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3 years ago

11

In a bag there are 3 red marbles, 2 yellow marbles, and 1 blue marble. What is the percent chance of a yellow marble being selec

ted on the first draw?
Group of answer choices

1 answer:

aksik [14]3 years ago

8 0

Answer:

33.333333% (the 3 goes on forever)

You might be interested in

What are the domain restrictions of q2−3q−4q2−7q−8?

ioda

Answer:

{xer}

Step-by-step explanation:

It would be greatly appreciated if you gave me brainlest

8 0

3 years ago

RUDIKE [14]

<em>The correct expressions are as follows:</em>

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $343$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}$

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall following formula about Exponents and Surds:

$\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }$

$\boxed { (a ^ b) ^ c = a ^ { b . c } }$

$\boxed {a ^ b \div a ^ c = a ^ { b - c } }$

$\boxed {\log a + \log b = \log (a \times b) }$

$\boxed {\log a - \log b = \log (a \div b) }$

<em>Let us tackle the problem!</em>

$\texttt{ }$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7^2)^{\frac{7}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7)^{2\times \frac{7}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5} + \frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{15}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{3}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{343}$

$\texttt{ }$

<em>From the results above, it can be concluded that the correct statements are:</em>

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $343$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}$

$\texttt{ }$

<h3>Learn more</h3>

Coefficient of A Square Root : brainly.com/question/11337634
The Order of Operations : brainly.com/question/10821615
Write 100,000 Using Exponents : brainly.com/question/2032116

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent , Perfect , Square , Factor.

4 0

4 years ago

Read 2 more answers

of the 650 students at WJH 4% will receive a perfect attendance award. How many students ill receive the award?

Keith_Richards [23]

Its 26
650 x 4 = 2600
2600 ÷ 100 = 26

7 0

3 years ago

Mr lance designed a class banner shaped like a the polygon shown. What is the name of the polygon?

Scorpion4ik [409]

Answer:

hexagon

Step-by-step explanation:

i g

7 0

3 years ago

Question1) Describe three scenarios that involve a real-world linear or exponential function. At least one must be exponential.

elixir [45]

Answer:

1) Let's suppose that you go in a straight line, in a car that moves at a constant speed of 80km/h.

Then the distance from your house (assuming that you start the drive in your house) can be modeled with a linear equation:

D(t) = 80km/h*t

where t is time in hours.

This will be a linear function.

2) Suppose that you have a population of some animal, that grows by 2% each month, and initially, there are 100 individuals of that animal.

Then the first month, the population is 100.

The second month the population increased by a 2%, then it will be:

100 + 100*0.02 = 100*(1.02)

The third month, the population will be 100*(1.02) + 0.2*100*(1.02) = 100*(1.02)^2.

and so on, this is an exponential relation, where the population as a function of the number of months, can be written as:

P(m) = 100*(1.02)^(m - 1)

3) Suppose that you have $100 saved, and each month you can save another $80, let's find a function that says the amount of money that you have saved as a function of the number of months. S(m)

The month number zero (before you started saving) you had $100 saved.

S(0) = $100.

One month after, you have saved $80 more, then you have:

S(1) = $100 + $80

Another month after, you have:

S(2) = $100 + $80 + $80 = $100 + 2*$80

And so on, you already can see the pattern, after m months, you will have:

S(m) = $100 + m*$80 saved.

5 0

3 years ago