MATHEMATICS COLLEGE

Is 27.14159 rational or irrational

Answers

Answer 1
Answer:

Answer:

It´s rational

Step-by-step explanation:

27,14159 = 2714159/100000

Rational

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A rectangle is shown. The length of the rectangle is labeled as 8 cm, and the width is labeled as 6 cm. What will be the perimeter and the area of the rectangle below if it is enlarged using a scale factor of 4.5? (5 points)Perimeter = 46 cm, area = 131.25 cm2
Perimeter = 126 cm, area = 972 cm2
Perimeter = 46 cm, area = 972 cm2
Perimeter = 126 cm, area = 131.25 cm2
Perimeter = 46 cm, area = 131.25 cm2
Perimeter = 126 cm, area = 972 cm2
Perimeter = 46 cm, area = 972 cm2
Perimeter = 126 cm, area = 131.25 cm2

Answers

Answer:

Perimeter = 126 cm, area = 972 cm2

Step-by-step explanation:

Rectangle perimeter:

  • P = 2(w + l)

Rectangle area:

  • A = wl

When scaled, the perimeter will change by same factor but the area by the square of same factor.

Applying to the given rectangle

Perimeter:

  • P = 2(6 + 8)(4.5) = 126 cm

Area:

  • A = 6*8*4.5² = 972 cm²

Correct choice is B

Answer:

Perimeter = 126cm

Area = 972cm^2

Step-by-step explanation:

6*4.5 = 27

8*4.5 = 36

Perimeter = 27*2+36*2= 126 cm

Area = 27*36 = 972 cm^2

Hope this helped!

Freddie is as half as tall as his mother .Freddie is one meter shorter than his father.
How tall is Freddie's mother?

Answers

Hi there! Freddie's mom is 1/2 a meter tall.

Answer:

0.25 1/2 of 1/2 is 0.25 sorry i just can't. studie peps. i want u to be this not this^﹏^

Step-by-step explanation:

-_--_--_--_--_-

) A patient drank 12 ounces of orange juice. How many milliliters did the patient drink?

Answers

Answer:

He drank 354.882 mills of orange juice

Step-by-step explanation: One ounce is equal to 29.5735 mills, so you multiply 29.5735 by 12

Answer: 355 Millileters

Write a function in terms of t that represents the situation.A population of 210,000 increases by 12.5% each year

y=

Answers

The exponential function is given by y = 210000(1.125)^t \n

exponential function

An exponential function is in the form:

y=ab^x

Where a is the initial value of y and b is the multiplier factor

Let y represent the population after t years.

Given an initial population of 210000, hence:

  • a = 210000

The population increases by 12.5% each year, hence:

  • b = 100% + 12.5% = 112.5% = 1.125

The exponential function is given by y = 210000(1.125)^t \n

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A municipal bond service has three rating categories (A, B, and C). Suppose that in the past year, of the municipal bonds issued throughout the United States, 70% were rated A, 20% were rated B, and 10% were rated C. Of the municipal bonds rated A, 50% were issued by cities, 40% by suburbs, and 10% by rural areas. Of the municipal bonds issued B, 60% were issued by cities, 20% by suburbs, and 20% by rural areas. Of the municipal bonds rated C, 90% were issued by cities, 5% by suburbs, and 5% by rural areas. a. If a new municipal bond is to be issued by a city, what is the probability that it will receive an A rating? b. What proportion of municipal bonds are issued by cities? c. What proportion of municipal bonds are issued by suburbs?

Answers

a) The probability that a new municipal bond issued by a city will receive an A rating is 0.625 or 62.5%.

b) 56% of municipal bonds are issued by cities.

c) The proportion of municipal bonds issued by suburbs is 0.325 or 32.5%.

Let's solve each part of the problem:

a. If a new municipal bond is to be issued by a city, what is the probability that it will receive an A rating?

Use conditional probability here.

Using conditional probability notation, we have:

P(A | City)

To calculate this, use the following formula:

P(A | City) = P(A and City) / P(City)

We are given:

- P(A) = 0.70 (probability of an A rating)

- P(B) = 0.20 (probability of a B rating)

- P(C) = 0.10 (probability of a C rating)

For bonds issued in cities:

- P(City | A) = 0.50 (probability that it's a city if it's rated A)

- P(City | B) = 0.60 (probability that it's a city if it's rated B)

- P(City | C) = 0.90 (probability that it's a city if it's rated C)

Now, let's calculate:

P(A and City) = P(A) * P(City | A)

P(City) = P(A) * P(City | A) + P(B) * P(City | B) + P(C) * P(City | C)

Substitute the values:

P(A and City) = 0.70 * 0.50

                      = 0.35

P(City) = (0.70 * 0.50) + (0.20 * 0.60) + (0.10 * 0.90)

          = 0.35 + 0.12 + 0.09

          = 0.56

Now, calculate the conditional probability:

P(A | City) = P(A and City) / P(City)

                = 0.35 / 0.56

                = 0.625

So, the probability is 0.625 or 62.5%.

b. What proportion of municipal bonds are issued by cities?

56% of municipal bonds are issued by cities.

c. What proportion of municipal bonds are issued by suburbs?

To find the proportion of municipal bonds issued by suburbs,  use a similar approach:

P(Suburb) = P(A) * P(Suburb | A) + P(B) * P(Suburb | B) + P(C) * P(Suburb | C)

We are given:

- P(Suburb | A) = 0.40

- P(Suburb | B) = 0.20

- P(Suburb | C) = 0.05

Now, calculate:

P(Suburb) = (0.70 * 0.40) + (0.20 * 0.20) + (0.10 * 0.05)

                 = 0.28 + 0.04 + 0.005

                 = 0.325

So, the proportion of municipal bonds issued by suburbs is 0.325 or 32.5%.

