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4 years ago

14

A bag contains 7 red marbles, 5 yellow marbles, 6 blue marbles, 4 green marbles, and 3 orange marbles. What is the probability o

f randomly selecting a yellow marble out of the bag?

2 answers:

Trava [24]4 years ago

8 0

Answer:

1/5

Step-by-step explanation:

baherus [9]4 years ago

6 0

The probability of randomly selecting a yellow marble out of the bag is $\frac{1}{5}$

Step-by-step explanation:

A bag contains:

7 red marbles
5 yellow marbles
6 blue marbles
4 green marbles
3 orange marbles

We need to find the probability of randomly selecting a yellow marble out of the bag

Probability is the ratio of number of favorable outcomes to the total number of possible outcomes P(A) = $\frac{n(A)}{n(outcoms)}$

∵ A bag contains 7 red marbles, 5 yellow marbles, 6 blue

marbles, 4 green marbles, and 3 orange marbles

- Add all the color to find the number of total marbles

∴ n(all) = 7 + 5 + 6 + 4 + 3 = 25

∵ There are 5 yellow marbles

∵ P(yellow) = $\frac{n(yellow)}{n(all)}$

∴ P(yellow) = $\frac{5}{25}$

- Divide up and down by 5 to simplify the fraction

∴ P(yellow) = $\frac{1}{5}$

The probability of randomly selecting a yellow marble out of the bag is $\frac{1}{5}$

Learn more:

You can learn more about the probability in brainly.com/question/9178881

#LearnwithBrainly

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Natalka [10]

Answer:

1. 3/5

2. 1.219841

3. 2 - f(2)/7

4. -1.9682

5. 1.502446

Step-by-step explanation:

1.

If we call

$\large f(x)=x^4-x-2$

then the second approximation to the root of the equation f(x)=0 would be

$\large x_2=x_1-\frac{f(x_1)}{f'(x_1)}$

$\large f(x_1)=f(1)=-2\\\\f'(x)=4x^3-1\Rightarrow f'(x_1)=f'(1)=3$

hence

$\large x_2=1-\frac{-2}{3}=1+\frac{2}{3}=\frac{5}{3}\approx1.666666$

2.

Here we have

$\large f(x)=x^4-2x^3+7x^2-9\\\\f'(x)=4x^3-6x^2+14x$

Let's start with

$\large x_1=1$

then

$\large x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{f(1)}{f'(1)}=1-\frac{-3}{12}=1+\frac{1}{4}=\frac{5}{4}=1.25$

$\large x_3=x_2-\frac{f(x_2)}{f'(x_2)}=1.25-\frac{f(1.25)}{f'(1.25)}=1.220343137$

$\large x_4=x_3-\frac{f(x_3)}{f'(x_3)}=1.220343137-\frac{f(1.220343137)}{f'(1.220343137)}=1.219841912$

$\large x_5=x_4-\frac{f(x_4)}{f'(x_4)}=1.219841912-\frac{f(1.219841912)}{f'(1.219841912)}=1.219841771$

Since the first 6 decimals of $\large x_4$ and $\large x_5$ are equal, the desired approximation is 1.219841

3.

If the line y = 4x − 1 is tangent to the curve y = f(x) when x = 2, then f'(2) = 4*2 - 1 = 7, so

$\large x_2=2-\frac{f(2)}{f'(2)}=2-\frac{f(2)}{7}$

4.

Here

$\large f(x)=x^5 + 8\\\\f'(x)=5x^4$

$\large x_2=-1-\frac{f(-1)}{f'(-1)}=-2.4$

$\large x_3=-2.4-\frac{f(-2.4)}{f'(-2.4)}=-1.9682$

5.

We want to find all the values x such that

$\large e^x=9-3x$

or what is the same, the x such that

$\large e^x+3x-9=0$

so, let f(x) be

$\large f(x)=e^x+3x-9$

and let's use Newton's method to find the roots of f(x).

Since

$\large f'(x)=e^x +3>0$

f is strictly increasing, and since

f(1) = e+3-9 = e - 6 < 0

and

$\large f(2)=e^2+6-9=e^2-3>0$

f has only one root in [1,2]

By using Newton's iterations starting with $\large x_1=1$

$\large x_2=1-\frac{f(1)}{f'(1)}=1.573899431$

$\large x_3=1.573899431-\frac{f(1.573899431)}{f'(1.573899431)}=1.503982961$

$\large x_4=1.503982961-\frac{f(1.503982961)}{f'(1.503982961)}=1.502446348$

$\large x_5=1.502446348-\frac{f(1.502446348)}{f'(1.502446348)}=1.50244564$

$\large x_6=1.50244564-\frac{f(1.50244564)}{f'(1.50244564)}=1.50244564$

Since $\large x_5=x_6$ then

x=1.50244564 is the desired root.

The answers as a comma-separated list would be

1, 1.573899431, 1.503982961, 1.502446348, 1.50244564

The answer rounded to six decimal places would be

1.502446

6 0

3 years ago