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marta [7]
3 years ago
13

A tile pattern decreases by 3 tiles. Figure 2 of the pattern has 4 tiles. Use y=mx+b

Mathematics
1 answer:
NeX [460]3 years ago
4 0

Answer:

tjjtjjttjjtjtjtj

Step-by-step explanation:

nnnnhnnhnhnhnhnhnhnhnhnn

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Multiply 12 by the sum of 8 and "t"
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Mathematically, that is 12(8+t). Distribute the 12, and you get 96 + 12t, or 12t + 96.
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The sum of the first and third of three consecutive integers is 140. find the three even integers​
Elina [12.6K]

Answer:

It would be 68,70,72

Step-by-step explanation:

The sum of first and third of 3 consecutive number is 140 as 68+72 is 140.

68 is the first

72 is the third

They are all even numbers as well

6 0
3 years ago
A sphere has a surface area of πcm^2
dimaraw [331]

Answer:

hope this helps you..........

5 0
3 years ago
You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c)   P(win first game)  = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

Standard deviation \sigma=\sqrt{variance} = \sqrt{0.3844}=0.62

8 0
3 years ago
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