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sesenic [268]
2 years ago
9

What is the range of the function y = x2? x ≥ 0 all real numbers y ≥ 0

Mathematics
1 answer:
Alla [95]2 years ago
3 0

Answer:

The range of x is all real numbers including 0 to infinity, making this x ≥ 0

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A certain bag of potting soil is 1/4 peat moss, and the rest is dirt, what part is dirt
AVprozaik [17]
If you do the math you'll get 3/4 of the bag is dirt.
4 0
3 years ago
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Use the properties and mental math to find the sum 8 1/5 + 3 2/5 + 5 4/5
Amanda [17]

Answer:

The answer is 17 2/5

Step-by-step explanation:

To find this, first add all the fractions together.

1/5 + 2/5 + 4/5 = 1 2/5

Now add all the whole numbers together.

8 + 3 + 5 = 16

Now add the two together.

16 + 1 2/5 = 17 2/5

6 0
3 years ago
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Find the volume of a right circular cone that has a height of 11.8 ft and a base with a radius of 7.3 ft. Round your answer to t
lora16 [44]

Answer:

The answer to your question is 658.2 ft³

Step-by-step explanation:

Data

Volume = x

height = 11.8 ft

radius = 7.3 ft

Formula

Volume of a cone = 1/3πr²h

Process

1.- Calculate the volume

Volume = 1/3(3.14)(7.3)²(11.8)

Volume = 1/3(3.14)(7.3)²(11.8)

Volume = 1/3(3.14)(53.29)(11.8)

Volume = 1/3(1974.5)

Rounded to the nearest tenth

Volume =  658.2 ft³

3 0
3 years ago
4. Which function represents exponential decay?
Leni [432]
Y=35(0.35) is exponential decay
8 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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