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adelina 88 [10]
3 years ago
9

I also need Help with e

Mathematics
1 answer:
kherson [118]3 years ago
7 0
They ate all the cheese .the end
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Subtract.<br><br> 1 2/5−(−3/5)<br><br> Enter your answer, in simplest form, in the box.
mars1129 [50]

Answer:

3/5

Step-by-step explanation:

you need to subtract the two fractions then you can divide them by 2 to make them in their simplest form!

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Wut is 100000000000 x 0
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3 years ago
2. A bag contains 25 paise coins and 50 paise Coins. The number of 25 paise coins is 6 times that of so paise Coins The total va
Karo-lina-s [1.5K]
<h2>~<u>Solution</u> :-</h2>

Here, it is given that the bag contains 25 paise coins and 50 paise coins in which, 25 paise coins are 6 times than that of 50 paise coins. Also, the total money in the bag is Rs. 6.

  • Hence, we can see that, here, we have been given the linear equation be;

Let the number of coins of 50 paise will be $ x $ and the number of coins of 25 paise will be $ 6x $ as given. . .

Hence,

x + 6x = 6

7x = 6

x =  \frac{6}{7}  \\

x = 1.1666667 \approx \: 2

  • Hence, the number of 50 paise coins will be <u>2</u>. And, 6 times of two be;

6(2)

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6 0
3 years ago
Answer if you know<br> DO NOT ANSWER IF YOU DO NOT KNOW THE ANSWERS<br><br> Thank you :)
ludmilkaskok [199]

 See below for pics of the answers

6 0
3 years ago
Suppose that 4 fair coins are tossed. Let Equals The event that exactly 2 coins show tails and Equal The event that at least 2 c
faust18 [17]

Answer:

a) P ( E | F ) = 0.54545

b) P ( E | F' ) = 0

Step-by-step explanation:

Given:

- 4 Coins are tossed

- Event E exactly 2 coins shows tail

- Event F at-least two coins show tail

Find:

- Find P ( E |  F )

- Find P ( E | F prime )

Solution:

- The probability of head H and tail T = 0.5, and all events are independent

So,

                    P ( Exactly 2 T ) = ( TTHH ) + ( THHT ) + ( THTH ) + ( HTTH ) + ( HHTT) + ( HTHT)  = 6*(1/2)^4 = 0.375

                    P ( At-least 2 T ) = P ( Exactly 2 T ) + P ( Exactly 3 T ) + P ( Exactly 4 T) = 0.375 + ( HTTT) + (THTT) + (TTHT) + (TTTH) + ( TTTT)

      = 0.375 + 5*(1/2)^4 = 0.375 + 0.3125 = 0.6875

- The probabilities for each events are:

                    P ( E ) = 0.375

                    P ( F ) = 0.6875

- The Probability to get exactly two tails given that at-least 2 tails were achieved:

                    P ( E | F ) = P ( E & F ) / P ( F )

                    P ( E | F ) = 0.375 / 0.6875

                    P ( E | F ) = 0.54545

- The Probability to get exactly two tails given that less than 2 tails were achieved:

                    P ( E | F' ) = P ( E & F' ) / P ( F )

                    P ( E | F' ) = 0 / 0.6875

                    P ( E | F' ) = 0                

6 0
3 years ago
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