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Alexeev081 [22]
3 years ago
13

Which set of data is correct for the quadratic relation f(x) = 5(x -

Mathematics
1 answer:
Burka [1]3 years ago
6 0

Answer:

Direction parabola opens upward.

Vertex of parabola is (27,-9).

Axis of symmetry is x=27.

Step-by-step explanation:

Note: Option sets are not correct.  

The vertex form of a parabola is

y=a(x-h)^2+k    ...(1)

where, (h,k) is vertex and x=h is the axis of symmetry.

If a<0, then parabola opens downward and if a>0, then parabola opens upward.

The given function is

f(x)=5(x-27)^2-9     ...(2)

On comparing (1) and (2), we get

a=5>0, so direction parabola opens upward.

h=27,k=-9, so vertex of parabola is (27,-9).

So, axis of symmetry is x=27.

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<A = <C -----> Congruent angles
<D = <B -----> Congruent angles
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The perimeter of a square is 4 times the length of one side of the square. Which graph represents this situation? *
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Answer:

A.

Step-by-step explanation:

Graph B must be incorrect as this is a directly proportional relationship. Likewise, graph C shows an inversely proportional relationship.

Between graph A and D, only graph A passes through the origin, which is true because if length is zero, perimeter is zero.

Hence the answer is Graph A.

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Step-by-step explanation:

3 0
3 years ago
The city has an average of 13 days of rainfall for April.
zhenek [66]

Using the Poisson distribution, we have that:

  • There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
  • There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
  • There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

For this problem, the mean is given as follows:

\mu = 13

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0

P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003

P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0
2 years ago
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Butoxors [25]

Answer:

A.

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Step-by-step explanation:

Let n be the no. of copies

The cost:

5 + 3n《59

3n《54

n《18

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