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frozen [14]
3 years ago
8

Madilyn has 2 triangular vegetable gardens. One measures

Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer:

Gamma

Step-by-step explanation:

g

You might be interested in
How many ways can we pick three distinct integers from {1, 2, ..., 15} such that their sum is divisible by 3?
gogolik [260]

Answer:

Clearly from 1 to 15 we can only have the sum 6,9,12 and 15 to be divisible by 3.

Now meaning we have 4! ( 4 factorial) which implies 4x3x2x1= 24 .

We have 24 ways to pick distinct integers that are divisible by 3.

Step-by-step explanation:

The first thing you need to do is to look for numbers that are divisible by 3. Then you take its factorial

5 0
3 years ago
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
Toya can run one lap of the track in 1 min. 3 sec., which is 90% of her younger sister niki's time. what is niki's time for one
Alex
Hello!

1 minute 3 seconds = 63 seconds

We must found out 63 is 90% of what number, using the percentage formula:

90 * 100 / 63 = 70

70 seconds = 1 minute 10 seconds

So, Niki's time for one lap is 1 minute 10 seconds (70 seconds).
5 0
3 years ago
Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3
dezoksy [38]

Answer:

Step-by-step explanation:

4/x + 4/(x²-9) = 3/(x - 3) 

4 / x + 4 / [( x - 3) ( x + 3 )] = 3 / ( x - 3 )  / * x ( x - 3 ) ( x + 3 )

Restrictions : x ≠ 0,  x ≠ - 3 ,  x ≠ 3;

4 ( x + 3 ) ( x - 3 ) + 4 x = 3 x ( x + 3 )

4 ( x² - 9 ) + 4 x = 3 x² + 9 x

4 x² - 36 + 4 x - 3 x² - 9 x = 0

x² - 5 x - 36 = 0

x² - 9 x + 4 x - 36 = 0

x ( x - 9 ) + 4 ( x - 9 ) = 0

( x - 9 ) ( x + 4 ) = 0

x - 9 = 0,  or : x + 4 = 0

Answer:

x = 9,  x = - 4

4 0
3 years ago
Which property justifies the following equation?
Leviafan [203]
1. . D . Identity , when you odd both properties it will always to be true on both sides
5 0
3 years ago
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