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Dmitrij [34]
3 years ago
6

A coin is tossed twice. Determine whether the events of getting tails twice are independent or dependent. Then identify the indi

cated probability.
Mathematics
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

Hi there!

They are independent. The first toss does <u>not</u> change the outcome of the sides. If you have a fair coin it's 50-50 weight.

1/2 × 1/2 = 1/4

There's a 25% chance of getting tails twice

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Which expression can be used to convert 100 USD to Japanese?
Montano1993 [528]

Answer:The expression that can be used to convert 100 USD to Japanese Yen is 100 USD = 100/Y Japanese Yen.

Step-by-step explanation:

3 0
3 years ago
Bob has a standard deck of playing cards. If he randomly draws a J, K, 2, and 2, what is the probability that the next card he d
nlexa [21]

Answer:

The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.


One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):


P(X=k)=(Kk)(N−Kn−k)(Nn)


where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.


P(X=2)=(42)(480)(522)


P(X=2)=61326=1221


In essence, this would give you the number of possible combinations of drawing two of the four ace cards in the deck (6, already enumerated by Ravish) over the number of possible combinations of drawing any two cards out of the 52 in the deck (1326). This is the way Ravish chose to solve the problem.


Another way is using simple probabilities and combinations:


P(X=2)=(4C1∗152)∗(3C1∗151)


P(X=2)=452∗351=1221


The chance of picking an ace for the first time (same as the chance of picking any card for the first time) is 1/52, multiplied by the number of ways you can pick one of the four aces in the deck, 4C1. This probability is multiplied by the probability of picking a card for the second time (1/51) times the number of ways to get one of the three remaining aces (3C1). This is the way Larry chose to solve the this.

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
Determine if the equation is linear. If so, graph the function.
musickatia [10]

Answer:

The equation is not linear

Step-by-step explanation:

You are given the equation

\dfrac{2}{x}+\dfrac{y}{4}=\dfrac{3}{2}

Express y in terms of x:

\dfrac{y}{4}=\dfrac{3}{2}-\dfrac{2}{x}\\ \\\dfrac{y}{4}=\dfrac{3x-4}{2x}\\ \\y=4\cdot \dfrac{3x-4}{2x}\\ \\y=\dfrac{6x-8}{x}

The linear function must of form

y=mx+b,

where m and b are real numbers.

Your function is not of this form, so this is not a linear function.

5 0
3 years ago
Solve.<br><br> 5x-2y&lt;10<br> I give Brainliest!
Anestetic [448]
<h2>Answer:</h2>

This is impossible to solve.

<h2>Step-by-step explanation:</h2>

For an equation or inequality to be solvable, there must be the same number of inequalities as variables. Here, there is an x and there is a y. This means that you need at least two inequalities to solve it.

You can, however, rearrange to get x or y on one side.

This can be done for x:

5x < 10 + 2y

x < 2 + 2/5y

Or it can be done for y:

5x < 10 + 2y

5x - 10 < 2y

2.5x - 5 < y

8 0
3 years ago
Help evaluating the question
serg [7]
X=1 so it is 1×1×1=1 so.... it is 19+1
5 0
3 years ago
Read 2 more answers
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