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2 years ago

7

Ellen has a bag with 3 red marbles and 2 blue marbles in it. She is going to randomly draw a marble from the bag 300 times, putt

ing the marble back in the bag after each draw.What is the best prediction for the number of times that Ellen will draw a blue marble?

2 answers:

sweet [91]2 years ago

7 0

Answer:

No. of times Ellen draws a blue marble = 120

Step-by-step explanation:

No. of red marbles = 3

No. of blue marbles = 2

$\text{Probability of drawing a blue marble = }\frac{\text{No. of possible outcomes}}{\text{Total no. of outcomes}}\\\\P(B)=\frac{2}{5}\\\\No. of draws (n) = 300\\\text{So, total probability = }n\times P(B)\\\\=100\times \frac{2}{5}=120.$

So, the required probability is 120.

Akimi4 [234]2 years ago

4 0

Answer:

The best prediction would be close to 120, not exactly 120.

Step-by-step explanation:

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2 years ago

angel is picking mountain blueberries for a delicious pie. she can pick 1/6 cup of blueberries in 2 minutes. if she needs 2 1/2

Vadim26 [7]

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3 years ago

Multiply. <br> 8 1/2 -2 2/3

MrRissso [65]

Answer: $-\frac{68}{3}$ or $-22\ 2/3$

Step-by-step explanation:

To solve this problem you must:

- Convert the mixed numbers to fractions, as following:

$8\ 1/2=\frac{((8*2)+1)}{2}=\frac{17}{2}$

$2\ 2/3=\frac{((3(2)+2)}{3}=\frac{8}{3}$

- Multiply both fractions. Mutilply the numerators and the denominators:

$(\frac{17}{2})(-\frac{8}{3})=-\frac{136}{6}$

- Simplify. Therefore, you obtain that the answer is:

$-\frac{68}{3}$ or $-22\ 2/3$

5 0

2 years ago

Read 2 more answers

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

2 years ago

A clothing store has the sign shown in the shop window.

kicyunya [14]

Pani would pay $59 if she gets a discount of $3 on each shirt and $3 on each jeans

Step-by-step explanation:

Given

Price of shirt = $12

Price of jeans = $19

Price of shirt after $3 discount = 12-3 = $9

Price of jeans after 3 discount = 19-3 = $16

So

The total money Pani has to pay will be:

$9(3)+16(2)\\=27+32\\=\$59$

Pani would pay $59 if she gets a discount of $3 on each shirt and $3 on each jeans

Keywords: Discount, percent

Learn more about discount at:

#LearnwithBrainly

8 0

3 years ago