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2 years ago

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Captain Jessica has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Luis and his merciless band of thiev

es. The Captain has probability 4/9 of hitting the pirate ship. The pirate only has one good eye, so he hits the Captain's ship with probability 1/4 .If both fire their cannons at the same time, what is the probability that both the pirate and the Captain hit each other's ships?

1 answer:

erik [133]2 years ago

6 0

Answer: Probability that both the pirate and the Captain hit each other's ships is $\frac{1}{9}$

Step-by-step explanation:

since we have given that

Probability that the captain hits the pirate ship is given by

$\frac{4}{9}$

Probability that the pirate hits the Captains's ship is given by

$\frac{1}{4}$

So, we have to find the "Probability that both the pirate and the Captain hit each other's ship" is given by

$P(\text{Captain hits})\times P(\text{ Pirates hit})\\\\=\frac{4}{9}\times \frac{1}{4}\\\\=\frac{1}{9}$

Hence, probability that both the pirate and the Captain hit each other's ships is $\frac{1}{9}$

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The school store marks up the price of all items by 30% before selling to

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Answer:

$65.325 or $65.33

Step-by-step explanation:

When looking for a percentage in a number, you want to multiply that number by the percentage. Remember that all percentages are to the hundreth power, or 0.00. So if you want to find 30% of 50.25, you multiply it by 0.30 or 0.3. After calculating that you will come up with 15.075 as the percentage. Since the store is marking the price up, you add it to the original price. So your equation is 50.25 + 15.075 =?. After calculating that, you get the answer 65.325. That is only half of your answer though because, when dealing with money, you always round to the nearest hundredth. So when you round 65.325, you get 65.33 which should be the correct answer.

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3 years ago

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

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2 years ago

55% of what number is 33?

EastWind [94]

Answer:

18.15

Step-by-step explanation:

55% = 55/100

33 = 33/1

33/1 x 55/100

33 x 55 = 1815

1 x 100 = 100

1815/100 = 18.15

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2 years ago

Amir is sorting his stamp collection. He has made a chart of what countries the stamps are from

natulia [17]

Answer:

Step-by-step explanation:

Given that Amir is sorting his stamp collection. He has made a chart of what countries the stamps are from

He got results as

France 1/3

Morocco 5/12

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To find out what fraction of Amir's stamps are from France, Morocco, or Spain, we have to simply add the three as there is common element between any two pairs

Hence answer is

$\frac{1}{3} +\frac{5}{12} +\frac{1}{6}\\ =\frac{11}{12}$

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3 years ago

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