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3 years ago

15

Use the drop down menus to complete each statement

2 answers:

maw [93]3 years ago

8 0

what can you explain your question a bit more

saw5 [17]3 years ago

4 0

Answer:

need more information please

Step-by-step explanation:

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Regina has a bag of marbles that contains 3 blue marbles, 4 red marbles, and 5 yellow marbles. She draws one marble, then replac

iVinArrow [24]

Answer:

Possible outcomes=12

$P(\text{Blue marble first})=\frac{1}{4}$

$P(\text{Yellow marble})=\frac{5}{12}$

$P(\text{ Blue then yellow})=\frac{5}{48}$

Step-by-step explanation:

We can see these two events are independent as Regina replaced marble after drawing one marble.

We can find possible outcomes by adding number of blue, red and yellow marbles.

$\text{Possible outcomes}=3+4+5=12$

Let us find probability of getting a blue marble on first draw.

$P(\text{Blue marble})=\frac{3}{(3+4+5)} =\frac{3}{12} =\frac{1}{4}$

Now we will find probability of getting yellow marble.

$P(\text{Yellow marble})=\frac{5}{(3+4+5)} =\frac{5}{12}$

We can find probability of getting a blue marble and then yellow marble by multiplying both probabilities.

$P(\text{ Blue then yellow})=\frac{1}{4}* \frac{5}{12} =\frac{5}{48}$

Therefore, probability of getting a blue marble and then yellow marble is $\frac{5}{48}$ .

6 0

4 years ago

Read 2 more answers

In the 30-60-90 triangle below, side s has a length of<br> hypotenuse has a length of

PSYCHO15rus [73]

Answer:

Side s has a length of Root 3

Hypotenuse has a length of 2

Hope it helps.............

8 0

3 years ago

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Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

so that the left side condenses into the derivative of a product.

$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

Integrate both sides, then solve for $v$ , then for $y$ :

$x^3v=\displaystyle\int3x^2\,\mathrm dx$

$x^3v=x^3+C$

$v=1+\dfrac C{x^3}$

$y^3=1+\dfrac C{x^3}$

$\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}$

6 0

3 years ago

Find the sum and product of the roots.

xz_007 [3.2K]

Answer:

sum= -3 product=-4

Step-by-step explanation:

2x^2-2x+8x-8

2x(x-1)+8(x-1)

(2x+8)(x-1)

x=-4 x=1

8 0

4 years ago

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Graph f(x) = –2|x + 4|<br><br> Khan academy

koban [17]

Answer:

i cant even see the graph

7 0

3 years ago

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