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2 years ago

Make v the subject of the formula in H=m(v²-u²)/2gx

Mathematics

1 answer:

goldfiish [28.3K]2 years ago

4 0

Answer:

$v=\sqrt{\dfrac{H\times 2gx}{m}+u^2}$

Step-by-step explanation:

The given formula is :

$H=\dfrac{m(v^2-u^2)}{2gx}$

We need to find the formula for v.

Cross multiplying the given formula.

$H\times 2gx=m(v^2-u^2)$

Dividing both sides by m. So,

$\dfrac{H\times 2gx}{m}=\dfrac{m(v^2-u^2)}{m}\\\\(v^2-u^2)=\dfrac{H\times 2gx}{m}$

Adding both sides by u².

$(v^2-u^2)+u^2=\dfrac{H\times 2gx}{m}+u^2\\\\v^2=\dfrac{H\times 2gx}{m}+u^2\\\\\text{So},\\\\v=\sqrt{\dfrac{H\times 2gx}{m}+u^2}$

Hence, this is the required formula.

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Read 2 more answers

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

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

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

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

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

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c: