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2 years ago

Can the union of two sets have a solution of a null set?why or why not?

Mathematics

1 answer:

sveticcg [70]2 years ago

3 0

Let's assume we have two sets

First set is A

second set is B

we know that

Union of two sets is the set that contains elements or objects that belong to either A or to B or to both.

So, it does not leave any element from any set

So, for union of two sets to become null , sets A and B won't have any element

For example:

set A={ }

set B= { x : for all real value of x ,x^2 +1=0 }

Since, set B won't have any elements

so,

A∪B=Ф

Yes, the union of two sets have a solution of a null set........Answer

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Stephanie can type around 70 words per minute. She has less than a week to type her final research paper. If the paper has to be

Lana71 [14]

The answer is 1.25 hours.

STEP 1.
Each page has about 350 words: 1 page - 350 words
15 pages is x words: 15 pages - x words
1 : 350 = 15 : x
x = 15 * 350 : 1 = 5250

So, she has to type at least 5250 words.

STEP 2:
Stephanie can type around 70 words per minute: 70 words - 1 minute
x words in 1 hour = 60 minute: x words - 60 minute

70 : 1 = x : 60
x = 70 * 60 : 1 = 4200

So, she types around 4200 words per hour.

STEP 3:
Now we have:
- she has to type at least 5250 words
- she types around 4200 words per hour

5250 : x = 4200 : 1
x = 5250 * 1 : 4200 = 1.25 hours

6 0

3 years ago

What are the measures of angles M and N? m m m m < M = 119° and m

NNADVOKAT [17]

Answer:

m<M = 113 deg

m<N = 61 deg

Step-by-step explanation:

In an inscribed quadrilateral, opposite angles are supplementary.

m<M + m<K = 180

m<N + m<L = 180

m<M + 67 = 180

m<M = 113

m<N + 119 = 180

m<N = 61

7 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

I have a few questions!

Sedaia [141]

Answer:

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Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

ASAPPP!!!! HELP I WILL GIVE 100POINTS AND BRAINLIST Question 9 (Essay Worth 10 points)

Luda [366]

Answer:

Part A)

The x-intercepts are (-1/4, 0) and (4, 0).

Part B)

The vertex is a maximum because the leading coefficient is negative.

The vertex is (1.875, 72.25).

Part C)

We can plot the zeros and the vertex, and connect them with a curve.

Step-by-step explanation:

The function given is:

$f(x)=-16x^2+60x+16$

Part A)

To find the x-intercepts of the function, set the function equal to 0 and solve for x:

$0=-16x^2+60x+16$

We can divide both sides by negative four:

$0=4x^2-15x-4$

Factor:

$0=(4x+1)(x-4)$

Zero Product Property:

$x-4=0\text{ or } 4x+1=0$

Solve for each case:

$\displaystyle x=-\frac{1}{4}\text{ or }x=4$

Hence, our zeros are:

$\displaystyle \left(-\frac{1}{4}, 0\right)\text{ and } \left(4, 0\right)$

Part B)

Note that the leading coefficient of our function is negative.

So, our function will be concave down.

Hence, our vertex will be the maximum.

The vertex is given by:

$\displaystyle \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$

In this case, a = -16, b = 60, and c = 16.

Find the x-coordinate of the vertex:

$\displaystyle x=-\frac{(60)}{2(-16)}=\frac{15}{8}=1.875$

Substitute this back into the function to find the y-coordinate:

$f(1.875)=-16(1.875)^2+60(1.875)+16=72.25$

Hence, our vertex is:

$(1.875, 72.25)$

Part C)

Since we already determined the zeros and the vertex, we can plot the two zeros and the vertex and draw a curve between the three points.

The graph is shown below. Again, to do this by hand, simply plot the three points and connect them with a parabola. If necessary, we can also find the y-intercept.

7 0

2 years ago