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2 years ago

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Gene is playing a game with a bag of marbles. 3 of the marbles are blue, 4 are green, and 7 are yellow. See below for awarded pr

izes. $2 green $0.5 yellow $4 blue What is the expected cost (or payout for Gene's game?

1 answer:

jek_recluse [69]2 years ago

5 0

Answer:

<h2><em>$23.5</em></h2>

Step-by-step explanation:

Gene is playing a game with a bag of marbles. If 3 of the marbles are blue, 4 are green, and 7 are yellow and awarded prices for the marbles are $2 green $0.5 yellow $4 blue, the expected payout for Gens game is expressed as shown;

If a blue marble costs $4, 3 blue marbles will cost 3*$4 = $12

If a green marble costs $2, 4 green marbles will cost 4*$2 = $8.0

If a yellow marble costs $0.5, 7 yellow marbles will cost 7*$0.5 = $3.5

Total payout for Gene's game will be the equivalent to $12+ $8 + $3.5 = $23.5.

<em>Hence Gene expected cost will be $21.5</em>

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One mole of iron (6 × 1023 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-c

MrMuchimi

Answer:

a) k $_{s}$ = 3.6 N/m ; b) 9.43 * 10¹²; c) 1.18 * 10¹¹; d) 75.08 N/m

Step-by-step explanation:

Length of iron bar = L = 2.7m;

side length of cross section = a = 0.07cm = 0.0007m;

x = 2.7 cm = 0.027m;

m = 100kg;

ρ = 7.87gm/cm³;

da = 2.28 * 10⁻¹⁰m;

a)

Fnet = F - mg

where Fnet = 0

So,

F = mg where F=k $_{s}$ x

k $_{s}$ = mg/x = 100*9.8/0.027

k $_{s}$ = 3.6 N/m

b)

Nchain = Aw/Aa =a²/(da)²

= (o.ooo7)²/(2.28 * 10⁻¹⁰)²

= 9.43 * 10¹²

c)

Nbond = L/da = 2.7/2.28 * 10⁻¹⁰

= 1.18 * 10¹¹

d)

Spring stiffness of wire = ksi = (Nbondk $_{s}$ )/Nchain

= [(1.18* 10¹¹)(6*10⁴)]/(9.43 * 10¹²)

= 75.08 N/m

8 0

2 years ago

Is this a percent increase or decrease? How do you know? 5 cups to 8 cups

kicyunya [14]

Answer:

increase

Step-by-step explanation:

Increase definition: become or make greater in size, amount, intensity, or degree.

Decrease definition: make or become smaller or fewer in size, amount, intensity, or degree.

5 0

2 years ago

(X^2+y^2+x)dx+xydy=0<br> Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

<em>is</em> exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of <em>x</em>, and dividing this by $N(x,y)=xy$ gives an expression independent of <em>y</em>. If we assume $\mu=\mu(x)$ is a function of <em>x</em> alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by <em>x</em> on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution <em>F</em> satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to <em>x</em> gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to <em>y</em> gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

Identify an equation in point-slope form for the line perpendicular to y = –4x – 1 that passes through (–2, 7).

Harlamova29_29 [7]

To get a perpendicular line, your slopes will equal -1 when multiplied...this means if we multiply - 4 x (1/4), we get out slope to use for the next step.

Y - Y1 = M(X - X1) passes through (-2, 7)
Y - 7 = 1/4(X - (-2))
Y - 7 = 1/4(X + 2)
Y - 7 = 1/4X + 1/2
+7 + 7
Y = 1/4X + 7.5

6 0

2 years ago

Read 2 more answers

Mathematics <br> Simple events likelihood experimental probability

saw5 [17]

Problem 1

Answer: Closer to 1

Explanation:

There are 20 gumballs total. Half of this is 20/2 = 10 gumballs. If there's more than 10 of one color, then the probability of getting that color is closer to 1, than it is to 0. Here we have 12 pink which is greater than 10, so that's why the answer is closer to 1.

================================================

Problem 2

Answer: Closer to 0

Explanation:

The amount of green (3) is less than 10, so that's why the probability is closer to 0 than it is to 1. We can see that 3/20 = 0.15 is less than 0.50

================================================

Problem 3

Answer: Closer to 0

Explanation:

We have a similar situation compared to problem 2. This time we have 5/20 = 0.25 which is less than 0.50

================================================

Problem 4

Answer: 48% chance; fairly likely

Explanation:

We have 12 green out of 25 total, so the probability of choosing green is 12/25 = 0.48 = 48%. While this probability is not over 50%, I still say it's fairly likely considering the other colors lead to smaller probabilities. For instance, purple has a chance of 6/25 = 0.24 = 24% and orange has a probability of 2/25 = 0.04 = 4%, both of which are smaller than 48%

3 0

2 years ago