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Komok [63]
2 years ago
13

Giving free brainliest

Mathematics
2 answers:
bagirrra123 [75]2 years ago
5 0

Step-by-step explanation:

Ok grate thank you

Zielflug [23.3K]2 years ago
3 0
THXXX luvv… :)))))))(()
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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
Charlie is standing 45 feet from the base of 100 foot tall tree. What is the angle of elevation from him to the kite stuck on to
galben [10]
Hey. Let me help you on this one.

In order for us to solve this problem easier, let's break it apart. We can also graph the triangle so we get visual of this problem.

I drew a quick graph for you to easier understand this question. As you can see, Charlie is standing below the tree 45 feet from it, and we need to find the unknown angle measure.

Let's form a SOHCAHTOA function which you have probably already learned in your Geometry course.

First of all let's identify our sides. The side opposite of the right triangle is a Hypotenuse, and the side that's 45 feet is adjacent, since it's adjacent to the unknown angle.

We know that we need to use Cos (Cosine) with negative power to it in order to solve this question, since we are looking for the unknown angle measure.

Let's form the function which Cos, Hypotenuse, and Adjacent side.

Cos^{-1} =45/100

Now, let's solve the left part of this function

&#10;Cos^{-1} =0.45

Awesome. You will need to use your calculator for this part. Find the value of 0.45 when negative Cosine is used on it.

Cos^{-1} =0.45
63.25631605

Awesome. Let's round this number to the closest tenth

63.3 is the rounded value we have discussed below.

Answer: 63.3 degrees is the unknown measure of the angle

5 0
3 years ago
Four less than five times a number is equal to 11
SVETLANKA909090 [29]
5*3=15
15-4= 11
hope I helped:)
7 0
3 years ago
It has been estimated that 1000 curies of a radioactive substance introduced at a point on the surface of the open sea would spr
Zanzabum
<span>1000 meters per second is the answer 

radius =25.23Km

(t)= days 

</span>
8 0
3 years ago
HELP PLSSS
Gwar [14]

Answer:

LINE D

Step-by-step explanation:

4 0
3 years ago
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