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ivann1987 [24]
3 years ago
6

6.00 as a mixed number or fraction in simplest form

Mathematics
1 answer:
Andrew [12]3 years ago
7 0
Okay, are there any choices?
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Please help best answer gets brainliest
Evgesh-ka [11]
There is no question? Add a question
6 0
3 years ago
Match the following items.
Pani-rosa [81]

This problem can be solved using probability, the equation of the probability of an event A is P(A)= favorable outcomes/possible outcomes. The interception of two probable events is P(A∩B)= P(A)P(B).

There are 12 black marbels, 10 red marbles, and 18 white marbels, all the same size. If two marbles are drawn from the jar without being replaced.

The total of the marbles is 40.

If two marbles are drawn from the jar without being replaced, what would the probability be:

1. of drawing two black marbles?

The probability of drawing one black marble is (12/40). Then, the probability of drawing another black marble after that is (11/39) due we drawing one marble before.

P(Black∩Black) = (12/40)(11/39) = 132/1560, simplifying the fraction:

P(Black∩Black) = 11/130

2. of drawing a white, then a black marble?

The probability of drawing one white marble is (18/40). Then, the probability of drawing then a black marble after that is (12/39) due we drawed one marble before.

P(White∩Black) = (18/40)(12/39) = 216/1560, simplifying the fraction:

P(White∩Black) = 9/65

3. of drawing two white marbles?

The probability of drawing one white marble is (18/40). Then, the probability of drawing another white marble after that is (17/39) due we drawed one marble before.

P(White∩White) = (18/40)(17/39) = 306/1560, simplifying the fraction:

P(White∩White) = 51/260

4. of drawing a black marble, then a red marble?

The probability of drawing one black marble is (12/40). Then, the probability of drawing then a red marble after that is (10/39) due we drawed one marble before.

P(Black∩Red) = (12/40)(10/39) = 120/1560, simplifying the fraction:

P(Black∩Red) = 1/13

4 0
3 years ago
In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
Serjik [45]

Complete Question:

In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.

Answer:

The change of coordinate matrix is :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

Step-by-step explanation:

Let U =  {D, E, F} be any vector with respect to Basis B

U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)

U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)

In Matrix form;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

To find D, E, F in (**) such that U = t²

D + 2E + F = 0.................(1)

E + 2F = 0.........................(2)

-3D -5E = 1........................(3)

Substituting eqn (2) into eqn (1 )

D=3F...................................(4)

Substituting equations (2) and (4) into eqn (3)

-9F+10F=1

F = 1

Put the value of F into equations (2) and (4)

E = -2(1) = -2

D = 3(1) = 3

Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

3 0
2 years ago
So I have a question, this : Insert 2 grouping symbols to make the equation true
kotykmax [81]
Between 4 and 6 is where u put the grouping symbols!
3 0
3 years ago
A research study investigated differences between male and female students. Based on the study results, we can assume the popula
oksian1 [2.3K]

Answer: 0.0548

Step-by-step explanation:

Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.

Let \overline{X} represents the sample mean GPA for each student.

Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:

P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z

hence, the required probability is 0.0548.

6 0
3 years ago
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