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3 years ago

7

Alice, Bernice, and Cheryl shared some stamps in the ratio 8 : 9 : 18. Alice received 184 stamps. Find the number of stamps that

Bernice and Cheryl each received.

1 answer:

faust18 [17]3 years ago

6 0

184/8=23
8=alice
9=bernice
18=cheryl

8:9:18=184:207:414

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The editor of a textbook publishing company is deciding whether to publish a proposed textbook. Information on previous textbook

kogti [31]

Answer:

34.86% probability that it will be huge success

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Receiving a favorable review.

Event B: Being a huge success.

Information on previous textbooks published show that 20 % are huge successes

This means that $P(B) = 0.2$

99 % of the huge successes received favorable reviews

This means that $P(A|B) = 0.99$

Probability of receiving a favorable review:

20% are huge successes. Of those, 99% receive favorable reviews.

30% are modest successes. Of those, 70% receive favorable reviews.

30% break even. Of those, 40% receive favorable reviews.

20% are losers. Of those, 20% receive favorable reviews.

Then

$P(A) = 0.2*0.99 + 0.3*0.7 + 0.3*0.4 + 0.2*0.2 = 0.568$

Finally

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.2*0.99}{0.568} = 0.3486$

34.86% probability that it will be huge success

4 0

3 years ago

Advanced conversions:

san4es73 [151]

Answer:

Step-by-step explanation:

1). ∵ 1 L = 1000 ml

∴ 4 L = 4000 ml

2). ∵ 1 cm = 10 mm

∴ 5 cm = 50 mm

3). ∵ 1 mg = 0.1 cg

∴ 5 mg = 0.5 cg

4). ∵ 1 ml = 0.001 L

∴ 25 ml = 0.025 L

5). ∵ 1 gm = 1000 mg

∴ 0.125 gm = 25 mg

6). ∵ 1 ml = 0.001 L

∴ 570 ml = 0.57 L

7). ∵ 1 mm = 0.1 cm

∴ 135 mm = 13.5 cm

8). ∵ 1 mg = 0.001 g

∴ 1820 mg = 1.82 g

9). ∵ 1 g = 100 cg

∴ 0.5 g = 50 cg

10). ∵ 1 ml = 0.001 L

∴ 0.2385 ml = 0.0002385 L

6 0

3 years ago

Multiply by 10s 3,721 times 30<br><br><br> I NEED DONE ASAP PLEASE

Rasek [7]

Answer:

111630

Step-by-step explanation:

4 0

3 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

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Calculate each or quotient 2 2/5 x3 1/3=

My name is Ann [436]

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You multiply the numbers

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