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masha68 [24]
3 years ago
8

The difference of y and 8.1 is 20. Flnd the value of y.

Mathematics
1 answer:
White raven [17]3 years ago
5 0

Answer:

y = 28.1

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Set Up Equation</u>

"Difference" is subtraction

"y" is y

"8.1" is 8.1

"is" is equal

"20" is 20

y - 8.1 = 20

<u>Step 2: Solve for </u><em><u>y</u></em>

  1. Add 8.1 to both sides:                    y = 28.1
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Problem 3.2.14a<br><br> Show that 2^2x+1 +1 is divisible by 3.
elena-s [515]

Answer:

The given expression is divisible by 3 for all natural values of x.

Step-by-step explanation:

The given expression is

2^{2x+1}+1

For x=1,

2^{2(1)+1}+1=2^{3}+18+1=9

9 is divisible by 3. So, the given statement is true for x=1.

Assumed that the given statement is true for n=k.

2^{2k+1}+1

This expression is divisible by 3. So,

2^{2k+1}+1=3n              .... (1)

For x=k+1

2^{2(k+1)+1}+1

2^{2k+2+1}+1

2^{(2k+1)+2}+1

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Using equation (1), we get

(3n-1)2^2+1

(3n)2^2-2^2+1

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4 0
3 years ago
The best player on a basketball team makes 80​% of all free throws. The​ second-best player makes 75​% of all free throws. The​
Artist 52 [7]

Answer:

Expected number of free throws in 60 attempts:

        Best player = 48

        2nd best player = 45

        3rd best player = 42

Step-by-step explanation:

Solution:-

- The probability that best player makes free throw, p1 = 0.8

- The probability that second-best player makes free throw, p2 = 0.75

- The probability that third-best player makes free throw, p3 = 0.70

- Total number of attempts made in free throws, n = 60.

- The estimated number of free throws that any player makes is defined by:

                      E ( Xi ) = n*pi

Where,           Xi = Player rank

                      pi = Player rank probability

- Expected value for best player making the free throws would be:

                    E (X1) = n*p1

                              = 60*0.8

                              = 48 free throws

- Expected value for second-best player making the free throws would be:

                    E (X2) = n*p2

                              = 60*0.75

                              = 45 free throws

- Expected value for third-best player making the free throws would be:

                    E (X3) = n*p3

                              = 60*0.70

                              = 42 free throws

8 0
3 years ago
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