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Ad libitum [116K]
3 years ago
12

The situation that is best modeled by a discrete graph is

Mathematics
1 answer:
Dominik [7]3 years ago
7 0

Answer:

Hope It helps you :) Cause the graph haves to show two unsame activities

Step-by-step explanation:

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D.) the initial height of the roller coaster

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Please help NO LINKS! OR FILES!!! I REPEAT NO LINKS! OR FILES! Thank you!
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bottom right.

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Mary can run 5 kilometers in 18.75 minutes.
trasher [3.6K]

Answer:

30 minutes

Step-by-step explanation:

Find the unit rate (time per SINGLE kilometer).

5/5=1 18.75/5=3.75

Multiply 3.75 (unit rate) by 8.

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8 0
3 years ago
Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
Mhanifa please help! Solve each proportion. Round your answer to the nearest hundreth
Marianna [84]

Answer:

  • 5) m/5 = 7/2 ⇒ m = 5*7/2 ⇒ m = 35/2
  • 6) 7/4 = r/5 ⇒ r = 5*7/4 ⇒ r = 35/4
  • 7) 7/6 = 5/x ⇒ x = 5*6/7 ⇒ x = 30/7
  • 8) 6/5 = 2/(5n) ⇒ 5n*6 = 5*2 ⇒ n = 10/30 ⇒ n = 1/3
4 0
3 years ago
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