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Anni [7]
3 years ago
14

Winning the jackpot in a particular lottery requires that you select the correct five numbers between 1 and 25 and, in a separat

e drawing, you must also select the correct single number between 1 and 31. Find the probability of winning the jackpot.
Mathematics
2 answers:
Flura [38]3 years ago
8 0

Answer: 2

Step-by-step explanation: rip kobe bryant

s2008m [1.1K]3 years ago
4 0

Answer: The probability of winning the lottery is 0.002%

Step-by-step explanation:

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What are the 2 equations for the coordinates (6,4)
JulijaS [17]

Answer:

x = 6

y = 4

I'm not sure what type of equations you are looking for but these are the most basic ones.

Step-by-step explanation:

Hope this helps! if you have any questions ask me in the comments.

8 0
3 years ago
Benito spent $1837 to operate his car last year. Some of these expenses are listed on the table below:
Dmitry_Shevchenko [17]

Answer:

$0.085 per mile or 8.5 cent per mile

Step-by-step explanation:

We know:

Total expenses = $1837

Gasoline expense = Total expenses - other expenses

= 1837 - 972 - 114 - 105 = $646

Benito drove in total of: 7600 miles

Cost of gasoline per mile = Money spent for gasoline / total miles

= $646 / 7600

= $0.085 per mile or 8.5 cent per mile

Hope this helped :3

3 0
3 years ago
Read 2 more answers
A farmer sees 56 of his cows out of the barn. He knows that he has 83 cows altogether. Let c represent the number of cows still
Delicious77 [7]
For this question you want to take 83 and subtract 56 which would give you 27. So 33 couldn't be the number of cows still in the barn.
4 0
4 years ago
Read 2 more answers
Thelocal library has a children’s section with picture books and chapter books. The ratio is 2:3. If there are a total of 500 bo
Aleksandr-060686 [28]

Answer:

300

Step-by-step explanation:

500-200=300

3 0
3 years ago
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

x=(y+6)^3

And solve for y

\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

And we have y = f^-1(x)

Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

4 0
1 year ago
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