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3 years ago

What is the probability that a 13-card bridge hand contains all cards from the same suit?

1 answer:

4 0

The total number of 13-card hands that can be dealt is 52C13. Four of these are of the same suit. Therefore the required probability is given by:
$\frac{4}{52C13}= 6.3\times 10^{-12}$

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If Jorge has 61 hours of work to do and he can work 7.5 hours a week, how many days will it take for him to finish it all.

mestny [16]

Answer:57 days

Step-by-step explanation:

3 0

3 years ago

Question 5 Multiple Choice Worth 1 points)

postnew [5]

Answer:

i have no clue but good luck

Step-by-step explanation:

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3 years ago

Find the two correct values of x in the equation 3x² = y when y = 75.

valentina_108 [34]

Answer:

5 and -5

Step-by-step explanation:

3x² = 75

x² = 75/3

x² = √25

x = 5 and - 5

3 0

2 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago

Can someone explain the process getting from:

spin [16.1K]

$\rm 5=e^{3b}$

The unknown b is stuck in the exponent position.
We can can fix that by using logarithms.
Log is the inverse operation of the exponential.

We'll take log of each side.
Log of what base tho?

Well, the base of our exponential is e,
so we'll take log base e of each side.

$\rm log_e(5)=log_e(e^{3b})$

We'll apply one of our log rules next:
$\rm \log(x^y)=y\cdot\log(x)$

This allows us to take the exponent out of the log,

$\rm log_e(5)=(3b)log_e(e)$

Another thing to remember about logs:
When the base of the log matches the inside of the log,
then the whole thing is simply 1,
$\rm log_{10}(10)=1$
$\rm log_5(5)=1$
$\rm log_e(e)=1$

So our equation simplifies to this,

$\rm log_e(5)=(3b)\cdot1$

As a final step, divide both sides by 3,

$\rm \frac13log_e(5)=b$

k, hope that helps!

6 0

3 years ago