879 challenges is not a significantly high result.

Step-by-step explanation:

For each challenge, there are only two possible outcomes. Either it was overturned, or it was not. The probability of a challenge being overturned is independent of any other challenge. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Significantly high:

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

If a value is more than 2.5 standard deviations above the mean, this value is considered significantly high.

25% of the challenges are successfully upheld with the call overturned.

This means that $p = 0.25$

879 challenges

This meas that $n = 879$

a. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is exactly 231.

This is P(X = 231). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 231) = C_{879,231}.(0.25)^{231}.(0.75)^{648} = 0.0209$

0.0209 = 2.09% probability that among the 879 challenges, the number of overturned calls is exactly 231.

b. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is 231 or more. If the 25% rate is correct, is 231 overturned calls among 879 challenges a result that is significantly high

The mean is:

$E(X) = np = 879*0.25 = 219.75$

The standard deviation is:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{879*0.25*0.75} = 12.84$

$219.75 + 2.5*12.84 = 251.85 > 231$

231 is less than 2.5 standard deviations above the mean, which means that 231 overturned calls among 879 challenges is not a significantly high result.

5 0

3 years ago

What is the distance between the points (4,9) and (1,3)?

qaws [65]

6 I think but don't take my word for it.

3 0

3 years ago

What is the slope of the line represented by the equation -2y=x-1?

ozzi

The slope is negative one half or -1/2

5 0

3 years ago

Read 2 more answers

A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on four numbers pays 8 to 1 (that is, if you bet $1 and one of t

Natasha_Volkova [10]

Answer:

Lose $0.05

Step-by-step explanation:

There are 38 possible spots on the roulette wheel (numbers 1 to 36, 0 and 00).

If the player can choose four numbers on single $1 bet, his chances of winning (W) and losing (L) are as follows:

$P(W) = \frac{4}{38} \\P(L) = 1-P(W) = 1-\frac{4}{38} \\P(L) = \frac{34}{38}$

The expected value of the bet is given by the probability of winning multiplied by the payout ($8), minus the probability of losing multiplied by the bet cost ($1)

$EV=\frac{4}{38}*\$8 -\frac{34}{38}*\$1\\EV= -\$0.05$

On each bet, the player is expected to lose 5 cents ($0.05).

4 0

3 years ago

A 1 newton force will stretch a spring 1 meter. The spring/mass system is damped by a force that is 8 times the instantaneous ve

sukhopar [10]

Answer:

Step-by-step explanation:

Given the mass is m =16kg, and 1N force will stretch the spring 1 m.

That is, F =1N,Z =1m. Now find the spring constant k:

F = kL = 1 = k(1) = k= 1N/m.

The damping force is 8times the instantaneous velocity, this means β = 8,

and the external force is f(t) = 0

Initially the object compressed 0.6m above equilibrium position,

with the downward velocity is 2m/s.

The differential equation for a spring mass system with

damping force and extemal force is: mx" + βxt + kx = f(t).

so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.

Now solve the DE:

The auxilary equation for the homogeneous equation is 16x"+8x'+x=0

solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.

Then the general solution for the homogenous system is: $x(t)=c_1e^{-t/4} +c_2te^{-t\4}$ .

Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:

$x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85$ .

Hence, $x(t) =-0.6e^{-t/4}+1.85te^{-t/4}$ .

5 0

3 years ago