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adelina 88 [10]
2 years ago
9

point (4,4),(4,0), and (0,0) are the vertices of the triangle the enclosed whole squares and half squares on the grid. determine

the area of the triangle by counting the whole squares and the half squares
Mathematics
1 answer:
Advocard [28]2 years ago
6 0

Answer:

8

Step-by-step explanation:

finding the area is h x l x 1/2. The h is 4 and the l is also 4. That multiplied is 16. Multiply that by 1/2 and you get 8

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the perimeter of a rectangle is 48 inches. the ration of the width the length is 3:5, as shown. find the width and the length of
aleksandrvk [35]

Answer:

Width= 18inches Length= 30inches

Step-by-step explanation:

3+5= 8

48/8=6

3*6=18

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Please help on this one?
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The inverse is where the x and y values are flipped, so the left side would have 6 7 8 9 and the right side would have 9 10 11 12 for the inverse.

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2 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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