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LenKa [72]
2 years ago
15

A soccer ball is kicked into the air from the ground. If the ball reaches a maximum height of 25 ft and spends a total of 2.5 s

in the air, which equation models the height of the ball correctly? Assume that acceleration due to gravity is –16 ft/s2.
Mathematics
1 answer:
kondaur [170]2 years ago
6 0
Hello,

g=9.81m/s²= 9.81/0.3048 =32.18.. ft/s² rounded to 32

h=-g/2*t²+vt+h0
if t=0,h=0 ==>h0=0

==>h=-16*t²+vt
50 ft in 2.5s ==> 25 ft in 1.25s

25=-16*1.25²+v*1.25
==>v=50/1.25=40

Equation h=-16t²+40*t
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Answer: Product = 13.8

Step-by-step explanation:

<u>As we have to use area model :</u>

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Area of rectangle is given by

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Now, we must write in standard form, word form and expanded form:

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13.8=1\times 10+3\times 1+8\times \frac{1}{10}

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According to New Jersey Transit, the 8:00 A.M. weekday train from Princeton to New York City has a 90% chance of arriving on tim
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Answer:

See explanation below.

Step-by-step explanation:

Assuming this problem: "According to New Jersey Transit, the 8:00 A.M. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let W = the number of days on which the train arrives late. "

For this case we can check if the binomial model can be used checking conditions:

1) We satisfy that we have independent trials and is assumed for this case

2) We have a fixed value for the trials n =6 and for the probability p = 0.9 on each trial

3) We have bernoulli experiments on each trial since we have success or failure for each case.

So then since all the conditions are satisfied we can assume that the binomial model can be used here.

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Let W the random variable of interest "the number of days on which the train arrives late", on this case we now that:

W \sim Binom(n=6, p=0.9)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We can find all the possible probabilities for all the possible values of X like this:

P(X=0)=(6C0)(0.9)^0 (1-0.9)^{6-0}=1x10^{-6}

P(X=1)=(6C1)(0.9)^1 (1-0.9)^{6-1}=5.4x10^{-5}

P(X=2)=(6C2)(0.9)^2 (1-0.9)^{6-2}=0.001215

P(X=3)=(6C3)(0.9)^3 (1-0.9)^{6-3}=0.01458

P(X=4)=(6C4)(0.9)^4 (1-0.9)^{6-4}=0.0984

P(X=5)=(6C5)(0.9)^5 (1-0.9)^{6-5}=0.354

P(X=5)=(6C5)(0.9)^6 (1-0.9)^{6-6}=0.531

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