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lisov135 [29]
3 years ago
13

A person is selected at random. If there are seven days in a week, what is the probability that the person was not born on a Tue

sday or Wednesday?
Mathematics
1 answer:
levacccp [35]3 years ago
7 0

Answer:

5 out of 7.

Step-by-step explanation:

There are seven days in one week. They are asking what is the probability of the person chosen not being born in a Tuesday or Wednesday. There are five other days, so there are five other options.

I hope I helped you!

You might be interested in
When 2/3 is added to a certain number, the sum is 3 greater than twice the number. part 1: set up an equation using n to represe
Gennadij [26K]
2/3 + n = 2n + 3 <== ur equation
2/3 - 3 = 2n - n
2/3 - 9/3 = n
-7/3 = n <== ur solution

or
2/3 + n = 2n + 3...multiply everything by common denominator of 3
2 + 3n = 6n + 9
2 - 9 = 6n - 3n
-7 = 3n
-7/3 = n
7 0
3 years ago
Solve the equation for the variable.<br><br> 15.25 – 3.8x = -26.75 +2.2x<br> x = [?]
nirvana33 [79]
X=-2.3

I hope this is helpful! Let me know if you need anymore help!
4 0
2 years ago
There are 10 true-false questions and 20 multiple choice questions from which to choose a five-question quiz how many ways can t
beks73 [17]

Answer:

In 68229 ways can the quiz be selected such that there is atleast three multiple choice questions

Step-by-step explanation:

Given:

Number of True or false questions= 10

Number of multiple choice questions= 20

To Find:

How many ways can 5 questions can be selected if there must be at least three multiple choice questions =?

Solution:

Combination

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.  

The question States there sholud be ATLEAST 3 multiple choice question,

So, we may have

(3 Multiple choice question and 2 true or false question) or

(4 Multiple choice question and 1 true or false question) or

(5 Multiple choice question and 0 true or false question)

Required Number of ways = (20C3 X10C2) +(20C4 X10C1) + (20C5 X10C0)

Required Number of ways =(\frac{20!}{20!(20-3)!}\times\frac{10!}{10!(10-2)!})+(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-1)!}) +(\frac{20!}{20!(20-4)!} \times \frac{10!}{10!(10-0)!})

Required Number of ways = ( 1140 x 42) + (4845 x 10) +(15504 x 1)

Required Number of ways = 47880+48450+15504

Required Number of ways = 68229

3 0
3 years ago
6. Solve the the following system of equations: 4y+2x=14 2y+2x=10
Inessa05 [86]
4y+2x=14 ==> multiply this 1st equation by (-1) => -4y-2x=-14
Now add it up to the 2nd equation
-4y-2x=-14
 2y+2x=10
--------------    And now add them up
-2y +0x=-4
==> -2y=-4==> 2y = 4 &    y=2                   
Last step: plug y into any equation & you will find x=3
3 0
4 years ago
What is the rate of change and initial value for the linear relation that includes the points shown in the table?
melomori [17]

Answer: I need the table to give the answer

Step-by-step explanation: copy and paste the table

5 0
3 years ago
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