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Final answer:

The probability that a municipal bond issued by a city will receive an A rating is 35%. The proportion of all municipal bonds issued by cities is 56%. The proportion of all municipal bonds issued by suburbs is 32.5%.

Explanation:

This question requires an understanding of probability and conditional probability.

a) To find the probability that a new municipal bond issued by a city will receive an A rating, we must first determine the likelihood that an A-rated municipal bond is issued by a city. Given that 50% of A-rated bonds are issued by cities and that 70% of all bonds receive an A rating, we can calculate this probability as (0.50)*(0.70) = 0.35, or 35%.

b) To find the proportion of municipal bonds issued by cities, we must add up the bonds issued by cities across all ratings. So, (0.70*0.50) + (0.20*0.60) + (0.10*0.90) = 0.35 + 0.12 + 0.09 = 0.56, or 56%.

c) To calculate the proportion of municipal bonds issued by suburbs, we do the same calculation as in part b) but for suburbs. So, (0.70*0.40) + (0.20*0.20) + (0.10*0.05) = 0.28 + 0.04 + 0.005 = 0.325, or 32.5%.

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Use the Gram-Schmidt process to find an orthonormal basis for the subspace of R4 spanned by the vectors (1, 0, 1, 1), (1, 0, 1, 0), (0, 0, 1, 1).

Answers

Answer:

$ e_1 = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}         $ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}        $ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{2}}}{\textbf{2}}\n\n0\end{pmatrix}

Step-by-step explanation:

we have to orthonormalize the vectors:

v_1 = \begin{pmatrix} 1 \n 0 \n 1 \n 1 \end{pmatrix}    v_2 = \begin{pmatrix} 1 \n 0 \n 1 \n 0 \end{pmatrix}      $ v_3 = \begin{pmatrix} 0 \n 0 \n 1 \n 1 \end{pmatrix}

According to Gram - Schmidt process, we have:

u_k = v_k - \sum_(j = 1) ^ {k - 1} proj_(uj) (v_k) where, $ proj_u (v) = (u . v)/(u . u)u

The normalized vector is: $ e_k = (u_k)/(√(u_k.u_k)) $

Now, the first step.

v_1 = \begin{pmatrix} 1 \n 0 \n 1 \n 1 \end{pmatrix} = u₁

Therefore, e₁ = $ (u_1)/(√(u_1.u_1)) $

$ = \begin{pmatrix}\frac{\sqrt{\textbf{3}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\n\n\frac{\sqrt{\textbf{3}}}{\textbf{3}}\end{pmatrix}

Now, we find e₂.

$ u_2 = v_2 - (u_1.v_2)/(u_1.u_1)u_1 $

$ = \begin{pmatrix} (1)/(3)\n \n 0 \n\n (1)/(3)\n\n (-2)/(3) \end{pmatrix}

Therefore, $ e_2 = (u_2)/(√(u_2.u_2)) $

$ e_2 = \begin{pmatrix}\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{6}}}{\textbf{6}}\n\n\frac{\sqrt{\textbf{-6}}}{\textbf{3}}\end{pmatrix}

To find e₃:

$ u_3 = v_3 - (u_1. v_3)/(u_1.u_1)u_1 - (u_2. v_3)/(u_2.u_2) u_2 $

$ = \begin{pmatrix} (-1)/(2) \n\n 0\n \n (1)/(2) \n\n 0 \n\end{pmatrix}

$ e_3 = (u_3)/(√(u_3.u_3)) $

$ e_3 = \begin{pmatrix}\frac{\sqrt{\textbf{-2}}}{\textbf{2}}\n\n\textbf{0} \n\n\frac{\sqrt{\textbf{2}}}{\textbf{2}}\n\n0\end{pmatrix}

So, we have the orthonormalized vectors $ e_1, e_2, e_3 $.

Hence, the answer.

Final answer:

To find the orthonormal basis using the Gram-Schmidt process, we calculate the first vector by dividing the first given vector by its magnitude and normalize it. Then, we subtract the projection of each subsequent vector onto the previously found orthonormal vectors and normalize the resulting vector.

Explanation:

To find an orthonormal basis for the subspace of R4 spanned by the given vectors using the Gram-Schmidt process, we will start by finding the first vector of the orthonormal basis. Let's call the given vectors v1, v2, and v3, respectively. The first vector of the orthonormal basis, u1, is equal to v1 divided by its magnitude, which is ||v1||. So, u1 = v1 / ||v1||. We can calculate ||v1|| as √(1^2 + 0^2 + 1^2 + 1^2) =  √3.

Therefore, u1 = (1/√3, 0/√3, 1/√3, 1/√3).

Now, we need to find u2, the second vector of the orthonormal basis. To do this, we subtract the projection of v2 onto u1 from v2, then divide the result by its magnitude. We calculate the projection of v2 onto u1 as proj_u1(v2) = u1 * dot(u1, v2), where dot(u1, v2) represents the dot product of u1 and v2.

Finally, we subtract proj_u1(v2) from v2 to get v2' = v2 - proj_u1(v2), and then normalize v2' to get u2 = v2' / ||v2'||.

We can repeat this process to find u3, the third vector of the orthonormal basis. Subtract proj_u1(v3) and proj_u2(v3) from v3, then normalize the result to get u3 = v3' / ||v3'||.

Therefore, the orthonormal basis for the subspace spanned by the given vectors is (1/√3, 0/√3, 1/√3, 1/√3), (0, 0, 0, 1), and (-1/√3, 0/√3, 1/√3, 1/√3).

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