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2 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]2 years ago

8 0

Answer:

33.333333% (the 3 goes on forever)

You might be interested in

Select the correct answer from each drop-down menu.

vazorg [7]

Answer:

Step 1: Distribute $-2$ to $5x$ and $8$

Step 2: Subtract from both sides of the equation $6x$

Step 3: Add to both sides of the equation $16$

Step 4: Divide both sides of the equation by $-16$

Step-by-step explanation:

Step 1: Apply the Distributive Property. Then you must distribute $-2$ to $5x$ and $8$

Then:

$-2(5x+8)=14+6x\\\\-10x-16=14+6x$

Step 2: You must apply the Subtraction property of Equality and subtract $6x$ from both sides of the equation. Then:

$-10x-16-6x=14+6x-6x\\\\-16x-16=14$

Step 3: You must apply the Addition property of Equality and add $16$ to both sides of the equation. Then:

$-16x-16+16=14+16\\\\-16x=30$

Step 4: You must apply the Division property of Equality and divide both sides by $-16$ . Then:

$\frac{-16x}{-16}=\frac{30}{-16}\\\\x=\frac{30}{-16}\\\\x=-\frac{15}{8}$

6 0

3 years ago

Wha bout dis one????/

PolarNik [594]

6-3=3, so m=3. Easy as pie

5 0

2 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers

Find the value of each trigonometric ratio!! Pls hurry!!!

Citrus2011 [14]

remember to use SOH CAH TOA. THE SOLUTION THAT I GOT CAME OUT AS 15/8

7 0

3 years ago

Can someone help meWhat is the missing numerator?

galina1969 [7]

Answer:

d 2

Step-by-step explanation:

x/7 +13/14 = 17/14

double the first one so all the denominators are the same

2x/14+13/14=17/14

then subtract the second fraction from the thrid

2x/14=4/14

then we can divide by two so we are back to 7 as the demoninator

x/7=2/7

5 0

2 years ago

Read 2 more answers