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zloy xaker [14]
3 years ago
11

Jessica is returning home from a trip. The function d(t)=458-60t models the distance, in miles, that Jessica is from her house t

hours after she began returning home. How far is Jessica from her home 5 minutes after she began returning home?
A.

5 miles


B.

158 miles


C.

300 miles


D.

453 miles
Mathematics
1 answer:
S_A_V [24]3 years ago
4 0

Answer: D. 453 miles .

Step-by-step explanation:

Given: The function d(t)=458-60t models the distance, in miles, that Jessica is from her house <em>t </em>hours after she began returning home.

At t = 5 minutes = \dfrac{5}{60} hour     [1 hour = 60 minutes],

Distance covered by Jessica in 5 minutes = d(\dfrac{5}{60})=458-60(\dfrac{5}{60})

=458-5=453\ miles

Jessica is 453 miles from her home after 5 minutes she began returning home.

Hence, the correct option is D. 453 miles .

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Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
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The probability is:

P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

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P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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P(X = 16) = C_{20,16}.(0.5)^{16}.(0.5)^{4} = 0.0046

P(X = 17) = C_{20,17}.(0.5)^{17}.(0.5)^{3} = 0.0011

P(X = 18) = C_{20,18}.(0.5)^{18}.(0.5)^{2} = 0.0002

P(X = 16) = C_{20,19}.(0.5)^{19}.(0.5)^{1} = 0

P(X = 17) = C_{20,20}.(0.5)^{20}.(0.5)^{0} = 0

Then:

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The probability that the student will get 15 correct questions in this test by guessing is 0.0207 = 2.07%.

You can learn more about the binomial distribution at brainly.com/question/24863377

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