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Gnom [1K]
3 years ago
11

A lottery has 60 numbers. To win the jackpot one needs to match all 7 numbers that are drawn by the machine. Is this a PERMUTATI

ON or a COMBINATION problem? What is the “chance” (or, more mathematically speaking, what is the probability) to hit the jackpot?
Mathematics
1 answer:
Vadim26 [7]3 years ago
4 0

Answer: Hence, our required probability is \dfrac{1}{386206920}

Step-by-step explanation:

Since we have given that

Numbers in a lottery = 60

Numbers to win the jackpot = 7 numbers

We need to find the probability to hit the jackpot:

So, our required probability is given by

P=\dfrac{^7C_7}{^{60}C_7}\\\\P=\dfrac{1}{386206920}

This is a combination problem as we need to select 7 numbers irrespective of any arrangements.

Hence, our required probability is [tex]\dfrac{1}{386206920}[/tex

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Which numbers are greater than 3x10^-7
USPshnik [31]
3x10^-7 = 3 / (10^7) = 3 / 10000000 = 0.0000003;
A number greater then 0.0000003 is 1.
6 0
3 years ago
A bank pays 5% interest compounded annually. What principal will grow to $12,000 in 10 years.
makvit [3.9K]
<h3>Answer: 7366.96 dollars</h3>

========================================================

Use the compound interest formula:

A = P(1+r/n)^(n*t)

where in this case,

A = 12000 = amount after t years

P = unknown = deposited amount we want to solve for

r = 0.05 = the decimal form of 5% interest

n = 1 = refers to the compounding frequency (annual)

t = 10 = number of years

-------

Plug all these values into the equation, then solve for P

A = P(1+r/n)^(n*t)

12000 = P(1+0.05/1)^(1*10)

12000 = P(1.05)^(10)

12000 = P(1.62889462677744)

12000 = 1.62889462677744P

1.62889462677744P = 12000

P = 12000/1.62889462677744

P = 7366.95904248911

P = 7366.96

6 0
3 years ago
Two subtracted from b is greater than -18
Ber [7]
(b-2)>-18
OR
-18+(b-2)
8 0
3 years ago
Which congruence theorem can be used to prove △ABC ≅ △DBC?
Tema [17]

Answer:

SAS

Step-by-step explanation:

Hope this helps :)

7 0
3 years ago
Please write the answer​
Travka [436]

\sf \dfrac{3^x - 5 \times 3^{(x-2)}}{3^{(x-3)}} \\ \\ \longrightarrow \sf \dfrac{ {3}^{x} }{ {3}^{(x - 3)}} - \frac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)}  \\ \\ \longrightarrow \sf {3}^{[x - (x - 3)]} - \dfrac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ {3}^{(x - 3)} } \times \dfrac{ {3}^{x}}{9} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ \bigg(\dfrac{ {3}^{x} }{ {3}^{3} }\bigg) } \times \dfrac{ {3}^{x}}{9}\\ \\ \longrightarrow \sf {3}^{3} - \dfrac{ ({3}^{3})( 5)}{ {3}^{x} } \times \dfrac{ {3}^{x}}{9}\\ \\ \sf \longrightarrow {3}^{3} - (5 \times 3) \\  \\ \longrightarrow \sf \: 27 - 15 \\  \\ \longrightarrow \leadsto{\underline{\boxed{\sf{ \pink{ 12}}}}}

6 0
2 years ago
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