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BigorU [14]
3 years ago
5

Is (1, 1) a solution to the inequality 6x + 6y < 15?

Mathematics
1 answer:
djyliett [7]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

Plug the given values in for x  and y to check (1,1) = (x,y)

6(1)+6(1) < 15

6+6 < 15

12<15

12 is less than 15 therefore (1,1) is a solution

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Use the table to identify the values of p and q that should be used to factor x2
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7. In a state lottery, a player must choose 8 of the numbers from 1 to 40. The lottery commission then performs an experiment th
Vlad1618 [11]

Answer:a) P(8 of the players numbers are drawn)=1.3×10^-8

b) P(7 of the players number are drrawn)=3.33×10^-c) P(at least 6 of the players number were drawn)=1.84×10^-4

Step-by-step explanation:

Players has 8 combinations of numbers from 1-40. The outcome S contains all the combinations of 8 out of 40

a) P(8 of the players numbers are drawn)= 1/40/8= 1.3×10^-8

There are one in hundred million chances that the draw numbers are precisely the chosen ones.

b) Number of ways of drawing 78 selected numbers from 1-40=8×(40-7)

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P(7 of the players number are drawn)=8×32/40 =3.33×10^-6.

There are approximately 300,000 chances that 7 of the players numbers are chosen

c) P(at least 6 players numbers are drawn)= 32/2×(8/6) ways to draw.

P(at least 6 players numbers are drawn)=P(all 8 chosen are drawn)+P(7 players numbers drawn)+P(6 chosen are drawn) = 1+ 8 x32/40/8 +[8\6 ×32/2]

P(at least 6 players numbers are drawn) = 1.84×10^-4.

There are approximately 5400chances that at least6 of the numbers drawn are chosen by the player.

5 0
3 years ago
HELP!!!
Naddika [18.5K]

Check the picture below, so it reaches the maximum height at the vertex, let's check where that is

h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}

5 0
2 years ago
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