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evablogger [386]

3 years ago

8

A bag contains 3 red marbles 4 blue marbles and 7 Green marbles if two marbles are drawn out of the bag what is the probability

to the nearest 1000th that both marbles drawn will be red

1 answer:

mr_godi [17]3 years ago

3 0

Answer:

0.368

Step-by-step explanation:

Number of red marbles= 3

Number of blue marbles= 4

Number of green marbles= 7

Total number of marbles= 3+4+7= 14

Let the first ball is taken out to be red

P(R)= $\frac{3}{14}$

than again a second ball is drawn, number of balls left= 14-1=13

Number of red= 2

P(R)= $\frac{2}{13}$

Total probability= $\frac{3}{14}$ + $\frac{2}{13}$

Total probability= 0.368

Hence, the correct answer is 0.368

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Please please help. Marking Brilliant for the correct answer

NISA [10]

Answer:

x = 19

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

(x-7+4) * 4 = 8^2

(x-3)*4 = 64

Divide each side by 4

x-3 =16

Add 3 to each side

x-3+3 = 16+3

x = 19

5 0

3 years ago

In the graph below, the hyperbola gets close to the red lines but never touches

soldier1979 [14.2K]

Answer:

Option A asymptote.

Step-by-step explanation:

Asymptotes are lines to which the graph approaches very it can come very close to it but will never touch it. Asymptote are the limits of a graph.

IN the given graph of Hyperbola the two blue lines representing the Hyperbola are coming too close to the red lines but not touching it .

The two red lines are called the Asymptotoes of the Hyprbola.It is the limit of the Hyperbola.

Step-by-step explanation:

7 0

3 years ago

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

$y'=1-2x$

It can be written as

$\frac{dy}{dx}=1-2x$

Use variable separable method to solve this differential equation.

$dy=(1-2x)dx$

Integrate both the sides.

$\int dy=\int (1-2x)dx$

$y=x-2(\frac{x^2}{2})+C$ $[\because \int x^n=\frac{x^{n+1}}{n+1}]$

$y=x-x^2+C$ ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

$-2=1-(1)^2+C$

$-2=1-1+C$

$-2=C$

The value of C is -2. Substitute C=-2 in equation (1).

$y=x-x^2-2$

Therefore the solution of the given initial value problems in explicit form is $y=x-x^2-2$ .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.

4 0

3 years ago

In a population distribution, a score of x=57 corresponds to z=-0.25 and a score of x=87 corresponds to z=1.25. Find the mean an

Serggg [28]

Here is dependence between scores and x-values:

$Z_i=\dfrac{X_i-\mu}{\sigma},$

where $\mu$ is the mean, $\sigma$ is standard deviation and i changes from 1 to 2.

1. When i=1, $Z_1=-0.25,\ X_1=57,$ then

$-0.25=\dfrac{57-\mu}{\sigma}.$

2. When i=2, $Z_2=1.25,\ X_2=87,$ then

$1.25=\dfrac{87-\mu}{\sigma}.$

Now solve the system of equations:

$\left\{\begin{array}{l}-0.25=\dfrac{57-\mu}{\sigma}\\ \\1.25=\dfrac{87-\mu}{\sigma}.\end{array}\right.$

$\left\{\begin{array}{l}-0.25\sigma=57-\mu\\ \\1.25\sigma=87-\mu.\end{array}\right.$

Subtract first equation from the second:

$1.25\sigma-(-0.25\sigma)=87-57,\\ \\1.5\sigma=30,\\ \\\sigma=20.$

Then

$1.25=\dfrac{87-\mu}{20},\\ \\87-\mu=25,\\ \\\mu=87-25=62.$

Answer: the mean is 62, the standard deviation is 20.

3 0

3 years ago

Read 2 more answers

Tara combines 7.49 ounces of strawberries with 6.26 ounces of blueberries to make fruit bowls. She pours the fruit equally into

kupik [55]

Answer:

3.33 ounces in each bowl

Step-by-step explanation:

7.49 + 6.26 = 13.75

subtract 0.43 and that equals 13.32

13.32 ÷ 4 = 3.33

I hope this helped :/

6 0

3 years ago