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3 years ago

What is 3.8 repeating as a fraction in simplest form

1 answer:

6 0

$\bf 3.8888888\overline{8}\\\\
-------------------------------\\\\
x=3.8888888\overline{8}\qquad thus\qquad 
\begin{array}{llcll}
10x&=&38.888888\overline{8}\\
&&\downarrow \\
&&35+3.8888888\overline{8}\\
&&\downarrow \\
&&35+x
\end{array}
\\\\\\
10x=35+x\implies 9x=35\implies x=\cfrac{35}{9}$

the idea being, you multiply the "x" by some power of 10 that moves the repeating part to the left of the decimal point and the split it like above.

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Pls I need help<br> Solve <br> x-4=⅓(6x-54)

Aleks04 [339]

Answer:

x=14

Step-by-step explanation:

x-4=⅓(6x-54)

x-4=2x-18

x+14=2x

x=14

5 0

3 years ago

Read 2 more answers

Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte

sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so $n = 11$

Each player with a 5% probability of injury, so $p = 0.05$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688$

$P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293$

$P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848$

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so $n = 2$

Each team has a 0.9848 probability of showing up to play, so $p = 0.9848$ .

This probability is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299$

2.99% probability that the cost will be paid by only one team

7 0

4 years ago

Can someone help me with this question?

KonstantinChe [14]

-3. Put the two equation equal to each other and solve.
4x+18=2x+12
-2x=-2x
2x+18=12
-18=-18
2x=-6
2x/2=-6/2
x=-3

8 0

3 years ago

Un dispositivo de seguridad biométrico erróneamente rechaza la admisión de 1 en 1000 personas autorizadas de una instalación que

Leona [35]

Usando probabilidad condicional, se encuentra que hay una probabilidad de 0.0186 = 1.86% que la persona esta realmente autorizada para ingresar.

------------------

Probabilidad Condicional

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

En que:

P(B|A) es la probabilidad de que ocurra el evento B, dado que A sucedió.
$P(A \cap B)$ es la probabilidad de que ocurran tanto A como B.
P(A) es la probabilidad de que ocurra el evento A.

En este problema:

Evento A: Persona rechazada.
Evento B: Autorizada.

La probabilidad de una persona ser rechazada es:

$\frac{1}{1000} = 0.001$ de 95% = 0.95(autorizada).
$\frac{1 000 000 - 1}{1 000 000} = 0.999999$ de 5% = 0.05(non autorizadas).

O sea:

$P(A) = 0.001(0.95) + 0.999999(0.05) = 0.05094995
$

La probabilidad de una persona estar autorizada y ser rechazada es:

$P(A \cap B) = 0.001(0.95)$

¿Cuál es la probabilidad que la persona esta realmente autorizada para ingresar?

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

$P(B|A) = \frac{0.001(0.95)}{0.05094995}$

$P(B|A) = 0.0186$

Probabilidad de 0.0186 = 1.86% que la persona esta realmente autorizada para ingresar.

Un problema similar es dado en brainly.com/question/18734831

3 0

3 years ago

Investigating Hollywood Movies

erik [133]

Answer:

a. Genre with highest mean is drama. Mean is 72.10

B. 10.46 is difference between means of comedy and horror

C. Horror has the lowest minimum value at 25.00

Action and comedy both have the highest maximum value at 93.00

Step-by-step explanation:

First create a table with the data provided like I did in the attachment. It would give a clearer picture of what your answers should be.

1. Which genre has the highest mean score?

From the table, we have the means as:

Action = 58.63

Comedy = 59.11

Drama = 72.10

Horror = 48.65

The genre with the highest score is drama at 72.10

2. Difference in mean score between comedy and horror

Mean of comedy = lc = 59.11

Mean of horror = ly = 48.65

Difference = lc-ly

= 59.11 - 48.65

= 10.46

3. The genre with lowest score is the genre with the lowest minimum value. This genre is horror and the lowest minimum is 25.00

The genre with highest minimum value is that whose maximum value is the highest out of all the genres. Both action and comedy have the highest maximum value at 93.00

7 0

3 years ago