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OverLord2011 [107]
1 year ago
5

Round of 1,546 to the nearest thousands

Mathematics
1 answer:
EleoNora [17]1 year ago
4 0

Answer:

1,546 ≈ 2,000

Hope this helps :)

1. To round to nearest thousands, first look at the hundreds place if the number is 1,2,3,4 then round down. If the number is 5,6,7,8,9 then round up.

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In a road-paving process, asphalt mix is delivered to the hopper of the paver by trucks that haul the material from the batching
Advocard [28]

Answer:

a) Probability that haul time will be at least 10 min = P(X ≥ 10) ≈ P(X > 10) = 0.0455

b) Probability that haul time be exceed 15 min = P(X > 15) = 0.000

c) Probability that haul time will be between 8 and 10 min = P(8 < X < 10) = 0.6460

d) The value of c is such that 98% of all haul times are in the interval from (8.46 - c) to (8.46 + c)

c = 2.12

e) If four haul times are independently selected, the probability that at least one of them exceeds 10 min = 0.1700

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 8.46 min

Standard deviation = σ = 0.913 min

a) Probability that haul time will be at least 10 min = P(X ≥ 10)

We first normalize/standardize 10 minutes

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (10 - 8.46)/0.913 = 1.69

To determine the required probability

P(X ≥ 10) = P(z ≥ 1.69)

We'll use data from the normal distribution table for these probabilities

P(X ≥ 10) = P(z ≥ 1.69) = 1 - (z < 1.69)

= 1 - 0.95449 = 0.04551

The probability that the haul time will exceed 10 min is approximately the same as the probability that the haul time will be at least 10 mins = 0.0455

b) Probability that haul time will exceed 15 min = P(X > 15)

We first normalize 15 minutes.

z = (x - μ)/σ = (15 - 8.46)/0.913 = 7.16

To determine the required probability

P(X > 15) = P(z > 7.16)

We'll use data from the normal distribution table for these probabilities

P(X > 15) = P(z > 7.16) = 1 - (z ≤ 7.16)

= 1 - 1.000 = 0.000

c) Probability that haul time will be between 8 and 10 min = P(8 < X < 10)

We normalize or standardize 8 and 10 minutes

For 8 minutes

z = (x - μ)/σ = (8 - 8.46)/0.913 = -0.50

For 10 minutes

z = (x - μ)/σ = (10 - 8.46)/0.913 = 1.69

The required probability

P(8 < X < 10) = P(-0.50 < z < 1.69)

We'll use data from the normal distribution table for these probabilities

P(8 < X < 10) = P(-0.50 < z < 1.69)

= P(z < 1.69) - P(z < -0.50)

= 0.95449 - 0.30854

= 0.64595 = 0.6460 to 4 d.p.

d) What value c is such that 98% of all haul times are in the interval from (8.46 - c) to (8.46 + c)?

98% of the haul times in the middle of the distribution will have a lower limit greater than only the bottom 1% of the distribution and the upper limit will be lesser than the top 1% of the distribution but greater than 99% of fhe distribution.

Let the lower limit be x'

Let the upper limit be x"

P(x' < X < x") = 0.98

P(X < x') = 0.01

P(X < x") = 0.99

Let the corresponding z-scores for the lower and upper limit be z' and z"

P(X < x') = P(z < z') = 0.01

P(X < x") = P(z < z") = 0.99

Using the normal distribution tables

z' = -2.326

z" = 2.326

z' = (x' - μ)/σ

-2.326 = (x' - 8.46)/0.913

x' = (-2.326×0.913) + 8.46 = -2.123638 + 8.46 = 6.336362 = 6.34

z" = (x" - μ)/σ

2.326 = (x" - 8.46)/0.913

x" = (2.326×0.913) + 8.46 = 2.123638 + 8.46 = 10.583638 = 10.58

Therefore, P(6.34 < X < 10.58) = 98%

8.46 - c = 6.34

8.46 + c = 10.58

c = 2.12

e) If four haul times are independently selected, what is the probability that at least one of them exceeds 10 min?

This is a binomial distribution problem because:

- A binomial experiment is one in which the probability of success doesn't change with every run or number of trials. (4 haul times are independently selected)

- It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure. (Only 4 haul times are selected)

- The outcome of each trial/run of a binomial experiment is independent of one another. (The probability that each haul time exceeds 10 minutes = 0.0455)

Probability that at least one of them exceeds 10 mins = P(X ≥ 1)

= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 1 - P(X = 0)

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 4 haul times are independently selected

x = Number of successes required = 0

p = probability of success = probability that each haul time exceeds 10 minutes = 0.0455

q = probability of failure = probability that each haul time does NOT exceeds 10 minutes = 1 - p = 1 - 0.0455 = 0.9545

P(X = 0) = ⁴C₀ (0.0455)⁰ (0.9545)⁴⁻⁰ = 0.83004900044

P(X ≥ 1) = 1 - P(X = 0)

= 1 - 0.83004900044 = 0.16995099956 = 0.1700

Hope this Helps!!!

7 0
3 years ago
Plz help ASAP!! Explain your answer! I will mark at brainliest!!! And don’t copy anybody else’s answer
Klio2033 [76]

Answer:

No, it is not a square

Step-by-step explanation:

If one wall is 19", that would mean the wall perpendicular to this wall is also 19" (in fact all of the walls would be 19"!) If this was a square, then the diagonal we draw at 20.62" would serve as the hypotenuse of a right triangle.  One wall would serve as a leg, and another wall as another leg.  If this is a square, then the Pythagorean's Theorem would be satisfied when we plug in the 2 wall measures for a and b, and the diagonal for c:

19^2+19^2=20.62^2

We need to see if this is a true statement.  If the left side equals the right side, then the 2 legs of the right triangle are the same length, and the room, then is a square.

361 + 361 = 425.1844

Is this true?  Does 722 = 425.1844?  Definitely not.  That means that the room is not a square.

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3 years ago
Pls HELP, im stuck(picture included)
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Answer is x = (355*r^-35)/7 hope led me know if you have any questions

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Samia created the following tables of values for a linear system. She concluded that there is no
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Compare and contrast subtracting integers and subtracting whole numbers
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