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Lisa [10]
3 years ago
9

In a standard deck of playing cards, there is a total of 56 cards, 4 of which are jacks. What is the formula for the probability

of drawing a single jack from the deck of cards?
Mathematics
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

the probability is  1/14 (7.14%)

Step-by-step explanation:

Since each card of the total of 56 equally likely to be chosen, then the probability is the ratio of the number of favourable outcomes divided by the total number of outcomes of the experiment. Thus the probability of getting a single Jack when there are 4 in 56 cards is

probability = number of favourable outcomes/ total number of outcomes = 4 Jacks / 56 different cards = 1/14 =0.0714 (7.14%)

thus the probability is  1/14 (7.14%)

You might be interested in
Which of the following equations represent nonlinear functions?
s2008m [1.1K]

y=5x^2+7 is Non-Linear Functions

Option B is correct option.

Step-by-step explanation:

We need to identify Non-Linear Functions from the equations given.

First we will define Non-Linear Functions

<u>Linear Functions</u>

A function having exponent of variable equal to 1 or of the form y=c, where c is constant is called linear function.

<u>Non-Linear Functions</u>

A function that has variable having power greater than 1 (i.e 2 or above) is called non-linear function.

So, from all the options given, only Option B has power greater than 1 i.e 2. All remaining options are linear functions.

So, y=5x^2+7 is Non-Linear Functions

Option B is correct option.

Keywords: Linear and Non-Linear Functions

Learn more about Linear and Non-linear functions at:

  • brainly.com/question/12363217
  • brainly.com/question/4326955

#learnwithBrainly

3 0
3 years ago
19. Use Gauss-Jordan elimination to solve the following system of equations. 3x + 5y = 7 6x − y = −8
MrRa [10]

\left[\begin{array}{cc}3&5\\6&-1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}7\\-8\end{array}\right]\\\left[\begin{array}{cc|c}3&5 &7\\ 6& -1 & -8\end{array}\right] \rightarrow \left[\begin{array}{cc|c}3&5& 7\\0&-11 & -22\end{array}\right] \rightarrow \left[\begin{array}{cc|c}3&5& 7\\0&1 & 2\end{array}\right]\\3x +5y = 7\\\\y = 2\\\therefore x = -1

6 0
3 years ago
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

5 0
3 years ago
HELP PLEASE I NEED IT NOW!!!
Ostrovityanka [42]
Google them? google will have all of them
3 0
3 years ago
Help me out please and dont send suspicious links
Lena [83]

Step-by-step explanation:

I think B or F is the best answer

7 0
3 years ago
Read 2 more answers
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