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2 years ago

A card is drawn randomly from a standard 52-card deck. Find the probability of the given event.

Mathematics

1 answer:

irakobra [83]2 years ago

8 0

Answer:

a). 1/13

There are 4 5s. so 4/52

b). 3/13

12 total jack, king, and queen. 12/52

c). 15/26

30 cards are not face cards so 30/52

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When solving negative one over five -1/5 (x − 25) = 7, what is the correct sequence of operations

aleksandr82 [10.1K]

The solution would be like this for this specific problem:

$\left( { - 5} \right)
\cdot \left( { - \frac{1}{5}} \right) \cdot \left( {x - 25} \right) = 7 \cdot
\left( { - 5} \right)$

$\left( { - 5} \right)
\cdot \left( { - \frac{1}{5}} \right) = 1$

The correct sequence of operations when solving negative one over five (x − 25) = 7 would be multiply each side by −5, and adding 25 to each side.

5 0

2 years ago

Hi good nighttt c::::

noname [10]

Step-by-step explanation:

oyasumi..................

3 0

2 years ago

Read 2 more answers

Taylor baked a cake to serve after dinner. Suppose she cut the cake into 8 equal size pieces and 6 people ate all the pieces.exp

REY [17]

All you need to do is 8 divided by 6. When you do so you get: 1.3 which could also be written as 1 3/8.

So each person gets 1 whole piece and 3/8 of another piece.

8 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago

Select the correct answer,

Inessa05 [86]

Answer:

B. It is too risky to use credit cards online, and online payment services have better security because of the increasing number of hackers that may steal money from your bank account.

Step-by-step explanation:

7 0

2 years ago