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horsena [70]
3 years ago
7

Carlos is driving on a straight section of highway from Ashford to Lincoln. Ashford is at mile marker 333 and Lincoln is at mile

marker 453. A rest stop is located along the highway 2/3 of the distance from Ashford to Lincoln. Assuming Carlos drives at a constant rate of 60 miles per hour, how long will it take him to drive from Ashford to the rest stop? Round your answer to the nearest hundredth.
It will take _______hours to drive from Ashford to the rest stop?
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
3 0

Distance between the two cities:

453 - 333 = 120 miles.

Rest area is 2/3 of the way:

120 x 2/3 = 240/3 = 80 miles.

Divide the miles to the rest stop by his speed:

80 miles/ 60 miles per hour = 1 and 1/3 hours as a fraction. 1.3333 as a decimal( round as needed.

( 1 hour and 20 minutes)

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3 = 3/1 = 15/5
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4 = 4/1 = 12/3
12/3 + 1/3 = 13/3
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Now you can divide 17/5 and 13/3.

Step 1- Turn the fraction you are dividing by into its reciprocal.
13/3 reciprocal equals 3/13

Step 2- Multiply.
17/5 • 3/13
17 • 3 = 51
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Answer: 51/65

Step 3- Simplify if possible.
51/65
This is already in simplest form so there is no need to simplify.

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Answer:

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f(x)=2x^2 + 3

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Read 2 more answers
Consider the probability that greater than 90 out of 149 software users will not call technical support. Assume the probability
Artemon [7]

Answer:

  0.1220

Step-by-step explanation:

You want the probability that greater than 90 out of 149 software users will not call tech support, given that the probability a given user will not call is 56%. You want the value using the normal approximation to the binomial distribution.

<h3>Normal approximation</h3>

For a sample size of n, and probability p of "success", the normal approximation to the binomial distribution can be used if ...

  min(pn, (1-p)n) > 5

Here, that is ...

  min(0.56·149, 0.44·149) > 5

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Then the parameters of the normal distribution are ...

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<h3>Continuity correction factor</h3>

The probability that X = N will occur can be considered to be ...

  P(N -0.5 < X < N +0.5)

Then the probability that X > N will be modeled using the normal distribution as ...

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For this problem, we want ...

  P(X > 90.5) . . . . using the normal distribution with µ = 83.44, σ = 6.0592

The attached calculator shows that value to be ...

  P(>90) ≈ 0.1220 . . . . . . rounded to 4 decimal places

__

<em>Additional comment</em>

A probability calculator says the actual binomial distribution result for this problem is 0.1217.

7 0
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