What are the factors that gravitational force law depend on?<br>

OMG I was playing robIox and a glitch happened and I couldnt be shot I could only be hit but OFC right when that happens my stup

hammer [34]

Would of said "You arnt my real dad you can't make me do anything that's why mom cheated on you"

8 0

3 years ago

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Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w

Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so $p = 0.6$

The week has 5 days, so $n = 5$

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

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

$P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102$

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

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

$P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768$

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

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

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

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

$P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456$

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

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

$P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592$

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so $P(X = 5)$ .

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

$P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778$

7.78% probability of spending 12 dollars on coffee over the course of a five day week

8 0

4 years ago

The graph below shows the solution to which system of inequalities?

nlexa [21]

Answer:

Option D.

Step-by-step explanation:

Consider option D. $y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3$

Take point (0,0)

On putting this point in inequation $y\leq \frac{3x}{4}+10$ , we get

$0\leq 10$ which is true . So, solution is region towards the origin i,e region below the line $y= \frac{3x}{4}+10$ including the line itself .

On putting (0,0) in inequation $y\leq \frac{-x}{2}-3$ , we get $0\leq -3$ which is false , so solution is region away from the origin i.e region below line $y= \frac{-x}{2}-3$ including the line itself .

So, common solution to both the inequations is the shaded part in the given figure .

In other words, we can say that the graph shown in the given figure represents system of equations: $y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3$

5 0

3 years ago

You and some of your friends go to Six Flags. There, tickets are only $2 per

lora16 [44]

Answer:

Yes, it graphs easily.

Step-by-step explanation:

First of all we can see that is proportional and is positive, due to no powers being present, and the slope being positive. If you want the equation we first have to remember it is in y=ax+b form, using the problem we can see that the admission, or constant is 5, and the slope is 2. Then we plug in the numbers and get y=2x+5, and seeing that there is a small admission fee, we are forced to start at 5$. Therefore the answer is Yes, it graphs easily.

5 0

3 years ago

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What’s the correct answer for this

Olenka [21]

Given 2 angle : 45 and 90 (right angle)
==> 180-45-90 = 45

5 0

3 years ago

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What are the factors that gravitational force law depend on?​

What are the factors that gravitational force law depend on?