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Norma-Jean [14]
3 years ago
9

Please help me .................​

Mathematics
1 answer:
Zinaida [17]3 years ago
6 0

Answer:

<em><u>Hi</u></em><em><u> </u></em><em><u>you</u></em>

Hope that will help you

<em>Nice</em><em> </em><em>time</em>

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Help confused thanks
sukhopar [10]
\left[\begin{array}{cc}Juice&Water\\1&3\\2&6\\3&9\\4&12\\\end{array}\right]

w=3j
36=3j
j=\frac{36}{3}=12cups
3 0
3 years ago
What is the sum of -1/11 and 5/11<br><br> 4/11<br><br> 6/11<br><br> -4/11<br><br> -6/11
kaheart [24]
<span>-1/11 + 5/11
= 5/11 - 1/11
= 4/11

Answer
</span><span>4/11</span>
6 0
3 years ago
Read 2 more answers
The sum of three consecutive even integers is 258. write an equation ad solve to find the integers
slavikrds [6]

The consecutive integers are 85,86,87

Explanation:

n

:

the first number

n

+

1

:

the second number

n

+

2

:

the third number

n

+

(

n

+

1

)

+

(

n

+

2

)

=

258

3

n

+

3

=

258

3

n

=

258

−

3

3

n

=

255

n

=

255

3

n

=

85

n

+

1

=

85

+

1

=

86

n

+

2

=

85

+

2

=

87

8 0
3 years ago
Read 2 more answers
Expand the given power by using Pascal’s triangle. (9a - 10b)^6
xxMikexx [17]

Answer:

531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6

Step-by-step explanation:

                     1                                n=0

                  1      1                            n=1

                 1   2   1                           n=2

                1  3  3  1                          n=3

              1  4  6  4   1                      n=4

             1  5 10 10 5  1                    n=5

           1 6 15 20 15 6 1                   n=6

This is where n is the exponent in

(x+y)^n.

(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6

Now we want to expand:

(9a-10b)^6 or we we can rewrite as (9a+(-10b))^6.

Let's replace x with (9a) and y with (-10b) in the expansion:

(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6

((9a)+(-10b))^6

=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6

Let's simplify a bit:

=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6

=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6

8 0
3 years ago
At a raffle, 1500 tickets are sold at $2 each. There are four prizes given for $500, $250, $150, and $75. You buy one ticket. Us
wolverine [178]

Answer:

<u>The expected value of every ticket is a loss of $ 1.35</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of tickets sold at the raffle = 1,500

Price of each ticket = $ 2

Total prizes = 1 * $ 500 + 1 * $ 250 + 1 * $ 150 + 1 * $ 75

2. Use a probability distribution table to calculate the expected value of your gain. Your expected value is_____ ?

Let's answer the question using a probability distribution for the gains, this way:

  • Probability of 1st prize of $ 498 (500 - ticket) = 1/1,500
  • Probability of 2nd prize of $ 248 (250 - ticket) = 1/1,500
  • Probability of 3rd prize of $ 148 (150 - ticket) = 1/1,500
  • Probability of 4th prize of $ 73 (75 - ticket) = 1/1,500
  • Probability of losing $ 2 (ticket) = 1,496/1,500

Now, we calculate the mean for all the tickets (winners and non-winners), this way:

Expected value = [(1,496 * -2) + (1 * 498) + (1 * 248) + ( 1 * 148) + ( 1 * 73)]/1,500

Expected value = [- 2,992 +498 + 248 + 148 + 73)/1,500

Expected value = -2,025/1,500 = - 1,35

<u>The expected value of every ticket is a loss of $ 1.35</u>

8 0
4 years ago
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